Chapter 8 - The Complete Response of RL and RC Circuits
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Chapter 8 - The Complete Response of RL and RC Circuits


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Finally: 
 
 
6400 6400
6400
 ( ) 400 ( ) 0.1 ( ) 400 (.40625 .09375) 0.1( 6400) (0.40625 )
 37.5 97.5 V
t t
t
dv t i t i t e e
dt
e
\u2212 \u2212
\u2212
= + = + + \u2212 
= \u2212 
 
 
 
P8.3-19 
Before the switch closes v(t) = 0 so ( ) ( )0 0 0v v+ = \u2212 = V . 
 
For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the 
capacitor, i.e. the part of the circuit to the left of the terminals a \u2013 b. 
 
Find Rt: 
 
 
 
 
( )
( )
12 10 2
6 
12 10 2t
R
+= = \u3a9+ + 
 
Write mesh equations to find voc: 
 
 
Mesh equations: 
 ( )
( )
1 1 2 1
2 1 2
12 10 6 0
6 3 18
i i i i
i i i 0
+ \u2212 \u2212 =
\u2212 + \u2212 = 
 
1 2
2 1
28 6
9 6 1
i i
i i 8
=
\u2212 = 
 
1 1
2
136 18 A
2
14 1 7= A
3 2 3
i i
i
= \u21d2 =
\u239b \u239e= \u239c \u239f\u239d \u23a0
 
 
Using KVL, 2 1
7 13 10 3 10 12 V
3 2oc
v i i \u239b \u239e \u239b \u239e= + = + =\u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0 
Then 
t
1 1 16 s 4 
24 4 s
R C\u3c4 \u3c4
\u239b \u239e= = = \u21d2 =\u239c \u239f\u239d \u23a0
1 
and 
( ) ( )( ) ( ) ( )oc oc 4 40 0 12 12 12 1 V for 0t t tv t v v e v e e t\u3c4\u2212 \u2212 \u2212= + \u2212 + = \u2212 + = \u2212 \u2265 
 
(checked: LNAP 7/15/04) 
 
 
 
P8.3-20 
Before the switch closes the circuit is at steady state so the inductor acts like a short circuit. 
We have 
 
 
 
 
( ) ( )
1 24 0.8 A
2 5 20 20
i t
\u239b \u239e= =\u239c \u239f\u239c \u239f+\u239d \u23a0&
 
so 
( ) ( )0 0 0.8i i+ = \u2212 = A 
After the switch closes, find the Thevenin equivalent circuit for the part of the circuit connected 
to the inductor. 
 
 
Using voltage division twice 
 
oc
20 1 24 7.2 V
25 2
v \u239b \u239e= \u2212 =\u239c \u239f\u239d \u23a0 
 
 
( ) ( )t 5 20 20 20 14 R = + =& & \u3a9 
oc
sc
t
7.2 0.514 A
14
v
i
R
= = = 
 
Then 
t
3.5 1 1 1s 4 
14 4 s
L
R
\u3c4 \u3c4= = = \u21d2 = 
and 
( ) ( )( ) ( )sc sc 4 40 0.8 0.514 0.514 0.286 0.514 A for 0t t ti t i i e i e e t\u3c4\u2212 \u2212 \u2212= + \u2212 + = \u2212 + = + \u2265 
 
(checked: LNAP 7/15/04) 
 
 
 
P8.3-21 
Before the switch closes the circuit is at steady state so the inductor acts like a short circuit. 
We have 
 
 
So 
( ) ( )2 2
400 240
400 100 400
i
R R
\u239b \u239e\u239b \u239e\u239c \u239f\u2212 = \u239c \u239f\u239c \u239f\u239c \u239f+ +\u239d \u23a0\u239d \u23a0&
 
 
From the given equation 
 
( ) ( )o 30 15 53.6 10 68.6 mAi e \u2212+ = + × = 
So 
( ) ( ) ( )2 2
400 240.0686 0 0
400 100 400
i i
R R
\u239b \u239e\u239b \u239e\u239c \u239f= + = \u2212 = \u239c \u239f\u239c \u239f\u239c \u239f+ +\u239d \u23a0\u239d \u23a0&
 
or 
( ) 22
2
400
0.0686 400 100 9600
400
R
R
R
\u239b \u239e+ + =\u239c \u239f\u239c \u239f+\u239d \u23a0
 
Solving gives 
2 199.988 200 R = \u3a9\ufffd 
 
Next, consider the steady state circuit after the switch closes. At steady state the inductor acts 
like a short circuit. 
 
