Chapter 8
Basic RL and
RC Circuits
1
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Applying KVL:
di R
 i 0
dt L
We
can
solve
for
the
natural
response

if we know the initial condition i(0)=I0:
i(t)=I0e-Rt/L for t>0
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2
Show that the voltage v(t) will
be -12.99 volts at t=200 ms.
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3
The time constant
τ=L/R
determines the rate of
decay.
i(t)  I0e
t /

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4
Applying KCL:
dv
1

v 0
dt RC
We can solve for the natural response
if we know the initial condition v(0)=V0
v(t)=V0e-t/RC for t>0
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5
The time constant is
τ=RC
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6
Show that the voltage v(t)
is 321 mV at t=200 μs.
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7
The time constant of a single-inductor circuit
will be τ=L/Req where Req is the resistance seen
by the inductor.
Example: Req=R3+R4+R1R2 / (R1+R2)
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8
The time constant of a single-capacitor circuit
will be τ=ReqC where Req is the resistance seen
by the capacitor.
Example: Req=R2+R1R3 / (R1+R3)
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9



The voltage on a capacitor or the current
through a inductor is the same prior to and
after a 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/τ.
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10
Find i1(t) and iL(t) for t>0.
Answer: τ=20 μs; i1=-0.24e-t/τ,iL=0.36e-t/τ for t>0
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11
The unit-step function u(t) is a convenient
notation to respresent change:
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12

The unit step models a double-throw switch.

A single-throw switch is open circuit for t<0,
not short circuit.
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13
Rectangular
pulse
Pulsed
sinewave:
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14

The two circuits shown
both have i(t)=0 for t<0
and are also the same for
t>0.

We now have to find both
the natural response and
and the forced response due
to the source V0
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15
The total response is the
combination of the
transient/natural response and
the forced response:
V0
Rt / L
i(t)  1  e
u(t)

R
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16
V0
Rt / L
i(t)  1  e
u(t)

R
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17
Show that
i(t)=25+25(1-e-t/2)u(t) A
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18
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19
vC=20 + 80e-t/1.2 V and i=0.1 + 0.4e−t/1.2 A
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20
vC=20 + 80e-t/1.2 V
i=0.1 + 0.4e−t/1.2 A
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21