ECE 2006 Fall, 2011
Homework #06 Solution
Problems in the textbook:
7.4, 7.6, 7.10, 7.11, 7.17,
Chapter 7, Problem 4.
The switch in Fig. 7.84 moves instantaneously from A to B at t=0. Find v for t>0.
5 kΩ
A
40 V
10 µF
B
+
_
+
v
2 kΩ
–
Figure 7.84
For Prob. 7.4.
Chapter 7, Solution 4.
For t<0, v(0-)=40 V.
For t >0. we have a source-free RC circuit.
−6
τ RC
0.02
=
= 2 x103 x10 x10=
−t /τ
−50 t
=
v(t) v=
(0)e
40e
V
Chapter 7, Problem 6.
The switch in Fig. 7.85 has been closed for a long time, and it opens at t = 0. Find
v(t) for t ≥ 0.
Figure 7.85
Chapter 7, Solution 6.
v(t)
vo = v(0) =
2
(24) = 4 V
10 + 2
v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 =
2
25
v( t ) = 4e −12.5t V
Chapter 7, Problem 10.
For the circuit in Fig. 7.90, find vo(t) for t >0. Determine the time necessary for the
capacitor voltage to decay to one-third of its value at t=0.
t=0
9 kΩ
+
60 V
3 kΩ
+
_
20µF
vo
–
Figure 7.90
For Prob. 7.10.
Chapter 7, Solution 10.
For t<0,
vo(0–) = [3/(3+9)](60) = 15 V
For t>0, we have a source-free RC circuit
−6
=
τ RC
= 3 x103 x20 x10=
0.06 s
vo(t) = 15e–16.667t V
Let the time be to.
5 = 15e–16.667to or e16.667to = 15/5 = 3
to = ln(3)/16.667 = 65.92 ms.
Chapter 7, Problem 11.
For the circuit in Fig. 7.91, find io for t >0.
3Ω
t=0
4H
io
24 V
4Ω
+
_
Figure 7.91
8Ω
For Prob. 7.11.
Chapter 7, Solution 11.
For t<0, we have the circuit shown below.
3Ω
4H
4Ω
24 V
8Ω
+
_
4H
io
8A
3//4= 4x3/7=1.7143
3Ω
4Ω
8Ω
1.7143
(8) 1.4118 A
=
1.7143 + 8
For t >0, we have a source-free RL circuit.
L
4
τ= =
= 1/ 3
R 4+8
(0)e− t / τ 1.4118e−3t A
=
io(t) io=
=
io(0 − )
Chapter 7, Problem 17.
Consider the circuit of Fig. 7.97. Find v0(t) if i(0) = 2 A and v(t) = 0.
Figure 7.97
Chapter 7, Solution 17.
i( t ) = i(0) e - t τ ,
τ=
14 1
L
=
=
R eq
4 16
i( t ) = 2 e -16t
v o ( t ) = 3i + L
di
= 6 e-16t + (1 4)(-16) 2 e-16t
dt
v o ( t ) = - 2 e -16t u ( t )V