epc tpc 4 lctrs 8 n 9(1)

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

1/80

Electrical Principles & Circuits

Topic 4:

Transient Analysis of DC Circuits

Lectures 8 & 9

Prepared by Foo Kok Sey Harry

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

2/80

S/No. a b c d

Q

S/No. a b c d S/No. a b c d

Q1 Q14 Q2 Q15

Q3 Q16

Q4 Q17

Q5 Q18

Q6 Q19

Q7 Q20 Q8 Q21

Q9 Q22

Q10 Q23

Q11 Q24

Q12 Q25

Q13

Common Test on Electrical Principles & Circuits

Correct answers to the S2 2012/2013 25 Multiple-Choice Questions (11h Dec)

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

3/80

TOPIC 4: LEARNING OBJECTIVES

Understandthe meanings of

capacitive and inductive time constants,

five (5) time constants,

charging and discharging cycles ofRC and RL circuits

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

4/80

C [s] = R [ ] C [F]

L[s] = L [H] /R [ ]

Capacitive & Inductive Time Constants

[second] [Ohm] [Farad]= x

[second] [Henry] [Ohm]=

Slide 28

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

5/80

Apply the exponential rise and decay

equations to analyse and solve RC DCtransient circuits.

5 C [s] = 5 { R [ ] C [F] }

vc(t) = Vs ( 1e t / c = R C)

ic(t) = Im et /

C= R C


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

6/80

Apply the exponential rise and decay

equations to analyse and solve RL DCtransient circuits.

5 L [s] = 5 { L [H]/ R [ ] }

vL(t) = Vs et /

L = L / R

iL(t) = Im ( 1 e t / L = L / R)


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

7/80

vc(t) = Vs ( 1et / c = R C)

i(t) = Imet / C = R C


iL(t) = Im ( 1 et /

L= L / R

)

vL(t) = Vs et /

L = L / R

Exponential Rise Equations

Exponential Decay Equations

Slide 28

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

8/80

InductanceL

Capacitance

C+

ResistanceR

It neitherdelivers powernor

stores energy but

only dissipates

power.

It storesenergy in its

electric field

when charged

It stores energyin its magnetic

field when

energised.

a resistor has no

transient state. It

limits I, dissipates P

and creates p.d.

acts as an open

circuit in direct

current (DC)

steady state.

acts as a short

circuit in direct

current (DC)

steady state.

TRANSIENT ANALYSIS OF DC CIRCUITSThe table below summarises the properties of threepassive

components commonly used in electrical circuits.

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

9/80

Magnetic Circuit Formulae Electric Circuit Formulae

Note: A capacitorCcan produce

aburst of _____________.

COMPARISON BETWEEN MAGNETIC & ELECTRICAL FORMULAE

Magnetic Circuit Formulae Electric Circuit Formulae

Reluctance m [A/Wb or 1/H]

Magnetic Flux= MMF / m [Wb]

Resistance R [Ohm, ]

Current I = V /R [A] ; Also

Current I = V G ; G = 1 /R [ 1]

Inductance L [Henry, H]

L = N2 / m = N2 A / l

L = N / I = R [ s ]

Capacitance C [Farad, F]

C = A /d = Q / V = I t / VC = /R [ s / ]

Note: A coil L can produce a

surge of very ______________.

Inductors

dissipatesome heatdue to the small coil resistances.

An inductor stores magnetic

energy in the magnetic field of

the coil when it is energised.

Capacitors cannot be charged

instantly because their

capacitances oppose the flow of

current. A capacitor stores

electrical energy in the electric

field of its plates.

An inductor storesmagnetic energy in the

magnetic field of the coil

when it is energised.

