ELEG 205Fall 2017

Lecture #12

Mark Mirotznik, Ph.D.Professor

The University of DelawareTel: (302)831-4221

Email: mirotzni@ece.udel.edu

Chapter 8: RL and RC Circuits

1. Source-free RL circuits (natural response)2. Source-free RC circuits (natural response)3. Driven RL circuits (forced response)4. Driven RC circuits (forced response)

Summary

Can voltage or current change instantaneously

Capacitor

Inductor

Symbol UnitsVoltage Current relationship

Integralform

Stored Energy

L

CFarads(F)

Henry(H) dt

tdiLtv )()( =)(

)(1)(

o

t

t

ti

tdtvL

tio

+

′′= ∫

dttdvCti )()( =

)(

)(1)(

o

t

t

tv

tdtiC

tvo

+

′′= ∫

2)(21)( tiLtE =

2)(21)( tvCtE =

What does it look like at DC

Opencircuit

Shortcircuit

Voltage – No

Current - Yes

Voltage – Yes

Current - No

Summary: RL and RC Natural Response

RL

RC

Response Time constant

τt

o eVtv−

⋅=)( RC=τ

RL

=ττt

o eIti−

⋅=)(

Example Problems

4 kΩ 6 kΩ

)(tic

1 µF

This should be recognized as a RC natural response problem(why?). Although I asked for the current through the capacitor ALWAYS FIRST SOLVE FOR CAPACITOR VOLTAGE. After you find the capacitor voltage then use that to find whatever I asked for! WHY?

1 kΩ10 V t=0

4 kΩ

Example Problems

4 kΩ

1 kΩ10 V t=0 6 kΩ

)(tic

STEP #1: Solve the circuit at t=0- when there is nothing changing

1 µF

4 kΩ

1 kΩ10 V 6 kΩ)0( −

cv+

-

Example ProblemsSTEP #1: Solve the circuit at t=0- when there is nothing changing

Vkk

kvc 24.34.25

4.210)0( =Ω+Ω

Ω⋅=−

4 kΩ 6 kΩ1 µF)0( −cv

+

-

1 kΩ10 V

4 kΩ

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

0)0()(1

>⋅=−+ tevtv

tRC

cc

C is 1 µF but which R do we use?

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

0)0()(1

>⋅=−+ tevtv

tRC

cc

C is 1 µF but which R do we use? Ω=⇒

Ω+

Ω+

Ω=

=

kRkkkR

RR

eqeq

eq

5.16

14

14

11

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

Vvvtevevtv

cc

tc

te

cc

24.3)0()0(0)0()0()( 66711500

16

==

>⋅=⋅=−+

−+⋅−

+ −

024.3)( 667 >⋅= − tetv tc

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

024.3)( 667 >⋅= − tetv tc

)(tic

How do we now find ic(t)?

Example Problems

4 kΩ 6 kΩ1 µF)(tvc

+

-

1 kΩ10 V

4 kΩ

STEP #2: Solve the circuit for t>0 (switch closed)

024.3)( 667 >⋅= − tetv tc

)(tic

( )

00022.0)(

0667124.3)()(

667

6676

>⋅−=

>⋅−⋅⋅=⋅=

−

−−

tAeti

teedt

tdvCti

tc

tcc

Chapter 8: RL and RC Circuits

1. Source-free RL circuits (natural response)2. Source-free RC circuits (natural response)3. Driven RL circuits (forced response)4. Driven RC circuits (forced response)

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

Situation: When the switch closes (t=0) the inductor starts charging up. Eventually the inductor will be fully charged and the current will stop changing.

t=0oV

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

0)()( =++− tVtVV LRoKVL:

oVdt

tdiLtiR =+⋅)()(

Use Ohm’s law and the law of the inductor to write this in terms of current, i(t)

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

)()( tiRVdt

tdiL o ⋅−=

dttiRV

Ldio

=⋅− )(

After some algebra

oVdt

tdiLtiR =+⋅)()(

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

dttiRV

Ldio

=⋅− )(

After some algebra

Integrate both sides ∫∫ =⋅−

dttiRV

Ldio )(

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

Integrate both sides ∫∫ =⋅−

dttiRV

Ldio )(

This results in [ ] tVRiVRL

oo =−−− )ln()ln(

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

This results in [ ] tVRiVRL

oo =−−− )ln()ln(

[ ] tLRVRiV oo −=−− )ln()ln(

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

[ ] tLRVRiV oo −=−− )ln()ln(

[ ] tLR

VRiV ee oo−−− =)ln()ln(

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

[ ] tLRVRiV oo −=−− )ln()ln(

[ ] tLR

VRiV ee oo−−− =)ln()ln(

tLR

o

o eV

RiV −=

−

Driven RL Circuits (forced response))(tVR+ -

RL )(tVL

+

-

)(ti

t=0oV

tLR

o

o eV

RiV −=

−

tLR

oo eRV

RVti

−−=)(

Driven RL Circuits (forced response)

tLR

oo eRV

RVti

−−=)(

time

Vo/R

i(t)

