ee101: sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · ee101: sinusoidal steady...
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![Page 1: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/1.jpg)
EE101: Sinusoidal steady state analysis
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
![Page 2: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/2.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 3: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/3.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 4: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/4.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 5: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/5.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 6: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/6.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 7: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/7.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 8: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/8.jpg)
Sinusoidal steady state
t=0
Vc
R
Vm cos ωt C
R (C V ′c ) + Vc = Vm cos ωt , t > 0 . (1)
The solution Vc (t) is made up of two components, Vc (t) = V(h)c (t) + V
(p)c (t) .
V(h)c (t) satisfies the homogeneous differential equation,
R C V ′c + Vc = 0 , (2)
from which, V(h)c (t) = A exp(−t/τ) , with τ = RC .
V(p)c (t) is a particular solution of (1). Since the forcing function is Vm cos ωt, we try
V(p)c (t) = C1 cos ωt + C2 sin ωt .
Substituting in (1), we get,
ωR C (−C1 sin ωt + C2 cos ωt) + C1 cos ωt + C2 sin ωt = Vm cos ωt .
C1 and C2 can be found by equating the coefficients of sin ωt and cos ωt on the left
and right sides.
M. B. Patil, IIT Bombay
![Page 9: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/9.jpg)
Sinusoidal steady state
(SEQUEL file: ee101_rc5.sqproj)
−0.2
0.2
t=0
R
C 0
time (ms) 0 2 4 6 8 10
2 kΩ
0.5µFVm cos ωt
Vm = 1Vf = 1 kHz
Vc
Vc (V)
* The complete solution is Vc (t) = A exp(−t/τ) + C1 cos ωt + C2 sin ωt .
* As t →∞, the exponential term becomes zero, and we are left withVc (t) = C1 cos ωt + C2 sin ωt .
* This is known as the “sinusoidal steady state” response since all quantities(currents and voltages) in the circuit are sinusoidal in nature.
* Any circuit containing resistors, capacitors, inductors, sinusoidal voltage andcurrent sources (of the same frequency), dependent (linear) sources behaves in asimilar manner, viz., each current and voltage in the circuit becomes purelysinusoidal as t →∞.
M. B. Patil, IIT Bombay
![Page 10: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/10.jpg)
Sinusoidal steady state
(SEQUEL file: ee101_rc5.sqproj)
−0.2
0.2
t=0
R
C 0
time (ms) 0 2 4 6 8 10
2 kΩ
0.5µFVm cos ωt
Vm = 1Vf = 1 kHz
Vc
Vc (V)
* The complete solution is Vc (t) = A exp(−t/τ) + C1 cos ωt + C2 sin ωt .
* As t →∞, the exponential term becomes zero, and we are left withVc (t) = C1 cos ωt + C2 sin ωt .
* This is known as the “sinusoidal steady state” response since all quantities(currents and voltages) in the circuit are sinusoidal in nature.
* Any circuit containing resistors, capacitors, inductors, sinusoidal voltage andcurrent sources (of the same frequency), dependent (linear) sources behaves in asimilar manner, viz., each current and voltage in the circuit becomes purelysinusoidal as t →∞.
M. B. Patil, IIT Bombay
![Page 11: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/11.jpg)
Sinusoidal steady state
(SEQUEL file: ee101_rc5.sqproj)
−0.2
0.2
t=0
R
C 0
time (ms) 0 2 4 6 8 10
2 kΩ
0.5µFVm cos ωt
Vm = 1Vf = 1 kHz
Vc
Vc (V)
* The complete solution is Vc (t) = A exp(−t/τ) + C1 cos ωt + C2 sin ωt .
* As t →∞, the exponential term becomes zero, and we are left withVc (t) = C1 cos ωt + C2 sin ωt .
* This is known as the “sinusoidal steady state” response since all quantities(currents and voltages) in the circuit are sinusoidal in nature.
* Any circuit containing resistors, capacitors, inductors, sinusoidal voltage andcurrent sources (of the same frequency), dependent (linear) sources behaves in asimilar manner, viz., each current and voltage in the circuit becomes purelysinusoidal as t →∞.
M. B. Patil, IIT Bombay
![Page 12: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/12.jpg)
Sinusoidal steady state
(SEQUEL file: ee101_rc5.sqproj)
−0.2
0.2
t=0
R
C 0
time (ms) 0 2 4 6 8 10
2 kΩ
0.5µFVm cos ωt
Vm = 1Vf = 1 kHz
Vc
Vc (V)
* The complete solution is Vc (t) = A exp(−t/τ) + C1 cos ωt + C2 sin ωt .
* As t →∞, the exponential term becomes zero, and we are left withVc (t) = C1 cos ωt + C2 sin ωt .
* This is known as the “sinusoidal steady state” response since all quantities(currents and voltages) in the circuit are sinusoidal in nature.
* Any circuit containing resistors, capacitors, inductors, sinusoidal voltage andcurrent sources (of the same frequency), dependent (linear) sources behaves in asimilar manner, viz., each current and voltage in the circuit becomes purelysinusoidal as t →∞.
M. B. Patil, IIT Bombay
![Page 13: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/13.jpg)
Sinusoidal steady state
(SEQUEL file: ee101_rc5.sqproj)
−0.2
0.2
t=0
R
C 0
time (ms) 0 2 4 6 8 10
2 kΩ
0.5µFVm cos ωt
Vm = 1Vf = 1 kHz
Vc
Vc (V)
* The complete solution is Vc (t) = A exp(−t/τ) + C1 cos ωt + C2 sin ωt .
* As t →∞, the exponential term becomes zero, and we are left withVc (t) = C1 cos ωt + C2 sin ωt .
* This is known as the “sinusoidal steady state” response since all quantities(currents and voltages) in the circuit are sinusoidal in nature.
* Any circuit containing resistors, capacitors, inductors, sinusoidal voltage andcurrent sources (of the same frequency), dependent (linear) sources behaves in asimilar manner, viz., each current and voltage in the circuit becomes purelysinusoidal as t →∞.
M. B. Patil, IIT Bombay
![Page 14: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/14.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 15: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/15.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 16: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/16.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜
= ReˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 17: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/17.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜
= ReˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 18: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/18.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜
= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 19: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/19.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 20: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/20.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 21: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/21.jpg)
Sinusoidal steady state: phasors
* In the sinusoidal steady state, “phasors” can be used to represent currents andvoltages.
* A phasor is a complex number,
X = Xm 6 θ = Xm exp(jθ) ,
with the following interpretation in the time domain.
x(t) = ReˆX e jωt
˜= Re
ˆXm e jθ e jωt
˜= Re
ˆXm e j(ωt+θ)
˜= Xm cos (ωt + θ)
* Use of phasors substantially simplifies analysis of circuits in the sinusoidal steadystate.
* Note that a phasor can be written in the polar form or rectangular form,X = Xm 6 θ = Xm exp(jθ) = Xm cos θ + j Xm sin θ .
The term ωt is always implicit.
