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Problem 7.1 Applying KVL to Fig. 7.1. 10/03/1999
Taking the derivative of each term,
or
Integrating,
or
Problem 7.2
where is the Thevenin equivalent at the capacitor terminals.
Problem 7.3
where
Problem 7.4
where ,
Problem 7.5
where ,
Problem 7.6 ,
+?+?+v = 1
+
?
+
?
+
v = 1
?
8 ?
1 V
0.5 V
10 ?
i2
i
i1
,
Problem 7.7
or
Problem 7.8 ,
15 ?10
15 ?
10 ?
4 ?
iT
io
i
+
v
?
10 mF
i.e. if , then
across the capacitor terminals.
By applying the current division principle,
Problem 7.10 Applying KCL to the RL circuit,
Differentiating both sides,
If the initial current is , then
,
Problem 7.11 When t 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 ? resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).
2 H
2 H
4 ?
(b)
3 ?
12 V
+
?
(a)
i(0-)
Since the current through an inductor cannot change abruptly,
When t 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
Hence,
Problem 7.12
where is the Thevenin resistance at the terminals of the inductor.
Problem 7.13
where ,
Problem 7.14
(a) and
(b) where and
Problem 7.15 ,
0.4 HReqi(t)+
0.4 H
Req
i(t)
+
vo(t)
?
,
i1i
i1
i2
? +
i
i1
i2
1 V
i/2
10 ?
40 ?
To find we replace the inductor by a 1-V voltage source as shown above.
But and
i.e.
Problem 7.18
Let p be the fraction
i.e.
2 HRth
2 H
Rth
Vth
+
?
Problem 7.20 ,
,
Using current division, the current through the 20 ohm resistor is
10/03/1999
1 ?3 ?2 ?+
1 ?
3 ?
2 ?
+
vo(0) = 2 V
?
+
vx
?
1/3 H
The Thevenin resistance at the inductor’s terminals is
,
Problem 7.22
(a)
=
=
=
=
Problem 7.23
v(t)-1t
v(t)
-1
t
2
3
0
1
4
2
1
i(t)-1t
i(t)
-1
t
2
3
0
1
4
1
Problem 7.26
Problem 7.27
Problem 7.28
Problem 7.29
(a) ,
(b) ,
Problem 7.30
, or
, or
Problem 7.31
Before t = 0,
After t = 0,
, ,
Before t = 0, , where is due to the 12-V source and is due to the 2-A source.
To get , transform the
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