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Chapter 8

Basic RL and

RC Circuits

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Applying KVL:

We can solve for the natural responseif we know the initial condition i(0)=I0:

i(t)=I0e-Rt/L

for t>0

di

dt

R

L i0

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Show that the voltage v(t) will

be -12.99 volts at t=200 ms.

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The time constant

=L/R

determines the rate of

decay.

i(t) I0et/

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Applying KCL:

We can solve for the natural response

if we know the initial condition v(0)=V0

v(t)=V0e-t/RC

for t>0

dvdt

1RC

v 0

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The time constant is

=RC

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Show that the voltage v(t)

is 321 mV at t=200 s.

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The time constant of a single-inductor circuit

will be =L/ReqwhereReq is the resistance seen

by the inductor.

Example: Req=R3+R4+R1R2/ (R1+R2)

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The time constant of a single-capacitor circuit

will be =ReqC whereReq is the resistance seen

by the capacitor.

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The voltage on a capacitor or the current

through a inductor is the sameprior to and

aftera switch at t=0. Resistor voltage (or current) prior to the

switch v(0-) can be different from the voltage

after the switch v(0+).

All voltages and currents in an RC or RL

circuit follow the same natural response e-t/.

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Find i1(t) and iL(t) for t>0.

Answer: =20 s; i1=-0.24e-t/,iL=0.36e

-t/for t>0

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The unit-step function u(t)is a convenient

notation to respresent change:

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The unit step models a double-throw switch.

A single-throw switch is open circuit for t

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Rectangular

pulse

Pulsedsinewave:

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The two circuits shown

both have i(t)=0 fort0.

We now have to find both

the natural response and

and theforced response due

to the source V0Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display. 15

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i(t)V

0

R1eRt/L u(t)

The total response is the

combination of the

transient/natural response andthe forced response:

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i(t) V0R

1eRt/L u(t)

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Show thati(t)=25+25(1-e-t/2)u(t) A

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vC=20 + 80e-t/1.2 V and i=0.1 + 0.4et/1.2A

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vC=20 + 80e-t/1.2 V

i=0.1 + 0.4et/1.2A

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