A capacitor storeselectrical energy in the

electric field of its

plates.

high voltageV large currentI

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

10/80

Comparison betweenCapacitance and Inductance Formulae

C : capacitance[Farad F= s/ ] ; Also

C =

Time constant c Rwhere c = capacitive

time constant

L : inductance[Henry H = s] ; Also

L=

Time constant L x Rwhere L = inductive

time constant

Inductance LL

Capacitance CC

+

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

11/80

Comparison between Capacitance and Inductance Formulae

Capacitance C

C = o r A /d(applicable only to one capacitor)

C = (n1) o r A /d

(for n-plate capacitors)

o: permittivity of free space [F/m]

r: relative permittivity of dielectric material

A : plate surface area of uniformly charged

capacitor [m2]d: plate distance [m]

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

12/80

Comparison between Capacitance and Inductance Formulae

Inductance L

m: magnetic reluctance [At/Wb]o: permeability of free space [H/m]

r: relative permeability of material

(absolute value)l: mean flux path or length [m]

L = N

2

/ m wherem = l/ o r A [At /Wb = 1 / s]

N = No. of coil turns ;

A = cross-sectional area of core [m2

]

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

13/80

Q26 MCQ Final Exam EPC S1 Sep 10

Question 26

What is the total quantity of charge QTstored on

the equivalent capacitance or each series connected

capacitor i.e. C1 or C2 or Ceq as shown in Figure

Q26?

(a) QT = 0.545 mC

(b) QT = 1.636 mC(c) QT = 2.084 mC

(d) QT = 2.357 mC

(b) QT = 1.636 mC

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

14/80

CurrentI= Vs /(RT = R12 + R3 + R4)

I= 24 V / (600 + 150 + 250) = 0.024 A

VR4 = IR4 = 0.024 Ax 250 = 6 V

Thus, Qs(total) = Ceq(s)Vceq(s) = 272.73 F x 6 V

= 1.636 mC Ans. Q26(b)

C6 =

1500 F

C2=750 F

C4 = 1500 FI

Vs = 24 V

C5 = 1500 F

C3 =

500F

R3 = 150 Figure Q26

C1 = 750 F

R1 =

1.2k

R2 =

1.2k

R4 =

250

VR4= 6.0 V

0.024 A


I

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

15/80


(b) C3456 = C3 + C4/ 3 = 500 F + 1500 F / 3C3456 = 1000 F ; Ceq(s) = C1 x C2 / C1 + C2

Ceq(s) = 375x1000 /375 +1000 [ F] = 272.73 F

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

16/80


Question 27

What are the supply voltage Vs

applied across the

capacitive network shown in Figure Q27 and the

total electrical energy Wc stored in the equivalent

capacitance? Assume the total quantity of charge

QT stored in the capacitive network to be 2.7 mC?

(a) Vs = 5.19Vand Wc 14.0 mJ

(b) Vs = 5.19Vand Wc = 7.5 mJ(c) Vs = 5.19Vand Wc = 7.0mJ

(d) Vs = 9.0V and Wc = 14.0 mJ(c) Vs = 5.19Vand Wc = 7.0 mJ

Q27 MCQ Fi l E EPC S1 S 10

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

17/80


C5 = 120 F

C1 =

120 F

C2 = 120 F

Vs = ?V

C4

C3 = 60 F and C4 = 120 F

C6 =

120 F

C3

Figure Q27

Cp(total) = C12p + C34s + C56p = (240 + 40 + 240) F = 520 F

QT(total) = 2.7 mC (given)

5.19 V

Therefore, Vs =QT/ Cp(total) = 2.7 mC/ 520 F

= 5.19 V

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

18/80

A resistor R

neither produces a

burst of currentnor a surge of

voltage.

An inductor

stores magneticenergy in the

magnetic field of

the coil when it is

energised.

i = C dv / dt

Inductance

LCapacitance

C+

Resistance

R

Summary of Main Points (Part 1)

A coil L can

produce a surge of

very high voltageV.

A capacitorstores electrical

energy in the

electric field of

its plates.

A capacitor C

can produce a

large burst ofcurrent from its

stored charge.

vind = L di / dtR = V / I

A resistor R canonly dissipate heat

and / or light energy

as it is a non-energy

storage component.

