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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