θ
Xm
Re (X)
X
Im (X)
M. B. Patil, IIT Bombay
![Page 22: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/22.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V
V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 23: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/23.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 24: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/24.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 25: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/25.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 26: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/26.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 27: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/27.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 28: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/28.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 29: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/29.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 30: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/30.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 31: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/31.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 32: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/32.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 33: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/33.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 34: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/34.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 35: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/35.jpg)
Phasors: examples
Frequency domainTime domain
v1(t)=3.2 cos (ωt+30) V V1 = 3.2 6 30 = 3.2 exp (jπ/6) V
i(t) = −1.5 cos (ωt + 60) A
= 1.5 cos (ωt− 2π/3)A
= 1.5 cos (ωt + π/3− π) A
I = 1.5 6 (−2π/3)A
v2(t) = −0.1 cos (ωt) V
= 0.1 cos (ωt + π) V
V2 = 0.1 6 π V
i2(t) = 0.18 sin (ωt) A
= 0.18 cos (ωt− π/2) A
I2 = 0.18 6 (−π/2) A
I3 = 1 + j 1 A
=√
2 6 45 A
i3(t) =√
2 cos (ωt + 45) A
M. B. Patil, IIT Bombay
![Page 36: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/36.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜= Re
ˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 37: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/37.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜= Re
ˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 38: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/38.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜
= Reˆ`
Vm1e jθ1 + Vm2e jθ2´
e jωt˜
= ReˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 39: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/39.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜
= ReˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 40: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/40.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜= Re
ˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 41: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/41.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜= Re
ˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 42: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/42.jpg)
Addition of phasors
Consider addition of two sinusoidal quantities:v(t) = v1(t) + v2(t)
= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
Now consider addition of the phasors corresponding to v1(t) and v2(t).
V = V1 + V2
= Vm1e jθ1 + Vm2e jθ2
In the time domain, V corresponds to v(t), with
v(t) = ReˆVe jωt
˜= Re
ˆ`Vm1e jθ1 + Vm2e jθ2
´e jωt
˜= Re
ˆVm1e j(ωt+θ1) + Vm2e(ωt+jθ2)
˜= Vm1 cos (ωt + θ1) + Vm2 cos (ωt + θ2)
which is the same as v(t).
M. B. Patil, IIT Bombay
![Page 43: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/43.jpg)
Addition of phasors
* Addition of sinusoidal quantities in the time domain can be replaced by additionof the corresponding phasors in the sinusoidal steady state.
* The KCL and KVL equations,Pik (t) = 0 at a node, andPvk (t) = 0 in a loop,
amount to addition of sinusoidal quantities and can therefore be replaced by thecorresponding phasor equations,P
Ik = 0 at a node, andPVk = 0 in a loop.
M. B. Patil, IIT Bombay
![Page 44: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/44.jpg)
Addition of phasors
* Addition of sinusoidal quantities in the time domain can be replaced by additionof the corresponding phasors in the sinusoidal steady state.
* The KCL and KVL equations,Pik (t) = 0 at a node, andPvk (t) = 0 in a loop,
amount to addition of sinusoidal quantities and can therefore be replaced by thecorresponding phasor equations,P
Ik = 0 at a node, andPVk = 0 in a loop.
M. B. Patil, IIT Bombay
![Page 45: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/45.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 46: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/46.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 47: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/47.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 48: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/48.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 49: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/49.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 50: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/50.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 51: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/51.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 52: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/52.jpg)
Impedance of a resistor
Ii(t)
Vv(t)
R Z
Let i(t) = Im cos (ωt + θ).
v(t) = R i(t)
= R Im cos (ωt + θ)
≡ Vm cos (ωt + θ).
The phasors corresponding to i(t) and v(t) are, respectively,
I = Im 6 θ, V = R × Im 6 θ.
We have therefore the following relationship between V and I: V = R × I.
Thus, the impedance of a resistor, defined as, Z = V/I, is
Z = R + j 0
M. B. Patil, IIT Bombay
![Page 53: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/53.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 54: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/54.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 55: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/55.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 56: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/56.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 57: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/57.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 58: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/58.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 59: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/59.jpg)
Impedance of a capacitor
C
v(t)
i(t) I
V
Z
Let v(t) = Vm cos (ωt + θ).
i(t) = Cdv
dt= −C ω Vm sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
i(t) = C ω Vm cos (ωt + θ + π/2).
In terms of phasors, V = Vm 6 θ, I = ωCVm 6 (θ+π/2).
I can be rewritten as,
I = ωCVm e j(θ+π/2) = ωCVm e jθ e jπ/2 = jωC`Vm e jθ
´= jωC V
Thus, the impedance of a capacitor, Z = V/I, is Z = 1/(jωC) ,
and the admittance of a capacitor, Y = I/V, is Y = jωC .
M. B. Patil, IIT Bombay
![Page 60: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/60.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 61: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/61.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 62: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/62.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 63: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/63.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 64: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/64.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 65: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/65.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 66: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/66.jpg)
Impedance of an inductor
v(t)
Li(t)
V
I Z
Let i(t) = Im cos (ωt + θ).
v(t) = Ldi
dt= −Lω Im sin (ωt + θ).
Using the identity, cos (φ+ π/2) = − sin φ, we get
v(t) = Lω Im cos (ωt + θ + π/2).
In terms of phasors, I = Im 6 θ, V = ωLIm 6 (θ+π/2).
V can be rewritten as,
V = ωLIm e j(θ+π/2) = ωLIm e jθ e jπ/2 = jωL`Im e jθ
´= jωL I
Thus, the impedance of an indcutor, Z = V/I, is Z = jωL ,
and the admittance of an inductor, Y = I/V, is Y = 1/(jωL) .
M. B. Patil, IIT Bombay
![Page 67: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/67.jpg)
Sources
Vsvs(t)Isis(t)
* An independent sinusoidal current source, is (t) = Im cos (ωt + θ), can berepresented by the phasor Im 6 θ (i.e., a constant complex number).
* An independent sinusoidal voltage source, vs (t) = Vm cos (ωt + θ), can berepresented by the phasor Vm 6 θ (i.e., a constant complex number).
* Dependent (linear) sources can be treated in the sinusoidal steady state in thesame manner as a resistor, i.e., by the corresponding phasor relationship.For example, for a CCVS, we have,v(t) = r ic (t) in the time domain.V = r Ic in the frequency domain.
M. B. Patil, IIT Bombay
![Page 68: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/68.jpg)
Sources
Vsvs(t)Isis(t)
* An independent sinusoidal current source, is (t) = Im cos (ωt + θ), can berepresented by the phasor Im 6 θ (i.e., a constant complex number).
* An independent sinusoidal voltage source, vs (t) = Vm cos (ωt + θ), can berepresented by the phasor Vm 6 θ (i.e., a constant complex number).
* Dependent (linear) sources can be treated in the sinusoidal steady state in thesame manner as a resistor, i.e., by the corresponding phasor relationship.For example, for a CCVS, we have,v(t) = r ic (t) in the time domain.V = r Ic in the frequency domain.