0

tLR

oo eRV

RVti

−−=)(

Driven RL Circuits (forced response)General Response (including initial conditions)

( )

eq

t

LLLL

RL

teiiiti

=

>∞−+∞=−+

τ

τ 0,)()0()()(

time

)(∞i

)0( +i)(tiL

( ) τt

eiiiti−+ ∞−+∞= )()0()()(

0

Driven RC Circuits (forced response)

R C)(tVc

+

-

Situation: When the switch closes (t=0) the capacitor starts charging up. Eventually the capacitor will be fully charged and the voltage will stop changing.

t=0oI

Driven RC Circuits (forced response)

R C)(tVc

+

-t=0oI

0)(1

>−=−

teRIRItvt

RCooc

time

R Io

Vc(t)

0

0)(1

>−=−

teRIRItvt

RCooc

Driven RC Circuits (forced response)General Response (including initial conditions)

( )

CR

tevvvtv

eq

t

cccc

10,)()0()()(

=

>∞−+∞=−+

τ

τ

time

)(∞cv

)0( +cv)(tvc

0

( ) 0,)()0()()( >∞−+∞=−+ tevvvtv

t

ccccτ

Summary: RL and RC Natural and Forced Response

RL

RC

NaturalResponse

Timeconstant

τt

cc evtv−+ ⋅= )0()( CReq=τ

eqRL

=ττt

LL eiti−+ ⋅= )0()(

ForcedResponse

( ) τt

cccc evvvtv−+ ∞−+∞= )()0()()(

( ) τt

LLLL eiiiti−+ ∞−+∞= )()0()()(

Example Problems

This should be recognized as a RL forced response problem. ALWAYS FIRST SOLVE FOR INDUCTOR CURRENT FIRST. After you find the inductor current then use that to find whatever I asked for!

10 H1 kΩ

10 V

t=0

5 kΩ10 kΩ

5 kΩ

5 V)(ti

Example Problems

10 H1 kΩ

10 V

t=0

5 kΩ10 kΩ

5 kΩ

5 V)(ti

eqRL

=τ( ) τt

LLLL eiiiti−+ ∞−+∞= )()0()()(

Essentially 3 circuit problems to solve

final value initial value time constant

Example Problems

10 H1 kΩ

10 V

t=0

5 kΩ10 kΩ

5 kΩ

5 V)(ti

eqRL

=τ( ) τt

LLLL eiiiti−+ ∞−+∞= )()0()()(

Problem #1: Solve for the initial condition (t=0-). Nothing is changing at this time so this is a steady-state problem. This means you can replace the inductor by a short circuit.

Example Problems

1 kΩ10 V

5 kΩ10 kΩ

5 kΩ

5 V)0( −

Li

Problem #1: Solve for the initial condition (t=0-). Nothing is changing at this time so this is a steady-state problem. This means you can replace the inductor by a short circuit.

1i 2i

21)0( iiiL +=−

mAkkk

iL 67.25

551

10)0( =Ω

+Ω+Ω

=−

Example Problems

1 kΩ10 V

5 kΩ10 kΩ

5 kΩ

5 V)(∞Li

Problem #2: Solve for the final condition (t=infinity). Nothing is again changing at this time so this is a steady-state problem. This means you can replace the inductor by a short circuit.

mAk

iL 0.15

5)( =Ω

=∞

Example Problems

1 kΩ10 V

5 kΩ10 kΩ

5 kΩ

5 V

Problem #3: Find the time constant. Do this at t=0+. To find the equivalent resistance kill all the sources.

10 H

eqRL

=τΩ=

Ω+ΩΩ⋅Ω

=

=

kkkkkR

HL

eq 33.3510510

10

333,310

=τ

Example Problems

1 kΩ10 V

5 kΩ10 kΩ

5 kΩ

5 V

Now put it all together.

10 H

333,310

=τ( ) τ

t

LLLL eiiiti−+ ∞−+∞= )()0()()(

mAiL 0.1)( =∞

mAiL 67.2)0( =−

Example Problems

1 kΩ10 V

5 kΩ10 kΩ

5 kΩ

5 V

Now put it all together.

10 H

( ) mAeti tL

3330.167.20.1)( −−+=

mAeti tL

33367.10.1)( −+=

Actives

Passives

Sources

Discretes

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