ee602 ac circuit

7/27/2019 Ee602 Ac Circuit

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CIRCUITANALYSIS

INTRODUCTION TOAC CIRCUITS


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CHAPTER GUIDELINE

Introduction to alternating current

Characteristics of Sinusoidal Function

Phasor representation of sinusoids

Impedance and reactance

Parallel and series combinations in the frequencydomain

Ohms law and Kirchoffs law in the frequency domain

Delta-Wye Transformation


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COURSE OUTCOMES (CO)

Understand the concept of ac Identify the characteristics of ac signal & sinusoidalfunctions

Understand the concept of phasor and its application

in ac circuits Understand the concept of impedance, admittance,reactance and susceptance

Understand the Ohms law and Kirchoffs law in accircuit and relate them with dc circuit

Construct an AC circuit to investigate the relationbetween sinusoidal signals and phasor


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PHASOR REPRESENTATION OF SINUSOIDS

Sinusoids are easily expressed in terms of phasors

A phasor is a complex number that represents the amplitude and phaseof a sinusoid.

It can be represented in one of the following three forms:

rzjrez

)sin(cos jrjyxz a. Rectangular

b. Polar

c. Exponential 22 yx

zofmagnituder

xy

zofphasethe

1tan

where

1j

cosrx

sinry

jj

1


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Mathematic operation of complex number:1. Addition

2. Subtraction

3. Multiplication

4. Division

5. Reciprocal

6. Square root

7. Complex conjugate

8. Eulers identity

)()( 212121 yyjxxzz

)()(212121 yyjxxzz

212121 rrzz

21

2

1

2

1 r

r

z

z

rz

11

2 rz

jrerjyxz

sincos je j

(Rectangular form)

(Rectangular form)

(Polar form)

(Polar form)


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EXAMPLE

Evaluate the following complex numbers:

a.

b.

Solution:

a. 15.5 + j13.67

b. 8.293 + j2.2

]605j4)1j2)([(5o

oo

3010j43

403j510


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Transform a sinusoid to and from the time domain to thephasor domain:

(time domain) (phasordomain)

)cos()(

tVtv m

mVV

Amplitude and phase difference are two principal concerns in thestudy of voltage and current sinusoids.

Phasor will be defined from the cosine function in all our proceeding

study. If a voltage or current expression is in the form of a sine, it will

be changed to a cosine by subtracting from the phase.


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PHASOR REPRESENTATION OF

SINUSOIDS

Time domain

representation

Phasor domain

representation

)cos( tVm

)sin( tVm

)cos( tIm

)sin( tIm

mV

o

mV 90

mI

omI 90

Sinusoid-phasor transformation


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EXAMPLE

Transform the following sinusoids to phasors:i = 6cos(50t 40o) A

v =4sin(30t + 50o) V

Solution:

a. I A

b. Since sin(A) = cos(A+90o);

v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V

Transform to phasor => V V

406

1404


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EXERCISE:

Find the sinusoids corresponding to these phasors:a.

b. V3010 V

Aj12)j(5 I

Solution:

a) v(t) = 10cos(t + 210o) V

b) Since

i(t) = 13cos(t + 22.62o) A

22.6213)

12

5(tan512j512 122I


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The differences between v(t) and V:

v(t) is instantaneous or time-domainrepresentationV is the frequency or phasor-domain representation.

v(t) is time dependent, V is not. v(t) is always real with no complex term, V is

generally complex.

Note: Phasor analysis applies only when frequency is constant; whenit is applied to two or more sinusoid signals only if they have

the same frequency.


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Relationship between differential, integral operation

in phasor listed as follow:

)(tv VV

dt

dvVj

vdt jV



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So now we know how to represent a voltage or current in the phasor or

frequency domain

But how can we apply this to circuits involving the passive elements R,

L and C?

transform the voltage-current relationship from the time domain to the

frequency domain for each element


Time

domain

Frequency

domain Timedomain

Frequency

domain

Time

domain

Frequency

domain


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Summary of voltage-current relationship

Element Time domain Frequency domain

R

L

C

Riv IRV

dt

diLv IV Lj

dt

dvCi

Cj

Cj

IV

VI



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EXAMPLE

If voltage v(t) = 12cos(60t +45o) is applied to a 0.1 Hinductor, calculate the current, i(t), through the inductor.