M. B. Patil, IIT Bombay
![Page 69: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/69.jpg)
Sources
Vsvs(t)Isis(t)
* An independent sinusoidal current source, is (t) = Im cos (ωt + θ), can berepresented by the phasor Im 6 θ (i.e., a constant complex number).
* An independent sinusoidal voltage source, vs (t) = Vm cos (ωt + θ), can berepresented by the phasor Vm 6 θ (i.e., a constant complex number).
* Dependent (linear) sources can be treated in the sinusoidal steady state in thesame manner as a resistor, i.e., by the corresponding phasor relationship.For example, for a CCVS, we have,v(t) = r ic (t) in the time domain.V = r Ic in the frequency domain.
M. B. Patil, IIT Bombay
![Page 70: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/70.jpg)
Sources
Vsvs(t)Isis(t)
* An independent sinusoidal current source, is (t) = Im cos (ωt + θ), can berepresented by the phasor Im 6 θ (i.e., a constant complex number).
* An independent sinusoidal voltage source, vs (t) = Vm cos (ωt + θ), can berepresented by the phasor Vm 6 θ (i.e., a constant complex number).
* Dependent (linear) sources can be treated in the sinusoidal steady state in thesame manner as a resistor, i.e., by the corresponding phasor relationship.For example, for a CCVS, we have,v(t) = r ic (t) in the time domain.V = r Ic in the frequency domain.
M. B. Patil, IIT Bombay
![Page 71: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/71.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 72: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/72.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 73: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/73.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 74: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/74.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 75: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/75.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 76: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/76.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 77: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/77.jpg)
Use of phasors in circuit analysis
* The time-domain KCL and KVL equationsP
ik (t) = 0 andP
vk (t) = 0 can bewritten as
PIk = 0 and
PVk = 0 in the frequency domain.
* Resistors, capacitors, and inductors can be described by V = Z I in the frequencydomain, which is similar to V = R I in DC conditions (except that we aredealing with complex numbers in the frequency domain).
* An independent sinusoidal source in the frequency domain behaves like a DCsource, e.g., Vs = constant (a complex number).
* For dependent sources, a time-domain relationship such as i(t) = β ic (t)translates to I = β Ic in the frequency domain.
* Circuit analysis in the sinusoidal steady state using phasors is therefore verysimilar to DC circuits with independent and dependent sources, and resistors.
* Series/parallel formulas for resistors, nodal analysis, mesh analysis, Thevenin’sand Norton’s theorems can be directly applied to circuits in the sinusoidal steadystate.
M. B. Patil, IIT Bombay
![Page 78: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/78.jpg)
RL circuit
I
Vm 6 0
R
jωL
0
1
−1
time (ms) 0 10 20 30
R = 1 Ω
L = 1.6 mH
vs(t) (V)
i(t) (A)
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
In the time domain, i(t) = Im cos (ωt − θ), which lags the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds after that of the source voltage.
For R = 1 Ω, L = 1.6 mH, f = 50 Hz, θ = 26.6, tlag = 1.48 ms.
(SEQUEL file: ee101 rl ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 79: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/79.jpg)
RL circuit
I
Vm 6 0
R
jωL
0
1
−1
time (ms) 0 10 20 30
R = 1 Ω
L = 1.6 mH
vs(t) (V)
i(t) (A)
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
In the time domain, i(t) = Im cos (ωt − θ), which lags the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds after that of the source voltage.
For R = 1 Ω, L = 1.6 mH, f = 50 Hz, θ = 26.6, tlag = 1.48 ms.
(SEQUEL file: ee101 rl ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 80: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/80.jpg)
RL circuit
I
Vm 6 0
R
jωL
0
1
−1
time (ms) 0 10 20 30
R = 1 Ω
L = 1.6 mH
vs(t) (V)
i(t) (A)
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
In the time domain, i(t) = Im cos (ωt − θ), which lags the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds after that of the source voltage.
For R = 1 Ω, L = 1.6 mH, f = 50 Hz, θ = 26.6, tlag = 1.48 ms.
(SEQUEL file: ee101 rl ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 81: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/81.jpg)
RL circuit
I
Vm 6 0
R
jωL
0
1
−1
time (ms) 0 10 20 30
R = 1 Ω
L = 1.6 mH
vs(t) (V)
i(t) (A)
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
In the time domain, i(t) = Im cos (ωt − θ), which lags the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds after that of the source voltage.
For R = 1 Ω, L = 1.6 mH, f = 50 Hz, θ = 26.6, tlag = 1.48 ms.
(SEQUEL file: ee101 rl ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 82: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/82.jpg)
RL circuit
I
Vm 6 0
R
jωL 0
1
−1
time (ms) 0 10 20 30
R = 1 Ω
L = 1.6 mH
vs(t) (V)
i(t) (A)
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
In the time domain, i(t) = Im cos (ωt − θ), which lags the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds after that of the source voltage.
For R = 1 Ω, L = 1.6 mH, f = 50 Hz, θ = 26.6, tlag = 1.48 ms.
(SEQUEL file: ee101 rl ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 83: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/83.jpg)
RL circuit
I
VR
VLVs
R
jωLVm 6 0
Re (V)
Im (V)
θ
VR
VL
Vs
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
VR = I× R = R Im ∠(−θ) ,
VL = I× jωL = ωImL ∠(−θ + π/2) ,
The KVL equation, Vs = VR + VL, can be represented in the complex plane by a“phasor diagram.”
If R |jωL|, θ → 0, |VR| ' |Vs| = Vm.
If R |jωL|, θ → π/2, |VL| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 84: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/84.jpg)
RL circuit
I
VR
VLVs
R
jωLVm 6 0
Re (V)
Im (V)
θ
VR
VL
Vs
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
VR = I× R = R Im ∠(−θ) ,
VL = I× jωL = ωImL ∠(−θ + π/2) ,
The KVL equation, Vs = VR + VL, can be represented in the complex plane by a“phasor diagram.”
If R |jωL|, θ → 0, |VR| ' |Vs| = Vm.
If R |jωL|, θ → π/2, |VL| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 85: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/85.jpg)
RL circuit
I
VR
VLVs
R
jωLVm 6 0
Re (V)
Im (V)
θ
VR
VL
Vs
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
VR = I× R = R Im ∠(−θ) ,
VL = I× jωL = ωImL ∠(−θ + π/2) ,
The KVL equation, Vs = VR + VL, can be represented in the complex plane by a“phasor diagram.”
If R |jωL|, θ → 0, |VR| ' |Vs| = Vm.
If R |jωL|, θ → π/2, |VL| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 86: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/86.jpg)
RL circuit
I
VR
VLVs
R
jωLVm 6 0 Re (V)
Im (V)
θ
VR
VL
Vs
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
VR = I× R = R Im ∠(−θ) ,
VL = I× jωL = ωImL ∠(−θ + π/2) ,
The KVL equation, Vs = VR + VL, can be represented in the complex plane by a“phasor diagram.”
