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DC TRANSIENT DC TRANSIENT ANALYSISANALYSIS

CHAPTERCHAPTER 5 5

2

ObjectivesObjectives Investigate the behavior of currents and voltages

when energy is either released or acquired by inductors and capacitors when there is an abrupt change in dc current or voltage source.

To do an analysis of natural response and step response of RL and RC circuit.

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Lecture’s contentsLecture’s contents

• 5-1 NATURAL RESPONSE OF RL CIRCUIT

• 5-2 NATURAL RESPONSE OF RC CIRCUIT

• 5-3 STEP RESPONSE OF RL CIRCUIT• 5-4 STEP RESPONSE OF RC CIRCUIT

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First – Order CircuitFirst – Order Circuit• A circuit that contains only sources, resistor and

inductor is called and RL circuit.• A circuit that contains only sources, resistor and

capacitor is called an RC circuit.• RL and RC circuits are called first – order circuits

because their voltages and currents are describe by first order differential equations.

vs

R

–+

Ci

–+

L

R

iVs

An RL circuit An RC circuit

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Review (conceptual)Review (conceptual)

• Any first – order circuit can be reduced to a Thévenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor.

In steady state, an inductor behave like a short circuit. In steady state, a capacitor behaves like an open

circuit.

LRNIN –+

VTh C

RTh

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• The natural response of an RL and RC circuit is its behavior (i.e., current and voltage ) when stored energy in the inductor or capacitor is released to the resistive part of the network (containing no independent sources)

• The steps response of an RL and RC circuits is its behavior when a voltage or current source step is applied to the circuit, or immediately after a switch state is changed.

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5-1 Natural Response of 5-1 Natural Response of an RL circuitan RL circuit

Consider the following circuit, for which the switch is closed for t<0, and then opened at t = 0:

The dc voltage V, has been supplying the RL circuit with constant current for a long time

LRo RIs

t = 0 i +

V

–

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Solving the circuitSolving the circuit

• For t ≤ 0, i(t) = Io

• For t ≥ 0, the circuit reduce to

• At t = 0, the inductor has initial current Io, hence i(0) = Io

• The initial energy stored in the inductor is,

LRo RIo

i +

v

–

202

1)0( LIw

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Cont.Cont.• Applying KVL to the circuit:

dtL

R

ti

tdi

tRidt

tdiL

tRidt

tdiL

tRitv

)(

)(

)()(

0)()(

0)()( (1)

(4)

(3)

(2)

•From equation (4), let say;

dvL

R

u

du (5)

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Cont.Cont.• Integrate both sides of equation (5);

• Therefore,

• hence, the current is

t

t

ti

ti oo

dvL

R

u

du)(

)((6)

tL

R

i

ti

)0(

)(ln (7)

tLRtLR eIeiti )/(0

)/()0()(

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Cont.Cont.• From the Ohm’s law, the voltage across the resistor R is:

• And the power dissipated in the resistor is:

• Energy absorb by the resistor is:

tLRIRtitv )/(0 Re)()(

tLRR Itivp )/(22

0 Re)(

)1(2

1 )/(220

tLReLIw

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Time Constant, Time Constant, ττ for RL for RL circuitcircuit• Time constant, τ determines the rate

at which the current or voltage approaches zero.

• The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36.8% of its initial current

• Natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in Fig. 5-1

• Time constant for RL circuit is

• And the unit is in seconds.

R

L

Figure 5-1

13

• The expressions for current, voltage, power and energy using time constant concept:

)1(2

1

Re

Re)(

)(

/220

/220

/0

/0

t

t

t

t

eLIw

Ip

Itv

eIti

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Switching timeSwitching time• For all transient cases, the following instants of switching times are

considered. t = 0- , this is the time of switching between -∞ to 0 or time before. t = 0+ , this is the time of switching at the instant just after time t = 0s

(taken as initial value) t = ∞ , this is the time of switching between t = 0+ to ∞ (taken as final

value for step response)• The illustration of the different instance of switching times is:

0t 0t

-∞ ∞

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Example 1Example 1

• For the circuit below, find the expression of io(t) and Vo(t). The switch was closed for a long time, and at t = 0, the switch was opened.

2H0.1Ω 10Ω20A

t = 0i0

+

V

–iL

40Ω

2Ω

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Solution :Solution :• When t < 0, switch is closed and the inductor is short circuit.

• When t > 0, the switch is open and the circuit become;

0.1Ω 10Ω20A

iL(0-)

40Ω

2Ω

Therefore; iL(0-) = 20A

2H10Ω

20A

io(0+)

+

vo(0+)

–iL(0+)

40Ω

2Ω Hence; iL(0+) = iL(0-) = 20A

Current through the inductor remains

the same (continuous)

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RT = (2+10//40) = 10Ω

So, time constant, sec

By using current division, the current in the 40Ω resistor is:

Using ohm’s Law, the Vo is,

Hence:

2.010

2

TR

L

Aii Lo 44010

10

Aeti to

54)(

VtVo 160404)(

tetV 50 160)(

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Example 2Example 2The switch in the circuit below has been closed for a long time.