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

19/80

v(t) = Vs (1 et/ )

R = l/ A =l/ARseries =R1 + R2 + R3+

Rp= 1/(1/R1 + 1/R2

+ 1/R3+)

C = or A / d

Cseries =1 / (1/C1 + 1/C2 +

1/C3+ )

Cp = C1 + C2 +C3

V = I R = I / G

P = V I = V2/R

P = I2 / R

i = Cdv / dt

Q = I t = C V

Vind = Ldi / dt

MMF = N I = =BA = H l ;

= l/or A

L = N2or A /lLseries =L1 + L2 + L3 +

Lp= 1/(1/L1 +1/L2

+ 1/L3+)

Summary of Fundamental Equations (Part 1)

i(t) = Im (1 et/ )

InductanceL

CapacitanceC+

I= V / R

ResistanceR iI

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

20/80

Time Constant c of RC DC Circuits

In a series RC circuit, thetime (t) required to

t C and t R

charge proportionala capacitor (C) is directly

to the capacitance (C) value and the resistance

(R) value.

C 1 / R

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

21/80

Recall:

Q = It and Q = CV;

Equating these two equations gives

where V / I = R (Ohms Law)

I t = C V t = C V / I = C R

Time Constant c of RC DC Circuits

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

22/80

t C and t R

= R C

R = 1 k ; C = 1000 F

1 = 1 k x 1000 x 10 6 F = 1 s

If R = 10 k ; C = 1000 F

2 = 10 s R

10 x as R 10 x

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

23/80

If R = 1 k ; C = 10 F

3 = 10 ms C

100 x as C 100 x

= R C

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

24/80

R 2 x as C 0.5x i.e.

C 1 / R = R C

R= 1 k ; C= 1000 F 1 = 1 k x 1000 x 10

6 F = 1 s

If R= 2 k ; C= 500 F 2 = 2 k x 500 x 10

6 F = 1 s

for the same time constant , C is

inversely proportional to R i.e. C 1 / R

Ch i C l f A El t l ti C it M d E

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

25/80

+

Capacitor C

RResistor

Switch S

+

DC

VoltageSupply

Vs

VR(t)

VC(t)

i(t): charging currentVR(t=0s) = Vs

Vs = VR(t) + VC(t)

Vs

= VC(t = 5 )

= R C

i(t=0s) = im = Vs/R

VC(t=0s) = 0 V

i(t) = R

Vs

et /

vC(t)

= Vs( 1 - e

t/

)

Charging Cycle of An Electrolytic Capacitor Made Easy

C iti Ti C t t

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

26/80

At 5 ; vc (t = 5 ) 10 V

If Vs= 10 V ; vc (t = ) = 6.3 V

One time constant c is the time required for a

capacitor to charge to about ____ of the

supply voltage Vsduring the ______________

63%

charging cycle .

Exponential rise curve for

vc (t) = Vs ( 1et / = RC)

5

0.63Vs

0

Voltage [V]

Time constant

Vs

0.63Vs

Capacitive Time Constant C

C i i Ti C

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

27/80

0

Voltage [V]

Time constant

Vs

0.37Vs

Conversely, it is the time required for

a capacitor todischarge to about _____ of itsavailable voltage during the _______________.37%discharging cycle

5

Capacitive Time Constant C

Exponential decay curve

forVR(t) = Vset / = RC

and i(t) = Im e t / = RC

Exponential Rise and Decay Curves of Charging

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

28/80

Exponential Rise and Decay Curves of ChargingCycle of Series RC DC Circuits

R C+V Vs

+i(t) Note: Vc = Vs when

C is fully charged andcurrent i = 0A.

VR(t)Vc(t)

5

0.63VsExponential rise curve forvc (t) = Vs ( 1e

t / = RC )

Slide 4

Slides 7, 30 43Note: c = = capacitive time constant

2 4 760

Voltage [V]

Time constant 3

Vs

0.37VsExponential decay curve for

VR(t)

= Vs

e t / = RC and

i(t) = Im et / = RC

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

29/80

Exponential Rise Curve Equation

vc (t) = Vs

Equation for Calculating the

Capacitive Voltage During Charging

NoteCapacitor has no charge at time t = 0s

and when switch S is open.