If R |jωL|, θ → 0, |VR| ' |Vs| = Vm.
If R |jωL|, θ → π/2, |VL| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 87: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/87.jpg)
RL circuit
I
VR
VLVs
R
jωLVm 6 0 Re (V)
Im (V)
θ
VR
VL
Vs
I =Vm∠0
R + jωL≡ Im∠(−θ),
where Im =Vmp
R2 + ω2L2, and θ = tan−1(ωL/R).
VR = I× R = R Im ∠(−θ) ,
VL = I× jωL = ωImL ∠(−θ + π/2) ,
The KVL equation, Vs = VR + VL, can be represented in the complex plane by a“phasor diagram.”
If R |jωL|, θ → 0, |VR| ' |Vs| = Vm.
If R |jωL|, θ → π/2, |VL| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 88: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/88.jpg)
RC circuit
I
1/jωCVm 6 0
R
0
−1
1
time (ms) 0 10 20 30
R = 1 Ω
C = 5.3 mF
vs(t) (V)
i(t) (A)
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
In the time domain, i(t) = Im cos (ωt + θ), which leads the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds before that of the source voltage.
For R = 1 Ω, C = 5.3 mF, f = 50 Hz, θ = 31, tlead = 1.72 ms.
(SEQUEL file: ee101 rc ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 89: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/89.jpg)
RC circuit
I
1/jωCVm 6 0
R
0
−1
1
time (ms) 0 10 20 30
R = 1 Ω
C = 5.3 mF
vs(t) (V)
i(t) (A)
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
In the time domain, i(t) = Im cos (ωt + θ), which leads the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds before that of the source voltage.
For R = 1 Ω, C = 5.3 mF, f = 50 Hz, θ = 31, tlead = 1.72 ms.
(SEQUEL file: ee101 rc ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 90: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/90.jpg)
RC circuit
I
1/jωCVm 6 0
R
0
−1
1
time (ms) 0 10 20 30
R = 1 Ω
C = 5.3 mF
vs(t) (V)
i(t) (A)
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
In the time domain, i(t) = Im cos (ωt + θ), which leads the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds before that of the source voltage.
For R = 1 Ω, C = 5.3 mF, f = 50 Hz, θ = 31, tlead = 1.72 ms.
(SEQUEL file: ee101 rc ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 91: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/91.jpg)
RC circuit
I
1/jωCVm 6 0
R
0
−1
1
time (ms) 0 10 20 30
R = 1 Ω
C = 5.3 mF
vs(t) (V)
i(t) (A)
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
In the time domain, i(t) = Im cos (ωt + θ), which leads the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds before that of the source voltage.
For R = 1 Ω, C = 5.3 mF, f = 50 Hz, θ = 31, tlead = 1.72 ms.
(SEQUEL file: ee101 rc ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 92: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/92.jpg)
RC circuit
I
1/jωCVm 6 0
R
0
−1
1
time (ms) 0 10 20 30
R = 1 Ω
C = 5.3 mF
vs(t) (V)
i(t) (A)
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
In the time domain, i(t) = Im cos (ωt + θ), which leads the source voltage since thepeak (or zero) of i(t) occurs t = θ/ω seconds before that of the source voltage.
For R = 1 Ω, C = 5.3 mF, f = 50 Hz, θ = 31, tlead = 1.72 ms.
(SEQUEL file: ee101 rc ac 1.sqproj)
M. B. Patil, IIT Bombay
![Page 93: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/93.jpg)
RC circuit
I
VR
VCVs
R
1/jωCVm 6 0
Re (V)
Im (V)
θ
VC
VR
Vs
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
VR = I× R = R Im ∠θ ,
VC = I× (1/jωC) = (Im/ωC) ∠(θ − π/2) ,
The KVL equation, Vs = VR + VC, can be represented in the complex plane by a“phasor diagram.”
If R |1/jωC |, θ → 0, |VR| ' |Vs| = Vm.
If R |1/jωC |, θ → π/2, |VC| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 94: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/94.jpg)
RC circuit
I
VR
VCVs
R
1/jωCVm 6 0
Re (V)
Im (V)
θ
VC
VR
Vs
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
VR = I× R = R Im ∠θ ,
VC = I× (1/jωC) = (Im/ωC) ∠(θ − π/2) ,
The KVL equation, Vs = VR + VC, can be represented in the complex plane by a“phasor diagram.”
If R |1/jωC |, θ → 0, |VR| ' |Vs| = Vm.
If R |1/jωC |, θ → π/2, |VC| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 95: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/95.jpg)
RC circuit
I
VR
VCVs
R
1/jωCVm 6 0
Re (V)
Im (V)
θ
VC
VR
Vs
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
VR = I× R = R Im ∠θ ,
VC = I× (1/jωC) = (Im/ωC) ∠(θ − π/2) ,
The KVL equation, Vs = VR + VC, can be represented in the complex plane by a“phasor diagram.”
If R |1/jωC |, θ → 0, |VR| ' |Vs| = Vm.
If R |1/jωC |, θ → π/2, |VC| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 96: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/96.jpg)
RC circuit
I
VR
VCVs
R
1/jωCVm 6 0 Re (V)
Im (V)
θ
VC
VR
Vs
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
VR = I× R = R Im ∠θ ,
VC = I× (1/jωC) = (Im/ωC) ∠(θ − π/2) ,
The KVL equation, Vs = VR + VC, can be represented in the complex plane by a“phasor diagram.”
If R |1/jωC |, θ → 0, |VR| ' |Vs| = Vm.
If R |1/jωC |, θ → π/2, |VC| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 97: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/97.jpg)
RC circuit
I
VR
VCVs
R
1/jωCVm 6 0 Re (V)
Im (V)
θ
VC
VR
Vs
I =Vm∠0
R + 1/jωC≡ Im∠θ,
where Im =ωCVmp
1 + (ωRC)2, and θ = π/2− tan−1(ωRC).
VR = I× R = R Im ∠θ ,
VC = I× (1/jωC) = (Im/ωC) ∠(θ − π/2) ,
The KVL equation, Vs = VR + VC, can be represented in the complex plane by a“phasor diagram.”
If R |1/jωC |, θ → 0, |VR| ' |Vs| = Vm.
If R |1/jωC |, θ → π/2, |VC| ' |Vs| = Vm.
