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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.
Example: Req=R2+R1R3/ (R1+R3)Copyright 2013 The McGraw-Hill Companies, Inc. Permission required forreproduction or display. 9
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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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