At t = 0, the switch is opened. Calculate i(t) for t > 0.

+ 2H

2Ω

12Ω

t = 0

16Ω

4Ω

i(t)

40V

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Solution :Solution :• When t < 0, the switch is closed and the inductor is short circuit to dc.

The 16Ω resistor is short circuit too.

calculate i1;

using current division, calculate i(0-)

A812//42

40i1

+

2Ω

12Ω

4Ω

i(0-)40V

i1

Ai 6412

12)i(0 1

-

Hence; i(0) = i (0-) = 6A

Since the current through an inductor cannot

change instantaneously

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• When t > 0, the switch is open and the voltage source is disconnect.

sec4

1

eqR

L

2H12Ω

16Ω

4Ω

i(t)i(0+) = i(0) = i (0-) = 6A

hence;

RT = Req = (12 + 4)// 16 = 8Ω

Time constant,

Thus, A6)0( )( 4-t/ teeiti

Because current through the inductor is

continuously

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5-2 Natural Response of 5-2 Natural Response of an RC Circuitan RC Circuit

• The natural response of RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor, C is released to the resistors, R.

• Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0:

C

RoRVo

t = 0+

+v–

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Solving the circuitSolving the circuit• For t ≤ 0, v(t) = Vo

• For t > 0, the circuit reduces to

• At t = 0, the initial voltage v(0) = Vo

• The initial value of the energy stored is

+

v

–

C

RoRVo

+

i

2

2

1)0( oCVw

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Cont.Cont. • Applying KCL to the RC circuit:

(1)

(3)

(4)

(5)

0)()(

R

tv

dt

tdvC

0)()(

RC

tv

dt

tdv

RC

tv

dt

tdv )()(

dtRCtv

tdv 1

)(

)(

0 Rc ii

(2)

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Cont.Cont. • From equation (5), let say:

• Integrate both sides of equation (6):

• Therefore:

dyRCx

dx 1 (6)

RC

t

V

tv

o

)(

ln (8)

ttv

Vdy

RCdu

xo 0

)( 11(7)

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Cont.Cont.• Hence,

• The voltage is:

• Using Ohm’s law, the current is:

• The power dissipated in the resistor is:

• The energy absorb by the resistor is:

RCto

RCt eVevtv //)0()(

RCto eR

V

R

tvti /)()(

RCtoR e

R

Vvitp /2

2

)(

)1(2

1 /22 RCto eCVw

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Time Constant, Time Constant, ττ for RC for RC circuitcircuit

• The time constant for the RC circuit equal the product of the resistance and capacitance,

• Time constant, sec

• The natural response of RC circuit illustrated graphically in Fig 5.2

RC

Figure 5.2

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• The expressions for voltage, current, power and energy using time constant concept:

)1(2

1)(

)(

)(

)(

/22

/22

/

/

to

to

to

to

eCVtw

eR

Vtp

eR

Vti

eVtv

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Example 3Example 3The switch has been in position a for a long time. At time t = 0,

the switch moves to b. Find the expressions for the vc(t), ic(t) and

vo(t).

t = 0+

+Vo

–

5kΩ

10kΩ

a b 18kΩ

0.1μF12kΩ60kΩ90V

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SolutionSolutionAt t < 0, the switch was at a. the capacitor behaves like an open

circuit as it is being supplied by a constant source.

At t > 0, the instant when the switch is at b.

+

+Vc(0-)

–

5kΩ

10kΩ90VVvc 6090

15

10)0(

18kΩ

0.1μF

12kΩ60kΩ +

Vo

–

+Vc(0+)

– vc(0+) = vc(0-) = 60V

the voltage across capacitor remains the same at this particular instant

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VVo 246030

12)0(

RT = (18 kΩ + 12 kΩ) // 60 kΩ = 20 kΩ

time constant, τ = RTC = 20kΩ x 0.1 μF = 2ms

Vc(t) = 60e-500t V

Using voltage divider rule,

Hence,

500t-c 3e-

dt

dvC (t)i

Vo(t) = 24e-500t V

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Example 4Example 4The switch in the circuit below has been closed for a long time, and it is opened at t = 0. Find v(t) for t ≥0. Calculate the initial energy stored in the capacitor.

t = 0

+

+v–

3Ω 1Ω

20mF9Ω20V

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SolutionSolutionFor t<0, switch is closed and capacitor is open circuit.