M. B. Patil, IIT Bombay
![Page 98: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/98.jpg)
Series/parallel connections
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1
Z2
Z1 = j× 100× 0.25 = j 25Ω
Z2 = −j/(100× 100× 10−6) = −j 100Ω
Z = Z1 + Z2 = −j 75Ω
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1 Z2
Z =Z1Z2
Z1 + Z2
=25× 100
−j 75
=(j 25)× (−j 100)
j 25− j 100
= j 33.3Ω
M. B. Patil, IIT Bombay
![Page 99: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/99.jpg)
Series/parallel connections
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1
Z2
Z1 = j× 100× 0.25 = j 25Ω
Z2 = −j/(100× 100× 10−6) = −j 100Ω
Z = Z1 + Z2 = −j 75Ω
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1 Z2
Z =Z1Z2
Z1 + Z2
=25× 100
−j 75
=(j 25)× (−j 100)
j 25− j 100
= j 33.3Ω
M. B. Patil, IIT Bombay
![Page 100: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/100.jpg)
Series/parallel connections
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1
Z2
Z1 = j× 100× 0.25 = j 25Ω
Z2 = −j/(100× 100× 10−6) = −j 100Ω
Z = Z1 + Z2 = −j 75Ω
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1 Z2
Z =Z1Z2
Z1 + Z2
=25× 100
−j 75
=(j 25)× (−j 100)
j 25− j 100
= j 33.3Ω
M. B. Patil, IIT Bombay
![Page 101: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/101.jpg)
Series/parallel connections
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1
Z2
Z1 = j× 100× 0.25 = j 25Ω
Z2 = −j/(100× 100× 10−6) = −j 100Ω
Z = Z1 + Z2 = −j 75Ω
A
B
A
B
(ω = 100 rad/s)
Z0.25 H
100 µF
Z1 Z2
Z =Z1Z2
Z1 + Z2
=25× 100
−j 75
=(j 25)× (−j 100)
j 25− j 100
= j 33.3Ω
M. B. Patil, IIT Bombay
![Page 102: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/102.jpg)
Impedance example
A
B
A
B
Obtain Z in polar form.
(ω = 100 rad/s)
j 10 ΩZ
10 ΩZ1
Z2
Z =10× j10
10 + j10=
j10
1 + j
=j10
1 + j× 1− j
1− j
Method 1:
Convert to polar form → Z = 7.07 6 45 Ω
=10 + j10
2= 5 + j5Ω
Method 2:
= 5√
2 6 (π/2− π/4) = 7.07 6 45 Ω
Z =10× j10
10 + j10=
100 6 π/2
10√
2 6 π/4
M. B. Patil, IIT Bombay
![Page 103: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/103.jpg)
Impedance example
A
B
A
B
Obtain Z in polar form.
(ω = 100 rad/s)
j 10 ΩZ
10 ΩZ1
Z2
Z =10× j10
10 + j10=
j10
1 + j
=j10
1 + j× 1− j
1− j
Method 1:
Convert to polar form → Z = 7.07 6 45 Ω
=10 + j10
2= 5 + j5Ω
Method 2:
= 5√
2 6 (π/2− π/4) = 7.07 6 45 Ω
Z =10× j10
10 + j10=
100 6 π/2
10√
2 6 π/4
M. B. Patil, IIT Bombay
![Page 104: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/104.jpg)
Impedance example
A
B
A
B
Obtain Z in polar form.
(ω = 100 rad/s)
j 10 ΩZ
10 ΩZ1
Z2
Z =10× j10
10 + j10=
j10
1 + j
=j10
1 + j× 1− j
1− j
Method 1:
Convert to polar form → Z = 7.07 6 45 Ω
=10 + j10
2= 5 + j5Ω
Method 2:
= 5√
2 6 (π/2− π/4) = 7.07 6 45 Ω
Z =10× j10
10 + j10=
100 6 π/2
10√
2 6 π/4
M. B. Patil, IIT Bombay
![Page 105: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/105.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 106: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/106.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 107: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/107.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 108: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/108.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 109: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/109.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 110: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/110.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 111: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/111.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 112: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/112.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 113: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/113.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 114: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/114.jpg)
Circuit example
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Z1
Z3
IC
Vs
Z2
Z4
Is
IL
Vs
Is
ZEQ
Z3 =1
j × 2π × 50× 2× 10−3= −j 1.6 Ω
Z4 = 2π × 50× 15× 10−3 = j 4.7 Ω
ZEQ = Z1 + Z3 ‖ (Z2 + Z4)
= 2 + (−j 1.6) ‖ (10 + j 4.7) = 2 +(−j 1.6)× (10 + j 4.7)
−j 1.6 + 10 + j 4.7
= 2 +1.6∠ (−90)× 11.05∠ (25.2)
10.47∠ (17.2)= 2 +
17.7∠ (−64.8)
10.47∠ (17.2)
= 2 + 1.69∠ (−82) = 2 + (0.235− j 1.67)
= 2.235− j 1.67 = 2.79∠ (−36.8) Ω
M. B. Patil, IIT Bombay
![Page 115: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/115.jpg)
Circuit example (continued)
Z1
Z3
IC
Vs Vs
Z2
Z4
Is
IL
Is
ZEQ
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Is =Vs
ZEQ=
10 ∠ (0)
2.79 ∠ (−36.8)= 3.58 ∠ (36.8) A
IC =(Z2 + Z4)
Z3 + (Z2 + Z4)× Is = 3.79 ∠ (44.6) A
IL =Z3
Z3 + (Z2 + Z4)× Is = 0.546 ∠ (−70.6) A
3
2
1
0
−110 2 3
Is
IL
IC
Re(I)
Im(I
)
M. B. Patil, IIT Bombay
![Page 116: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/116.jpg)
Circuit example (continued)
Z1
Z3
IC
Vs Vs
Z2
Z4
Is
IL
Is
ZEQ
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Is =Vs
ZEQ=
10 ∠ (0)
2.79 ∠ (−36.8)= 3.58 ∠ (36.8) A
IC =(Z2 + Z4)
Z3 + (Z2 + Z4)× Is = 3.79 ∠ (44.6) A
IL =Z3
Z3 + (Z2 + Z4)× Is = 0.546 ∠ (−70.6) A
3
2
1
0
−110 2 3
Is
IL
IC
Re(I)
Im(I
)
M. B. Patil, IIT Bombay
![Page 117: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/117.jpg)
Circuit example (continued)
Z1
Z3
IC
Vs Vs
Z2
Z4
Is
IL
Is
ZEQ
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Is =Vs
ZEQ=
10 ∠ (0)
2.79 ∠ (−36.8)= 3.58 ∠ (36.8) A
IC =(Z2 + Z4)
Z3 + (Z2 + Z4)× Is = 3.79 ∠ (44.6) A
IL =Z3
Z3 + (Z2 + Z4)× Is = 0.546 ∠ (−70.6) A
3
2
1
0
−110 2 3
Is
IL
IC
Re(I)
Im(I
)
M. B. Patil, IIT Bombay
![Page 118: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/118.jpg)
Circuit example (continued)
Z1
Z3
IC
Vs Vs
Z2
Z4
Is
IL
Is
ZEQ
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Is =Vs
ZEQ=
10 ∠ (0)
2.79 ∠ (−36.8)= 3.58 ∠ (36.8) A
IC =(Z2 + Z4)
Z3 + (Z2 + Z4)× Is = 3.79 ∠ (44.6) A
IL =Z3
Z3 + (Z2 + Z4)× Is = 0.546 ∠ (−70.6) A
3
2
1
0
−110 2 3
Is
IL
IC
Re(I)
Im(I
)
M. B. Patil, IIT Bombay
![Page 119: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/119.jpg)
Circuit example (continued)
Z1
Z3
IC
Vs Vs
Z2
Z4
Is
IL
Is
ZEQ
iL
is
iC
15 mH2 mF
10 Ω2 Ω
10 6 0 Vf = 50 Hz
Is =Vs
ZEQ=
10 ∠ (0)
2.79 ∠ (−36.8)= 3.58 ∠ (36.8) A
IC =(Z2 + Z4)
Z3 + (Z2 + Z4)× Is = 3.79 ∠ (44.6) A
IL =Z3
Z3 + (Z2 + Z4)× Is = 0.546 ∠ (−70.6) A
3
2
1
0
−110 2 3
Is
IL
IC
Re(I)
Im(I
)
M. B. Patil, IIT Bombay
![Page 120: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/120.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 121: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/121.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 122: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/122.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 123: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/123.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 124: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/124.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 125: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/125.jpg)
Sinusoidal steady state: power computation
V
I Z
Let v(t) = Vm cos (ωt + θ), i.e., V = Vm ∠ θ ,
i(t) = Im cos (ωt + φ), i.e., I = Im ∠φ .
