R-L AND R-C CIRCUITS

TRANSIENTS IN R-L AND R-C CIRCUITS

• One capacitor and one resistor• The source and resistor may be equivalent to a

circuit with many resistors and sources.

R+

-Cvs(t)

+

-

vc(t)

+ -vr(t)

FIRST-ORDER RC CIRCUITS

R

1

C

2

K

E

RvEi c

c

KVL around the loop: EvRi Cc

EvRdtdvC c

c

EAev RCt

C

Initial condition 0)0()0( CC vv

)1()1( t

RCt

C eEeEv

dtdvCi c

c t

eRE

Switch is thrown to 1

RCCalled time constant

Transient Response of RC Circuits

EA

FIRST-ORDER RC CIRCUITS

)1( t

C eEv

/tc eE

dtdv

0

0

t

ct

c

dtdv

EEdtdv

RCTime Constant

R

1

C

2

K

E

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msV(2)

0V

5V

10V

SEL>>

RC

R=2kC=0.1F

FIRST-ORDER RC CIRCUITS

Switch to 2

R

1

C

2

K

E

RCt

c Aev

Initial condition Evv CC )0()0(

0 Riv cc

0dtdvRCv c

c

// tRCtc EeEev

/tc e

REi

Transient Response of RC Circuits

cc

dvi Cdt

FIRST-ORDER RC CIRCUITS

RCTime Constant

R

1

C

2

K

E

R=2kC=0.1F

Time

0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0ms 8.0msV(2)

0V

5V

10V

SEL>>

t

RCt

C EeEetv

)(

E

dtdv

t

C 0

0

t

C

dtdv

E

FIRST-ORDER RC CIRCUITS

Time

0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms 5.5ms 6.0msV(2) V(1)

0V

2.0V

4.0V

6.0V

FIRST-ORDER RC CIRCUITS

Ideal Linear Inductor

i(t) +

-

v(t)

Therestof

thecircuit

Ldt

tdiLdt

dtv )()(

t

dxxvL

ti )(1)(

)()( titi LLdtdiLiivP

)(21)( 2 tLiLidipdttwL Energy stored:

• One inductor and one resistor• The source and resistor may be equivalent to a

circuit with many resistors and sources.

FIRST-ORDER RL CIRCUITS

Switch to 1

R

1

L

2

K

E

dtdiLvL

KVL around the loop: EviR L

iRdtdiLE

Initial condition 0)0()0(,0 iit

called time constant

RL /

Transient Response of RL Circuits

FIRST-ORDER RL CIRCUITS

/

/

/

1

)1(

)1()1(

ttLRt

LR

L

tR

ttLR

EeeLR

RELe

RE

dtdL

dtdiLv

eEiRv

eREe

REi

Time constant

• Indicate how fast i (t) will drop to zero.• It is the amount of time for i (t) to drop to zero if it

is dropping at the initial rate .0t

t

dtdi

FIRST-ORDER RL CIRCUITS

Switch to 2

tLR

Aei

dtLR

idi

iRdtdiL

0

Initial conditionRE

It 0,0

/ttLR

eREe

REi

Transient Response of RL Circuits

R

1

L

2

K

E

0

0

0

0

0

: 0:

1

ln

i t t

I

i t tI

t ti I i t

Rdi dti L

Ri tL

tLR

Iti

0

)(ln tLR

eIti

0)(

FIRST-ORDER RL CIRCUITS

FIRST-ORDER RL CIRCUITS

R

1

L

2

K

E

Transient Response of RL Circuits

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)

0A

2.0mA

4.0mA

SEL>>

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)

0A

2.0mA

4.0mA

SEL>>

Input energy to L

L export its energy , dissipated by R

   Initial

Value （ t = 0）

Steady Value (t

)

time constant

RL Circuits

Source(0

state) Source-

free(0

input)

RCCircuits

Source(0

state) Source-

free(0

input)

00 iREiL

REi 0 0i

00 v Ev

Ev 0 0v

RL /

RL /

RC

RC

SUMMARY

The Time Constant

• For an RC circuit, = RC• For an RL circuit, = L/R• -1/ is the initial slope of an exponential with an

initial value of 1• Also, is the amount of time necessary for an

exponential to decay to 36.7% of its initial value

SUMMARY

SUMMARY

• How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. – IL(t), inductor current– Vc(t), capacitor voltage

• Find these two types of the values before the change and use them as the initial conditions of the circuit after change.

About Calculation for The Initial Value

iC iL

i

t=0

+

_

1A

+

-vL(0+

)

vC(0+)=4V

i(0+)

iC(0+) iL(0+)

1 3/ / 2R R

20 8V 4V2 2Cv

8V0 2A2 2

i

40 2A 1A4 4Li

0 0C Cv v

0 0L Li i

Method 1

(Analyzing an RC circuit or RL circuit)

Simplify the circuit

2) Find Leq(Ceq), and = Leq/Req ( = CeqReq)

1) Thévenin Equivalent.(Draw out C or L)

Veq , Req

3) Substituting Leq(Ceq) and to the previous solution of differential equation for RC (RL) circuit .

Method 2

(Analyzing an RC circuit or RL circuit)

3) Find the particular solution.

1) KVL around the loop the differential equation

4) The total solution is the sum of the particular and homogeneous solutions.

2) Find the homogeneous solution.

Examples

Method 3 (step-by-step)

(Analyzing an RC circuit or RL circuit)

1) Draw the circuit for t = 0- and find v(0-) or i(0-)2) Use the continuity of the capacitor voltage, or inductor current, draw

the circuit for t = 0+ to find v(0+) or i(0+)3) Find v(), or i() at steady state4) Find the time constant t

– For an RC circuit, t = RC– For an RL circuit, t = L/R

5) The solution is: /)]()0([)()( teffftf

Given f(0+) ， thus A = f(0+) – f(∞)

t

effftf

)]()0([)()(

Initial Steady

t

Aeftf

)()(In general,

Examples

vC (0)= 0, Find vC (t) for t 0. i1

6k

R1

R2 3k +

_ E

C=1000PF pf

i2 i3 t=0

9V

Method 3:

0

3K0 0, 9V 3V6K 3K

t

c c c c

c c

v t v v v e

v v

Apply Thevenin theorem :

6

1

6

2 10

1 1 2K6K 3K

2K 1000pF 2 10

3 3 V

Th

Th

t

c

R

R C

v t e

s

Examples

vC

R2 3k

+

_ v

U

I

t=0

6V

C=1000PF

R1=10k

R1=20k

+ - P3.2 vC (0)= 0, Find vC (t) for t 0.

Apply Thevenin’s theorem :

s

0 0

10K6V 4.615V10K 3K

c

c

v

v

6

1

6

2.31 10

1 1 30 K10K 3K 13

30 K 1000pF 2.31 1013

4.615 4.615 V

Th

Th

t

c

R

R C

v t e

THANK YOU