( 1 e t/ )= R C

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

30/80

I1 S

RC DC Charging Circuit

Ri

2

R2

R1

Vs C

Capacitive charging time constant c = (Ri + R1) C

Transient capacitive voltage vc(t) = Vs( 1et/

c= R C)

vc(t)ii

i (charging current)

R

Note :R1 and C are connected to the DC energy

sourceVsduring the charging cycle.

Slide 28

( )

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

31/80

Note : DC Energy Source Vsis disconnectedduring

the discharging cycle.

Capacitive discharging time constant d = (R1 + R2) C

Transient capacitive voltage vc(t) = Vs et/

d= R C)

I1

S

RC DC Discharging Circuit

Ri

2

R2

R1

Vs

C vC(t)

I

i

i (discharging current)

R

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

32/80

Exponential decay curve for :

vc(t) = Vset / = RC and

i(t) = Im et / = RC

Exponential Decay Curve of Voltage across Capacitor and itsCurrent during Discharging Cycle of Series RC DC Circuits

Voltage V / Current i

Time constant 0

Vs/ Im

5

0.37Vs

E ti f C l l ti th C iti

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

33/80

Exponential Decay Curve Equation

vc (t)= Vs

Exponential Decay Curve Equationic (t) = Im

Equation for Calculating the CapacitiveVoltage During Discharging

Equation for Calculating the Charging /

Discharging Current

e t/ = R C

= R Ce t/

Charging Holding and Discharging Cycles

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

34/80

Charging, Holding and Discharging Cycles

f

Discharging cycleCharging cycle

d

Vs

R

C

constant

> 5 (holding time)

e

V[ V ] /I[A] / Q[mC]

5 5 0

c

b

a

t [s/ms/ s]

Not drawn to scale

Q28 MCQ Fi l E EPC S1 S 10

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

35/80


Question 28

How long will it take the fully charged capacitor

as shown in Figure Q28 to be fully discharged

when switch S is placed at position 2. Assume

that all the electrical components are in goodoperating conditions.

(a) 12.65s(b) 8.54s

(c) 4.75s

(d) 2.53s

(a) 12.65s


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

36/80


vc = ?V

i1 S

Figure Q28

Ri =0.9

2

R2 =

2.2k

R1 = 330

Vs = 12V C = 1000 F

Discharging capacitive time constant

d = (R1 + R2) C = (2.2 k + 0.33k ) x 1000 F = 2.53 s

When switch S is placed at position 2, it will take about5 d= 5 x 2.53s =12.65 s for the charged capacitor to be

almost completely discharged.

Answer Q28(a) is correct.

Exponential Rise and Decay Curves of Series

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

37/80

Time constant 2 3 4 5 76

0.63 Qm

Current I [ mA, A ]

0

Im= Vs/ R

0.37 Im


Qc (t) = CVs ( 1et / = RC)

Exponential decay curve for

i(t) = (Vs / R) et / = RC

Qm

= CVs

At t = 0s ; Q = 0C; Vc = 0V and Im = Vs/ R

As t ; Q CVs ; Vc Vs; VR 0V; I 0A

Exponential Rise and Decay Curves of SeriesRC DC Circuits (Quantity of Charge Q)

Exponential Rise and Decay Curves of Charge Q

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

38/80

0.37Qm


Qc(t) = Qme t / = RC

and i(t) = Im et / = RC

Time constant

2 3 45

76

0

Qm

0.63Qm


Qc (t) = Qm ( 1et / = RC)

R C+V

Vs

i(t)VR(t)

Qc(t)

Exponential Rise and Decay Curves of Charge Qof Series RC DC Charging / Discharging Circuits


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

39/80

Q1 MCQ Final Exam EPC S1 Sep 10Question 1

Which one of the following statements about the charging and

discharging cycles of a 1200 resistor connected in series with a

470 F capacitor across a 12 V DC energy source shown in Figure Q1is false? Assume that the capacitor has yet to be charged.

(a)Reducing the resistance R by 10 times and increasing the

capacitance C by the same number of times, means that resistance R isinversely proportional to the capacitance C.