The instantaneous power absorbed by Z is,
P(t) = v(t) i(t)
= Vm Im cos (ωt + θ) cos (ωt + φ)
=1
2Vm Im [cos (2ωt + θ + φ) + cos (θ − φ)] (1)
The average power absorbed by Z is
P =1
T
Z T
0P(t) dt, where T = 2π/ω.
The first term in Eq. (1) has an average value of zero and does not contribute to P.Therefore,
P =1
2Vm Im cos (θ − φ) gives the average power absorbed by Z.
M. B. Patil, IIT Bombay
![Page 126: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/126.jpg)
Average power for R, L, C
General formula:V
I Z P =1
2Vm Im cos (θ − φ)
V = Vm 6 θ , I = Im 6 φ
V
I R
For I = Im 6 α, V = R Im 6 α ,
V = R I
P =1
2(R Im) Im cos (α− α) =
1
2I2m R =
1
2V2
m/R
V
I L
For I = Im 6 α, V = ωL Im 6 (α + π/2) ,
P =1
2Vm Im cos [(α + π/2)− α] = 0
V = jωL I
V
IC
For V = Vm 6 α, I = ωC Vm 6 (α + π/2) ,
P =1
2Vm Im cos [α− (α + π/2)] = 0
I = jωCV
M. B. Patil, IIT Bombay
![Page 127: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/127.jpg)
Average power for R, L, C
General formula:V
I Z P =1
2Vm Im cos (θ − φ)
V = Vm 6 θ , I = Im 6 φ
V
I R
For I = Im 6 α, V = R Im 6 α ,
V = R I
P =1
2(R Im) Im cos (α− α) =
1
2I2m R =
1
2V2
m/R
V
I L
For I = Im 6 α, V = ωL Im 6 (α + π/2) ,
P =1
2Vm Im cos [(α + π/2)− α] = 0
V = jωL I
V
IC
For V = Vm 6 α, I = ωC Vm 6 (α + π/2) ,
P =1
2Vm Im cos [α− (α + π/2)] = 0
I = jωCV
M. B. Patil, IIT Bombay
![Page 128: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/128.jpg)
Average power for R, L, C
General formula:V
I Z P =1
2Vm Im cos (θ − φ)
V = Vm 6 θ , I = Im 6 φ
V
I R
For I = Im 6 α, V = R Im 6 α ,
V = R I
P =1
2(R Im) Im cos (α− α) =
1
2I2m R =
1
2V2
m/R
V
I L
For I = Im 6 α, V = ωL Im 6 (α + π/2) ,
P =1
2Vm Im cos [(α + π/2)− α] = 0
V = jωL I
V
IC
For V = Vm 6 α, I = ωC Vm 6 (α + π/2) ,
P =1
2Vm Im cos [α− (α + π/2)] = 0
I = jωCV
M. B. Patil, IIT Bombay
![Page 129: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/129.jpg)
Average power for R, L, C
General formula:V
I Z P =1
2Vm Im cos (θ − φ)
V = Vm 6 θ , I = Im 6 φ
V
I R
For I = Im 6 α, V = R Im 6 α ,
V = R I
P =1
2(R Im) Im cos (α− α) =
1
2I2m R =
1
2V2
m/R
V
I L
For I = Im 6 α, V = ωL Im 6 (α + π/2) ,
P =1
2Vm Im cos [(α + π/2)− α] = 0
V = jωL I
V
IC
For V = Vm 6 α, I = ωC Vm 6 (α + π/2) ,
P =1
2Vm Im cos [α− (α + π/2)] = 0
I = jωCV
M. B. Patil, IIT Bombay
![Page 130: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/130.jpg)
Average power: example
Find the average power absorbed.
V
50Ω j25ΩI
Given: I = 2 6 45 A
Method 1:
V = (50 + j25)× 2 ∠ 45
= 55.9 ∠ 26.6 × 2 ∠ 45
= 111.8 ∠ (45 + 26.6)
P =1
2× 111.8× 2× cos (26.6) = 100 W .
Method 2:
No average power is absorbed by the inductor.
⇒ P = PR (average power absorbed by R)
=1
2I 2m R =
1
2× 22 × 50
= 100 W .
M. B. Patil, IIT Bombay
![Page 131: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/131.jpg)
Average power: example
Find the average power absorbed.
V
50Ω j25ΩI
Given: I = 2 6 45 A
Method 1:
V = (50 + j25)× 2 ∠ 45
= 55.9 ∠ 26.6 × 2 ∠ 45
= 111.8 ∠ (45 + 26.6)
P =1
2× 111.8× 2× cos (26.6) = 100 W .
Method 2:
No average power is absorbed by the inductor.
⇒ P = PR (average power absorbed by R)
=1
2I 2m R =
1
2× 22 × 50
= 100 W .
M. B. Patil, IIT Bombay
![Page 132: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/132.jpg)
Average power: example
Find the average power absorbed.
V
50Ω j25ΩI
Given: I = 2 6 45 A
Method 1:
V = (50 + j25)× 2 ∠ 45
= 55.9 ∠ 26.6 × 2 ∠ 45
= 111.8 ∠ (45 + 26.6)
P =1
2× 111.8× 2× cos (26.6) = 100 W .
Method 2:
No average power is absorbed by the inductor.
⇒ P = PR (average power absorbed by R)
=1
2I 2m R =
1
2× 22 × 50
= 100 W .
M. B. Patil, IIT Bombay
![Page 133: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/133.jpg)
Average power: example
Find the average power absorbed.
V
50Ω j25ΩI
Given: I = 2 6 45 A
Method 1:
V = (50 + j25)× 2 ∠ 45
= 55.9 ∠ 26.6 × 2 ∠ 45
= 111.8 ∠ (45 + 26.6)
P =1
2× 111.8× 2× cos (26.6) = 100 W .