 
From the given equation 
 
( ) ( )
t
lim 0.015i i t\u2192\u221e\u221e = = 
 
Apply KCL at node a to get 
 
a a
a
24
0.015 0 18 V
400 100
v v
v
\u2212+ + = \u21d2 = 
 
Apply KCL at node b to get 
 
b b
1
24
0.015
200
v v
R
\u2212+ = 
 
Next, vb = va = 18 V so 
 
 
1
1
18 6 0.015 80 
200
R
R
\u2212+ = \u21d2 = \u3a9 
Finally, consider the circuit after the switch closes but before the circuit reaches steady state. 
Replace the part of the circuit connected to the inductor by its Thevenin equivalent circuit. 
 
 
 
 
 
 
oc
400 200 24 2.057 V
500 280
v \u239b \u239e= \u2212 =\u239c \u239f\u239d \u23a0 
 
 
 
( ) ( )t 400 100 80 200 80 57 137 R = + = + = \u3a9& & 
 
From the given equation 1 548\u3c4 = so 
t
1 0.25 H
137 548
L L L
R
= = \u21d2 = 
(checked LNAPTR 7/24/04) 
 
 
P8.3-22 
Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short 
circuit so we have 
 
 
 
( ) ( )0 0 4i i+ = \u2212 = A 
 
After t = 0, we can replace the part of the circuit connected to the inductor by its Norton 
equivalent circuit. 
 
 
Using superposition, the short circuit current is 
given by 
 
( ) ( )sc
8 3 82 4 3.75 A
8 5 3 3 8 5
i
\u239b \u239e \u239b \u239e+= + =\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f+ + + +\u239d \u23a0 \u239d \u23a0
 
 
 
 
t 8 3 5 16 R = + + = \u3a9 
so 
2 10.125 s 8 
16 s
\u3c4 \u3c4= = \u21d2 =
1 
 
The inductor current is given by 
 
( ) ( )( ) ( )L sc sc 8 80 4 3.75 3.75 3.75 0.25 A for 0t t ti t i i e i e e t\u3c4\u2212 \u2212 \u2212= + \u2212 + = \u2212 + = \u2212 \u2265 
 
(checked: LNAP 7/15/04) 
 
 
P8.3-23 
 
The voltage across the 6 \u3a9 resistor may or may not 
be continuous. The inductor current will be 
continuous. We\u2019ll use the inductor to solve this 
problem. 
 
( ) ( ) 32 A for 0
6
tv ti t e t\u2212= = \u2212 \u2265 
 
As t , the switch is closed and the circuit 
reaches steady state. The inductor acts like a short 
circuit. 
\u2192 \u221e
 
Using current division 
 
( ) 1
1
4
6
R
i
R
\u239b \u239e\u221e = \u239c \u239f\u239c \u239f+\u239d \u23a0
 
 
 
 
 
 
 
Form the equation for i(t), . Thus ( ) 2 Ai \u221e =
 
1
1
1
2 4 6 
6
R
R
R
\u239b \u239e= \u21d2\u239c \u239f\u239c \u239f+\u239d \u23a0
= \u3a9 
 
For t < 0, the switch is open and the circuit is at 
steady state. The inductor acts like a short circuit. 
 