(b) Calculating the capacitive time constant gives C = 564ms and at

the instantaneous time t1 = 1.41s the equivalent number of time

constants is 2.5 C.

(c)Charging the 470 F electrolytic capacitor to its full capacity at time

t3 means that the time taken for it to be fully charged is about 2.82s.

(d) Discharging the fully charged capacitor from time t4 to t5 will

result in the quantity of charge stored on it dropping from 0.0564C to

about 2.075mC.

(a) Reducing the resistance R by 10 times and increasing the

capacitance C by the same number of times, means that resistanceR is inversely proportional to the capacitance C.


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

40/80

t 5t 4t 3t 1t 2

2.5 c=2.5 x 0.564s =1.41s


Discharging cycleCharging cycle

constant

5 = 2.82s

> (holding time)

5

f

0

Figure Q1 e

c

d

t = 0.564s

= 0.564s

b

a

t [ s ]

V [V]Q [ mC ]

Not drawn to scale

Vs = 12V

R = 1.2 k

C = 470 FQm = C Vs=470 F x 12 V =

5.64 mC

c = R C =1200 x 470 F= 564 ms = 0.564s

Charging time constant, c =

d, discharging time constant

= R x C


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

41/80


RecallQm = C Vs = 470 F x 12 V = 5.64 mC

RecallQ

d

= Qm et /

d where t = d and Qm =5.64 mC

Now,Q

d= 5.64mC ( e

d / d ) = 5.64 mC ( e 1)

Maximum quantity of charge Qm

Quantity of charge Q at 1 time constant

= 2.075 mC (at one discharging timeconstantfrom t4 and t5)

E ti l Ri d D C f

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

42/80

Exponential Rise and Decay Curves of

Series RC DC Circuits

It can be deduced from the exponential

rise and decay curves of the respective

series RC DC circuits that the electrolyticcapacitor C isalmost fully charged, while

the _______ across the resistance Rhas

decreased exponentially to almost zerovolt at _____________________.

Slides 4 & 28

voltage

five time constants (5s )

T h th t it i b t 100%

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

43/80

DC supply voltage Vs = 30 V,

resistance R = 10 k , and

capacitance of electrolyticcapacitor C = 50 F

(capacitor C has no charge Q[Vc = 0 V; Q = nil ] on it before the

charging process.

To show that capacitor is about 100%

charged at 5 time constants (5 cs)


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

44/80

Voltage across charging capacitor att = 5 is Vc (5 ) = Vm ( 1e

t/ ).

Substituting the values gives

Vc= 30V x 0.99326 29.8V


charged at 5 time constants (5 cs)

Vc= 30V ( 1e5 / )

= 30V (1e 5)

= 30V(10.00674)


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

45/80

Vs = 30V

R = 1.2 k

C = 470 F


charged at 5 time constants (5 c s)

Voltage across charging capacitor at

t = 5 is Vc (5 ) = Vm ( 1et/ ).

Vc= 30V ( 1e 5 / )= 30V (1e 5)

= 30V (1 0.00674) 29.8V Ch i C l f RC DC Ci it

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

46/80

Voltage across

charging

capacitor C

t = 0 t = c t = i.e. >>5

Vc (t) 0V 0.63Vs Vs

Vc(t) = Vs( 1et/ c ) [V]

Charging Cycle of a RC DC Circuit

Exponential Rise Equation

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

47/80

Charging Cycle of a RC DC Circuit

t = 0s Vc(t)

= 0V

t = Vc(t) = 0.63 Vs

t = i.e. 5 Vc(t) = Vs

Vc(t) = Vs( 1et/ = R

1C )

Voltage Curves of a RC DC Circuit

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

48/80

Voltage Curves of a RC DC Circuit

OFFOFF ON

100%

100%

63%

37%

VR +

t

c=RC 5c c=RC

Discharging

cycle

Charging

cycle

37%

100%

100%

037%

R

C

+V

Vs

+

(0V)

VR

Vc

Vs +

0 t1 t2 t3t

Vc +

0 t

Slide 4 & 32

i(t)