Atti

domaintimetothisConverting

VLj

o

o

o

o

)4560cos(2)(

,

452906

4512

VI


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EXERCISE

If voltage v = 10 cos (100t+30o) is applied to a 50F

capacitor, calculate the current through the capacitor.

Answer: 50 cos (100t + 120o) mA


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IMPEDANCE AND REACTANCE

The impedance Z of a circuit is the ratio of the phasor

voltage V to the phasor current I, measured in ohms .

where R = resistance

X = reactance

The admittance Y is the reciprocal of impedance,

measured in siemens (S).

jXRIVZ

V

I

ZY

1

Positive X is for LNegative X is for C


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RY

1

LjY

1

CjY

Impedances and admittances of passive elements

Element Impedance Admittance

R

L

C

RZ

LjZ

CjZ

1

IMPEDANCE AND REACTANCE


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OHMSS LAW AND KIRCHOFFS LAW IN THE FREQUENCY

DOMAIN

N series connected shown in figure impedances

Applying KVL:

V = V1 + V2+ + VN = I(Z1+Z2++ZN)

similar to the series connection of resistances

If N = 2 as shown in figure below,

1 2

VI

Z Z

1 1 2 2, V Z I V Z I

1 2

1

1 2 1 2

2

Z ZV V, V = V

Z Z Z Z


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DOMAIN

N Parallel connected impedances

Voltage across each impedances is same

Applying KCL:

1 2

1 2

1 2

1 2

.....

1 1 1......

1 1 1 1......

....

N

N

eq N

eq N

I I I I

V Z Z Z

I

Z V Z Z Z

Y Y Y Y


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DOMAIN

N Parallel connected impedances

If N = 2 as shown in figure below,

1 2

1 2 1 2 1 2

1 1 2 2

2 1

1

1 2 1 2

1 1

1/ 1/

eq

eq

eq

2

Z Z1Z =

Y Y Y Z Z Z Z

V IZ I Z I Z

Z ZI I, I = I

Z Z Z Z


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DELTA-WYE TRANSFORMATION

The delta-to-wye and wye-to-delta transformations that we applied to

resistive circuits are also valid for impedances

1 2 2 3 3 1

1

1 2 2 3 3 1

2

1 2 2 3 3 13

3

a

b

Z Z Z Z Z ZZ

Z

Z Z Z Z Z ZZ

Z

Z Z Z Z Z ZZ

Z

Y conversion: - Y conversion:

1

2

3

( )

( )

( )

b c

a b c

c a

a b c

a b

a b c

Z ZZ

Z Z Z

Z ZZ

Z Z Z

Z ZZ

Z Z Z


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EXAMPLE:Find the input impedance of the circuit shownbelow. Assume that the circuit operates at =50 rad/s.

1 3

2 3

3

2 2

1 110

50 2 10

1 13 3 (3 2)

50 10 10

8 8 ( 50 0.2) (8 10)

(3 2)(8 10)1 ( 2 || 3) 10

11 8(44 14)(11 8)

10 10 3.22 1.0711 8

3.22 11.07in

Z jj C j x x

Z jj C j x x

Z j L j x j

j jZin Z Z Z j

jj j

j j j

Z j


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EXERCISE:

Determine vo(t) in the circuit shown below.

Tips: to do the analysis in frequency domain, we must first transform the time-

domain circuit to the phasor domain equivalent

Answer: vo(t) = 17.15 coz (4t +15.96o)V


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SOLUTION

3

1

2

2

1 2

20cos(4 15 ) 20 15

1 1104 10 10

5 4 5 20

60

25 2025 || 20 100

25 20

100(20 15 )

60 100(0.8575 30.96 )(20 15 ) 17.15 15.96

( ) 17.15cos(4 15.

o o

x

o

o s

o o o

o

v t Vs V

mFj C j x x

H j L j x j

Z

j xjZ j j j

j j

Z jV V

Z Z jV

v t t

96 )o V


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Calculate vo in the circuit shown below.

EXERCISE

Answer: vo(t) = 7.01 cos (10t 60o)V

ee602 ac circuit

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