Method 2:
No average power is absorbed by the inductor.
⇒ P = PR (average power absorbed by R)
=1
2I 2m R =
1
2× 22 × 50
= 100 W .
M. B. Patil, IIT Bombay
![Page 134: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/134.jpg)
Average power: example
Find the average power absorbed.
V
50Ω j25ΩI
Given: I = 2 6 45 A
Method 1:
V = (50 + j25)× 2 ∠ 45
= 55.9 ∠ 26.6 × 2 ∠ 45
= 111.8 ∠ (45 + 26.6)
P =1
2× 111.8× 2× cos (26.6) = 100 W .
Method 2:
No average power is absorbed by the inductor.
⇒ P = PR (average power absorbed by R)
=1
2I 2m R =
1
2× 22 × 50
= 100 W .
M. B. Patil, IIT Bombay
![Page 135: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/135.jpg)
Maximum power transfer
ZLVTh
IZTh
Let ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 136: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/136.jpg)
Maximum power transfer
ZLVTh
IZThLet ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 137: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/137.jpg)
Maximum power transfer
ZLVTh
IZThLet ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 138: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/138.jpg)
Maximum power transfer
ZLVTh
IZThLet ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 139: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/139.jpg)
Maximum power transfer
ZLVTh
IZThLet ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 140: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/140.jpg)
Maximum power transfer
ZLVTh
IZThLet ZL = RL + jXL, ZTh = RTh + jXTh, and I = Im ∠φ .
The power absorbed by ZL is,
P =1
2I 2mRL
=1
2
˛VTh
ZTh + ZL
˛2RL
=1
2
|VTh|2(RTh + RL)2 + (XTh + XL)2
RL .
For P to be maximum, (XTh + XL) must be zero. ⇒ XL = −XTh.
With XL = −XTh, we have,
P =1
2
|VTh|2(RTh + RL)2
RL ,
which is maximum for RL = RTh.
Therefore, for maximum power transfer to the load ZL, we need,
RL = RTh, XL = −XTh, i.e., ZL = Z∗Th.
M. B. Patil, IIT Bombay
![Page 141: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/141.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 142: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/142.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 143: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/143.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 144: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/144.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 145: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/145.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 146: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/146.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 147: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/147.jpg)
Maximum power transfer: example
A
B
ZL
j3Ω4 Ω
-j6Ω16 6 0 V
A
B
ZLVTh
ZTh
ZTh = (−j 6) ‖ (4 + j 3) = 5.76− j 1.68 Ω .
For maximum power transfer, ZL = Z∗Th = 5.76 + j 1.68 Ω ≡ RL + j XL .
VTh = 16 ∠ 0 × −j 6
(4 + j 3) + (−j 6)= 19.2 ∠(−53.13) .
I =VTh
ZTh + ZL=
VTh
2 RL.
P =1
2I 2mRL =
1
2
„19.2
2 RL
«2
× RL =1
2
(19.2)2
4 RL= 8 W .
M. B. Patil, IIT Bombay
![Page 148: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/148.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Consider a periodic current i(t) passing through R.
The average power absorbed by R is,
P1 =1
T
Z t0+T
t0
[i(t)]2 R dt ,
where t0 is some reference time (we will take t0 to be 0).
The average power absorbed by R in the DC case is,
P2 = I 2eff R.
Ieff , the effective value of i(t), is defined such that P1 = P2, i.e.,
I 2eff R =
1
T
Z T
0[i(t)]2 R dt ,
Ieff =
s1
T
Z T
0[i(t)]2 dt .
M. B. Patil, IIT Bombay
![Page 149: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/149.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Consider a periodic current i(t) passing through R.
The average power absorbed by R is,
P1 =1
T
Z t0+T
t0
[i(t)]2 R dt ,
where t0 is some reference time (we will take t0 to be 0).
The average power absorbed by R in the DC case is,
P2 = I 2eff R.
Ieff , the effective value of i(t), is defined such that P1 = P2, i.e.,
I 2eff R =
1
T
Z T
0[i(t)]2 R dt ,
Ieff =
s1
T
Z T
0[i(t)]2 dt .
M. B. Patil, IIT Bombay
![Page 150: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/150.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Consider a periodic current i(t) passing through R.
The average power absorbed by R is,
P1 =1
T
Z t0+T
t0
[i(t)]2 R dt ,
where t0 is some reference time (we will take t0 to be 0).
The average power absorbed by R in the DC case is,
P2 = I 2eff R.
Ieff , the effective value of i(t), is defined such that P1 = P2, i.e.,
I 2eff R =
1
T
Z T
0[i(t)]2 R dt ,
Ieff =
s1
T
Z T
0[i(t)]2 dt .
M. B. Patil, IIT Bombay
![Page 151: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/151.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Consider a periodic current i(t) passing through R.
The average power absorbed by R is,
P1 =1
T
Z t0+T
t0
[i(t)]2 R dt ,
where t0 is some reference time (we will take t0 to be 0).
The average power absorbed by R in the DC case is,
P2 = I 2eff R.
Ieff , the effective value of i(t), is defined such that P1 = P2, i.e.,
I 2eff R =
1
T
Z T
0[i(t)]2 R dt ,
Ieff =
s1
T
Z T
0[i(t)]2 dt .
M. B. Patil, IIT Bombay
![Page 152: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/152.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Ieff =
s1
T
Z T
0[i(t)]2 dt .
Ieff is called the root-mean-square (rms) value of i(t) because of the operations(root, mean, and square) involved in its computation.
If i(t) is sinusoidal, i.e., i(t) = Im cos (ωt + φ),
Ieff =
s1
T
Z T
0I 2m cos2(ωt + φ) dt = Im
s1
T
Z T
0
1
2[1 + cos (2ωt + 2φ)] dt
= Im
r1
T
1
2T = Im/
√2 .
Similarly, Veff = Vm/√
2 .
M. B. Patil, IIT Bombay
![Page 153: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/153.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Ieff =
s1
T
Z T
0[i(t)]2 dt .
Ieff is called the root-mean-square (rms) value of i(t) because of the operations(root, mean, and square) involved in its computation.
If i(t) is sinusoidal, i.e., i(t) = Im cos (ωt + φ),
Ieff =
s1
T
Z T
0I 2m cos2(ωt + φ) dt = Im
s1
T
Z T
0
1
2[1 + cos (2ωt + 2φ)] dt
= Im
r1
T
1
2T = Im/
√2 .
Similarly, Veff = Vm/√
2 .
M. B. Patil, IIT Bombay
![Page 154: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/154.jpg)
Effective (rms) values of voltage/current
R Ri(t)
v(t) Veff
Ieff
time-varying v and i constant v and i
Ieff =
s1
T
Z T
0[i(t)]2 dt .
Ieff is called the root-mean-square (rms) value of i(t) because of the operations(root, mean, and square) involved in its computation.