Switch S

Di h i C l f

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

49/80

Vc(t) = Vs et/(

d= RC)

Discharging Cycle of a

RC DC Circuit

Note: R = R1

+ R2

t = 0s Vc (t) = Vst = d Vc (t) = 0.37 Vs

t = i.e.>> 5

Vc (t) = 0V

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

50/80

2VR1(t)

VR2(t)vc(t)

R1

C

+

R2

+

Vs

1

Discharging Cycle of a

RC DC Circuit

Note: R = R1 + R2

S

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

51/80

Tutorial on RC DC Circuits

Q1. After what time is the charging

process of a capacitor of 33 F

completed. The DC supply voltage Vsacross the capacitor C is 60V and the

series resistor R is 10k .

Ans.: t = 5c = 1.65s; c = 0.33s

Solutions to Tutorial on Transient

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

52/80

Solutions to Tutorial on Transient

RC DC Circuits

Q1. The charging process of anelectrolytic capacitor is completed in

about five (5) capacitive time constantsi.e. 5 c .

Capacitive time constant c

= R C

c= 10 x 103 x 33 x 106F = 0.33s

Thus, 5 c= 5 x 0.33s = 1.65 s

Charging Cycle of a RL DC Circuit Made Easy

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

53/80

When a DC supply voltage isfirst applied to a

series RL circuit, VLequals Vs , VR equalszero and current I is zero.

iL = I1 S

RL DC Charging Circuit

Ri2

R2

R1

L VLVs


VR1

iL = I = 0A

VLVs

At t = 0 s

= 0 V

= Vs

Exponential Rise and Decay Curves of Current I

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

54/80

0.37 Vs


vL(t) = Vs et / = L/R


Exponential Rise and Decay Curves of Current Iand VL of Series RL DC Charging / Discharging

Circuits

Im

0.63 Im

Exponential rise curve foriL (t) = Im ( 1e

t / = L/R)


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

55/80

After five time constants (5 s) , VL equals

zero, VR equals Vs and current I is

maximum (steady state values).

VL

iL1 S


Ri2

R2

R1

Vs

L

= I = Im

VR1

= 0 V

= Vs

At t = 5


The time (t) for current build up in a

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

56/80

The time (t) for current build-up in a

series RL circuit isdirectly proportional

( ) to the amount of inductance (L) inthe circuit, i.e. t L.

Inductive time constant L = L /RR = 100 ; L = 1 H

L1 = 1 H /100 = 10 msR = 100 ; L = 0.1 H

L2 = 0.1 H / 100 = 1 ms

L 10x 10x

R kept constant

However the time for current build up

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

57/80

However, the time for current build-up

in a series R L circuit is inversely

proportionalto the amount ofresistanceR in the circuit, i.e. t 1/ R

L = L/R R = 100 ; L = 1 HL1 = 1 H /100 = 10 ms

If R' = 10 ; L = 1 H

L2 = 1 H /10 = 0.1s

R 10x 10x L kept constant

Hence inductive time constant

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

58/80

Hence, inductive time constant

L = L /R[ s ] [ H ] [ ]

t L R kept constant

t 1/ R L kept constant

L =

(constant)

L

R

L =

(constant)L

R Exponential Rise and Decay Curves of a Series RL

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

59/80

DC Circuit

Current I


Im

0.37ImExponential decay curve

for iL(t) = Im e

t /

andVL(t) = Vset /

0.63Im


iL (t) = Im ( 1e t / )and VR(t) = Vs ( 1 - e

t / )

Note: = inductive time constant

R L+V

Vs

+i(t)

VR(t) VL(t)

Exponential Rise Curve of Charging Current Flowing

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

60/80

Current I [A]


Im

Exponential Rise Curve of Charging Current Flowingin Series RL DC Circuits

0.63Im

Exponential rise curve & equation

for iL (t) = Im ( 1et / = L/R )

I

Equation for Calculating the Inductive

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

61/80


iL (t)

= Im

( 1e )


VR (t) = Vs (1 et/ = L / R

)