+

+vc(0-)

–

3Ω 1Ω

9Ω20V Vvc 15)20(39

9)0(

For t>0, the switch is open and the RC circuit is

+vc(0+)

–

1Ω

20mF9Ω

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vc(0) = vc(0+) = vc(0-) = 15 V

Req = 1+9 = 10Ω

Time constant, τ = ReqC = 0.2s

Because;

So ;

Because;

Voltage across the capacitor; v(t) = 15e-5t V

Initial energy stored in the capacitor is;

wc(0) = 0.5Cvc2 =2.25J

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SummarySummary

No RL circuit RC circuit

1

2 Inductor behaves like a short circuit when being supplied by dc source for a long time

Capacitor behaves like an open circuit when being supplied by dc source for a long time

3 Inductor current is continuous

iL(0+) = iL(0-)

Voltage across capacitor is continuous

vC(0+) = vC(0-)

RCR

L

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5-3 Step Response of RL 5-3 Step Response of RL CircuitCircuit

• The step response is the response of the circuit due to a sudden application of a dc voltage or current source.

• Consider the RL circuit below and the switch is closed at time t = 0.

• After switch is closed, using KVL

dt

diLtRiVs )(

RVs

t = 0 +

v(t)

–

+

i

L

(1)

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Cont.Cont.• Rearrange the equation;

(4)

(3)

(2)

R

Vti

L

R

L

VtRi

dt

tdi ss )()()(

dtR

Vi

L

Rdi s

RVti

didt

L

R

s

)(

)(

00 )(

ti

s

t

RVu

dudv

L

R(5)

)(

)()(ln

0 RVI

RVtit

L

R

s

s

(6)•Therefore:

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Cont.Cont.• Hence, the current is;

• Or may be written as;

Where i(0) and i(∞) are the initial and final values of i, respectively.

• The voltage across the inductor is;

• Or;

tLRso

s eR

VI

R

Vti )/()(

tLRos eRIVtv )/()()(

/)]()0([)()( tL eiiiti

dt

diLtvL )(

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Example 5Example 5

The switch is closed for a long time at t = 0, the switch opens.

Find the expressions for iL(t) and vL(t).

2Ω10V

t = 0

+

3Ω1/4H

i

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Solution Solution When t < 0, the 3Ω resistor is short – circuit, and the inductor acts like short circuit.

2Ω10V +

iL(0-) AiL 52/10)0(

So; i(0) = i(0+) = i(0-) = 5A

Because inductor current cannot change

instantaneously

When t< 0, the switch is open and the both resister are in series.

2Ω10V +

3Ω

1/4H

iL(0+)

40

sR

L

R

T

T

20

1

5)32(

Time constant;

2Ω10V +

3Ω

iL(∞)

When t = ∞, the inductor acts as short circuit again.

ARVi TsL 2/)(

Thus: iL(t) = i(∞) +[i(0) – i(∞)]e-t/τ = 2 + 3e-20t A

And the voltage is:dt

diLtvL )( = -15e-20t V

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5-4 Step Response of RC 5-4 Step Response of RC CircuitCircuit

• Consider the RC circuit below. The switch is closed at time t = 0

• From the circuit;

• Division of Equation (1) by C gives;

R

v

dt

dvCI cc

s

Rt = 0 +

vc(t)

–i

Is C

(1)

RC

v

dt

dv

C

I ccs (2)

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Cont.Cont.

• Same mathematical techniques with RL, the voltage is:

• Or can be written as:

• And the current is:

• Or can be written as:

RCtsosc eRIVRItv /)()(

RCtos e

R

VIti /)(

v(t) = v(∞) + [v(0) – v(∞)]e-t/τ

dt

dvCti )(

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Example 6Example 6 The switch has been in position a for a long time. At t = 0, the

switch moves to b. Find Vc(t) for t > 0 and calculate its value at

t=1s and t=4s

3kΩ

t = 0+

+Vc

–

5kΩ

a b 4kΩ

0.5mF24V +

30V

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SolutionSolution

VVc 158

524)0(

VVVV cc 15)0()0()0(

When t<0, the switch is at position A. The capacitor acts like an open circuit.

3kΩ

+

+Vc (0-)

–

5kΩ24V

Using voltage division:

When t <0, the switch is at position B.

Since voltage across the capacitor remains

same.

sRC 2And the time constant is:

4kΩ

+ 30V0.5mF

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4kΩ

+ 30V

+Vc(∞)

–

At t = ∞, the capacitor again behaves like an open circuit.

VVc 30)( Hence;

Since, v(t) = v(∞) + [v(0) – v(∞)]e-t/τ

So; vc(t) = 30-15e-0.5t V

At , t = 1s, Vc(t) = 20.9VAt , t = 4s, Vc(t) = 28 V

And;