If i(t) is sinusoidal, i.e., i(t) = Im cos (ωt + φ),
Ieff =
s1
T
Z T
0I 2m cos2(ωt + φ) dt = Im
s1
T
Z T
0
1
2[1 + cos (2ωt + 2φ)] dt
= Im
r1
T
1
2T = Im/
√2 .
Similarly, Veff = Vm/√
2 .
M. B. Patil, IIT Bombay
![Page 155: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/155.jpg)
Apparent power, real power, and power factor
V
ZI
V = Vm 6 θ
I = Im 6 φ
The average (“real”) power absorbed by Z is,
P =1
2Vm Im cos (θ − φ) =
Vm√2
Im√2
cos (θ − φ)
= Veff Ieff cos (θ − φ) (Watts)
Apparent power is defined as Papp = Veff Ieff (Volt-Amp).
Power factor is defined as P.F . =Average power
Apparent power= cos (θ − φ).
(θ − φ) > 0: i(t) lags v(t), the P. F. is called a lagging P. F.
(inductive impedance)
(θ − φ) < 0: i(t) leads v(t), the P. F. is called a leading P. F.
(capacitive impedance)
M. B. Patil, IIT Bombay
![Page 156: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/156.jpg)
Apparent power, real power, and power factor
V
ZI
V = Vm 6 θ
I = Im 6 φ
The average (“real”) power absorbed by Z is,
P =1
2Vm Im cos (θ − φ) =
Vm√2
Im√2
cos (θ − φ)
= Veff Ieff cos (θ − φ) (Watts)
Apparent power is defined as Papp = Veff Ieff (Volt-Amp).
Power factor is defined as P.F . =Average power
Apparent power= cos (θ − φ).
(θ − φ) > 0: i(t) lags v(t), the P. F. is called a lagging P. F.
(inductive impedance)
(θ − φ) < 0: i(t) leads v(t), the P. F. is called a leading P. F.
(capacitive impedance)
M. B. Patil, IIT Bombay
![Page 157: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/157.jpg)
Apparent power, real power, and power factor
V
ZI
V = Vm 6 θ
I = Im 6 φ
The average (“real”) power absorbed by Z is,
P =1
2Vm Im cos (θ − φ) =
Vm√2
Im√2
cos (θ − φ)
= Veff Ieff cos (θ − φ) (Watts)
Apparent power is defined as Papp = Veff Ieff (Volt-Amp).
Power factor is defined as P.F . =Average power
Apparent power= cos (θ − φ).
(θ − φ) > 0: i(t) lags v(t), the P. F. is called a lagging P. F.
(inductive impedance)
(θ − φ) < 0: i(t) leads v(t), the P. F. is called a leading P. F.
(capacitive impedance)
M. B. Patil, IIT Bombay
![Page 158: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/158.jpg)
Apparent power, real power, and power factor
V
ZI
V = Vm 6 θ
I = Im 6 φ
The average (“real”) power absorbed by Z is,
P =1
2Vm Im cos (θ − φ) =
Vm√2
Im√2
cos (θ − φ)
= Veff Ieff cos (θ − φ) (Watts)
Apparent power is defined as Papp = Veff Ieff (Volt-Amp).
Power factor is defined as P.F . =Average power
Apparent power= cos (θ − φ).
(θ − φ) > 0: i(t) lags v(t), the P. F. is called a lagging P. F.
(inductive impedance)
(θ − φ) < 0: i(t) leads v(t), the P. F. is called a leading P. F.
(capacitive impedance)
M. B. Patil, IIT Bombay
![Page 159: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/159.jpg)
Apparent power, real power, and power factor
V
ZI
V = Vm 6 θ
I = Im 6 φ
The average (“real”) power absorbed by Z is,
P =1
2Vm Im cos (θ − φ) =
Vm√2
Im√2
cos (θ − φ)
= Veff Ieff cos (θ − φ) (Watts)
Apparent power is defined as Papp = Veff Ieff (Volt-Amp).
Power factor is defined as P.F . =Average power
Apparent power= cos (θ − φ).
(θ − φ) > 0: i(t) lags v(t), the P. F. is called a lagging P. F.
(inductive impedance)
(θ − φ) < 0: i(t) leads v(t), the P. F. is called a leading P. F.
(capacitive impedance)
M. B. Patil, IIT Bombay
![Page 160: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/160.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 161: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/161.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 162: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/162.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 163: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/163.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 164: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/164.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 165: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/165.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 166: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/166.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 167: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/167.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 168: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/168.jpg)
Power factor: examples
V
ZI
1. Given: V = 120 ∠ 0 V (rms), I = 2 ∠ (−36.9) A (rms).Find Papp , P.F., and P.
Papp = 120× 2 = 240 V-A.
P.F. = cos (0 − (−36.9)) = 0.8 lagging (since I lags V).
P = Papp × P.F. = 192 W .
2. Given: P = 50 kW , P.F. = 0.95 (lagging), V = 480 ∠ 0 V (rms).Find I.
Veff × Ieff × P.F = 50× 103
Ieff =50× 103
480× 0.95= 109.6 A.
Since P.F. is 0.95 (lagging), I lags V by cos−1(0.95) = 18.2.
⇒ I = 109.6 ∠ (−18.2) A (rms).
M. B. Patil, IIT Bombay
![Page 169: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/169.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 170: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/170.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 171: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/171.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 172: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/172.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 173: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/173.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.
The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 174: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/174.jpg)
Why is power factor important?
Vs VL ZL
IL
R = 0.1Ω
IC
C
Consider a simplified model of a power system consisting of a generator (Vs ),transmission line (R), and load (ZL).
The load is specified as P = 50 kW , P.F.= 0.6 (lagging), VL = 480 ∠ 0 V (rms).Note: lagging power factors are typical of industrial loads (motors).
P = 50× 103 W = |VL| × |IL| × P.F.⇒ |IL| =50× 103
480× 0.6= 173.6 A (rms).
Power loss in the transmission line Ploss = |IL|2R = (173.6)2 × 0.1 = 3 kW .
If the load power factor was 0.95 (lagging), we would have
|IL| =50× 103
480× 0.95= 109.6 A (rms), and Ploss = (109.6)2 × 0.1 = 1.2 kW .
Thus, a higher power factor can substantially reduce transmission losses.The effective power factor of an inductive load can be improved by connecting asuitable capacitance in parallel.
M. B. Patil, IIT Bombay
![Page 175: EE101: Sinusoidal steady state analysissequel/ee101/ee101_phasors_1.pdf · EE101: Sinusoidal steady state analysis M. B. Patil mbpatil@ee.iitb.ac.in sequel Department of Electrical](https://reader031.vdocuments.pub/reader031/viewer/2022021512/5af294277f8b9a95468b637c/html5/thumbnails/175.jpg)
Power computation: home work
I1
Is
I2
j5Ω−j2Ω
2Ω 10Ω
10 6 0 V
* Find I1, I2, Is .
* Compute the average power absorbed by each element.
* Verify power balance.
(SEQUEL file: ee101 phasors 2.sqproj)
M. B. Patil, IIT Bombay