Using current division 
 
( ) ( )2
60 4
6 6
i
R
\u239b \u239e\u239c \u239f\u2212 = \u239c \u239f+ +\u239d \u23a0
 
 
From the equation for i(t), so ( )0 1i \u2212 = A
 
( ) 22
61 4 12 
6 6
R
R
\u239b \u239e\u239c \u239f= \u21d2\u239c \u239f+ +\u239d \u23a0
= \u3a9 
 
 
 
 
 
For t > 0 the Thevenin impedance of the part of the 
circuit connected to the inductor is 
 
t 6 6 12 R = + = \u3a9 
The time constant is 
t 12
L L
R
\u3c4 = = 
From the equation for i(t), 1 13 
s\u3c4 = . 
Then 
1 4 H
3 12
L L= \u21d2 = 
 
 
 
(checked: LNAP 7/26/04) 
 
 
 
 
Section 8-4: Sequential Switching 
 
P8.4-1 
Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to 
get: 
 
Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc = 
5 V and Rt = 4 \u3a9 so 4 0.05 0.2 s\u3c4 = × = . Therefore 
 
5( ) ( (0) ) 5 5 V for 0 1.5 soc oc
t tv t v v v e e t\u3c4\u2212 \u2212= + \u2212 = + < < 
 
At t =1.5 s, ( )0.05 1.5(1.5) 5 5 = 5 Vv e\u2212= + . For 1.5 s < t, voc = 10 V and Rt = 8 \u3a9 so 
8 0.05 0.4 s\u3c4 = × = . Therefore 
( ) ( )1.5 2.5 1.5( ) ( (1.5) ) 10 5 V for 1.5 s oc oc t tv t v v v e e t\u3c4\u2212 \u2212 \u2212 \u2212= + \u2212 = \u2212 < 
Finally 
( )
5
2.5 1.5
5 5 V for 0 1.5 s
( )
for 1.5 s 10 5 V
t
t
e t
v t
te
\u2212
\u2212 \u2212
\u23a7 + < <\u23aa= \u23a8 <\u2212\u23aa\u23a9
 
 
P8.4-2 
Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get: 
 
 
 
Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 
A and Rt = 6 \u3a9 so 12 2 s
6
\u3c4 = = . Therefore 
0.5( ) ( (0) ) 2 A for 0 1.5 ssc sc
t ti t i i i e e t\u3c4\u2212 \u2212= + \u2212 = + < < 
 
At t =1.5 s, ( )0.5 1.5(1.5) 2 = 2.47 Ai e\u2212= + . For 1.5 s < t, isc = 3 A and Rt = 8 \u3a9 so 12 1.5 s
8
\u3c4 = = . 
Therefore 
( ) ( )1.5 0.667 1.5( ) ( (1.5) ) 3 0.53 V for 1.5 s sc sc t ti t i i i e e t\u3c4\u2212 \u2212 \u2212 \u2212= + \u2212 = \u2212 < 
Finally 
( )
0.5
0.667 1.5
2 A for 0 1.5 s
( )
for 1.5 s 3 0.53 A
t
t
e t
i t
te
\u2212
\u2212 \u2212
\u23a7 + < <\u23aa= \u23a8 <\u2212\u23aa\u23a9 
 
 
P8.4-3 
At t = 0\u2212: 
 
KVL : 52 18 (12 8) 0
 (0 ) =104 39 A
6 2 A (0 )
6 2L L
i i
i
i i i
\u2212
+
\u2212 + + =
\u21d2
\u239b \u239e\u2234 = = =\u239c \u239f+\u239d \u23a0
 
For 0 < t < 0.051 s 
 
 
t
( ) (0) where L L
ti t i e L R\u3c4 \u3c4\u2212= = 
t
6
6 12 2 6 
( ) 2 AtL
R
i t e\u2212
= + = \u3a9
= 
 
66 12( ) ( ) 6 A
6
t
Li t i t e
\u2212+\u239b \u239e\u2234 = =\u239c \u239f\u239d \u23a0 
 
For t > 0.051 s 
 
 
 
( )
( ) ( 0.051)
6 (.051)
14 ( 0.051)
14 ( 0.051)
( ) (0.051) 
(0.051) 2 1.473 A
( ) 1.473 A
( ) 1.473 A
R L t
L L
L
t
L
t
L
i t i e
i e
i t e
i t i t e
\u2212 \u2212
\u2212
\u2212 \u2212
\u2212 \u2212
=
= =
=
= =
 
 
P8.4-4 
At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 \u3bcF 
capacitor is 3 V. The charge stored by the capacitor