Equation for Calculating the InductiveCurrent During Charging Cycle

Equation for Calculating the Voltage

across the Resistor While Charging

t/ = L / R

t/ = L/ R

iL1 S

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

62/80

vL

iL1 S


Ri2

R2

R1

Vs

L

iL (t) = Im (1 et/

L= L / R)

VR1 (t)= Vs (1 et/

L= L / R)

Note: Charging inductive time constant L = L / (Ri + R1)

iL1

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

63/80

Note : DC Energy Source Vs is disconnected during the discharging cycle.

vL

L

S

RL DC Discharging Circuit

Ri

2

R2

R1

Vs

L

iL(t)

= Im

e t/ d = L / R

VL (t) = Vs et/

d= L / R

Discharging inductive time constant d = L / (R1 + R2 )

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

64/80

Series RL DC Exponential Decay Curve

Equation During Charging Cycle

Series RL DC Exponential Decay Curve

Equation During Discharging Cycle

vL (t) = Vs et/ = L / R

iL (t) = Im et/ = L / R

Charging, Holding and Discharging Cycles

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

65/80

g g, g g g y

Discharging cycle

f

5

Not drawn to scale

a

e

constant

> (holding time)

R L

Vs

0 t [s/ms/ s]

c

b

I [ A ] / V [V]

Charging cycle

d5

Exponential Rise and Decay Curves of a

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

66/80


0

0.63Im

Im

0.37Im

Exponential rise curve foriL (t) = Vs / R [( 1e

t / = L/R)]

Exponential decay curve

VL(t) = Vse t / = L / R

Note: Inductive time constant = L / R

At t = 0s ; I = 0A; VR = 0V and VL = Vs

As t ; I Vs/ R ; VL 0V ; VR Vs

t

Series RL DC Circuit

Current Curves of a RL DC Circuit

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

67/80

100%

63%37%

37%

IR

IL

+

+

100%

100%

OFF

0 t1

0

0

t

t3

t

5 L

L= RL

OFF

Discharging

cycle

ON

t2Charging

cycle

63%

100%

Im +

t

t t

R

L

+V

Vs

+

i(t)

VR

VLL=R L

Tutorial on RL DC Circuits

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

68/80


Q1.

When does the current of an inductor of

coil resistance 15 and inductance 1.9H

reach 65% of its final value after beingswitched on?

Ans.: t = 0.133s

Q1 Using the exponential rise

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

69/80

Q1. Using the exponential rise

current equation iL = Im(1 et/ L )

where the inductive charging time

constant = L = L / R =

L = 1.9H / 150.1267s ;

t = time when transient charging

inductive current iLreaches 65% of itsfinal (maximum) value after the coil

has been energised at to = 0s


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

70/80

Hence, iL = 0.65 Im = Im(1e

t / =L/R

)

Simplifying it gives,

1 /e t/ = 0.35 which yields

= e t/ =L/R = 10.65 = 0.35

e t/ =L/R = 1 / 0.35

iL



7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

71/80

Taking natural logarithm on both sides

gives ln e t / = ln 2.857

t / =L / R (ln e1=1) =ln 2.7183 =1.0498

Therefore, t = x 1.0498t = 0.1267s x 1.0498 = 0.133s


Q2

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

72/80

Q2.

What is the inductance L of a choke whose

resistance is 35 if the current flowing in itreaches of its maximum value of 1A

0.5s after it has been switched on ?

Ans.: L = 0.36067s ; L = 12.62H

R i ti DC Ci itC iti DC Ci it

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

73/80

Resistive DC CircuitsCapacitive DC Circuits

1 / Cs =

1 / C1 + 1 / C2 + 1 / C3

Vs = V1 + V2 + V3

(KVL: known as

Kirchhoffs Voltage

Law (KVL) or general

EMF equation)

C 1

+

+ +

V1 V2 V3

C 2 C 3

Rs

= R1

+ R2

+ R3

Vs = V1 + V2 + V3

(KVL) Also

Vs = I (R1 + R2 + R3)Voltage Divider Rule:

V1=Vs [R1 /(R1 + R2+ R3)

Q: Quantity of charge common I: Current is common

R2R1 R3

V1 V2 V3

Series Connection

I

Resistive DC CircuitsC iti DC Ci it

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

74/80

Resistive DC CircuitsCapacitive DC Circuits

Parallel

Connections

VC1 = VC2 = VC3

Cp= C1 + C2 + C3

+ C 2

+

C 3

+ Vs

C 1

1 / Rp= 1 / R1 + 1 / R2 + 1 / R3

GT = G1 + G2 + G3

Rp = 1 / GT

Current Divider Rule:

I1 = IT[ Rp / R1 ] where

IT= I1 + I2 + I3

Vs

IT

R1

R3

R2I2

I3

+

I1

VR1 = VR2= V

R3= V

s

Summary of Fundamental Equations (Part 1)C it

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

75/80

v(t) = Vs (1 et/ )

R = l/ A =l/AR

series=

R1 + R2 + R3+

Rp= 1/(1/R1 + 1/R2

+ 1/R3+)

C = or A / d

Cseries

=

1 / (1/C1 + 1/C2 +

1/C3+ )

Cp = C1 + C2 +C3

V = I R = I / G

P = V I = V2/R

P = I2 / R

i = Cdv / dt

Q = I t = C V

Vind = Ldi / dt

MMF = N I =

= BA = H l

=l/or A

L = N2or A /l

Lseries

=

L1 + L2 + L3 +

Lp= 1/(1/L1 +1/L2

+ 1/L3+)

i(t) = Im (1 et/ )

Inductance

LCapacitance

C+

I= V / R

Resistance

R iI

Capacitive DC Circuits Resistive DC Circuits

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

76/80

Capacitance C = A / d

where absolute

permittivity = o r ;

o = permittivity of free

space (vacuum) ;

r= relative permittivityof dielectric material ;

d = plate distance and A =

plate surface area ;C = o r A /d; Also

C=Q /V where Q = I t

and Q = C V

Resistance R = V / I

where R = 1 /Conductance

G ; Also R = l/ A

(a conductor) and R = 2l/A

(line resistance of circuit)

Since Q = I t ;R = V ( Q / t ) i.e.

R = V t / Q Hence,

R w/c = R20[ 1 20 ]

where 20= temperaturecoefficient of resistance at

20C ; = temperature

difference = w20C

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

77/80

Capacitive DC Circuits Resistive DC Circuits

Electrical energy of a

capacitor Wc ;

(Work done = Q V)

Wc = (CV2);

Also Wc = (QV)

as Q = C V

Electrical energy

W(Work done)

W = VI t ; V = I R

and P = VI

W = P t ;

W = I2 R t ;

W = (V2/R) t

RC DC Transient Formulae RL DC Transient Formulae

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

78/80

Charging Cycle (Switching ON)

Vc(t)

= Vs

( 1et / c)

i(t) = Im et / = R C

Time constant = R C

Maximum current Im

= Vs

/ R

i(t) = Im ( 1et / L)

V L (t) = Vs et / = L / R

Time constant = L / R

Maximum currentIm = Vs / R

i(t)

C

+

R

Vs

i(t)

VR

Vc

+

1

ON0

OFF

Slide 4 i(t)

Charging Cycle (Switching ON)

R

L

Vs

i(t)

VR

VL

+

0OFF

1ON

Slide 28


7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

79/80


Source-free circuit (switch S at

position 0)

Vc(t) = Vmet / = RC

= capacitive discharging

time constant

Source-free circuit (switch S at

position 0)

i(t) = Im et / = L/R

= inductive discharging

time constant

Discharging cycle (switching OFF) Discharging cycle (switching OFF)

i(t)

R

L

Vs

i(t)

VR

VL

+

0

OFF1ON

i(t)

+

R

Vs

i(t)

VR

Vc

+

1ON

0OFF

C

Slide 15 Slide 33

7/29/2019 EPC Tpc 4 Lctrs 8 n 9(1)

80/80

End of Topic 4:

Transient Analysis of

DC Circuits

epc tpc 4 lctrs 8 n 9(1)

Documents