Homework #10 Solutions P14.6-4
vc 12i L
2
diL dt
8 and i L
C
d vc dt
Taking the Laplace transform yields
Vc s
12 I L s
2 sI L s
iL 0
8 s
vc (0) 0, iL (0) 0
IL s
C sVc s
vc 0
Solving yields
Vc s s s2 4C 6s
72 s s 3 24
3t
C 2
a s b s 3
Homework #7 Solutions P8.3-15 At steady-state, immediately before t = 0:
i 0
10 10 40
12 16 40|10
0.1 A
After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be
so I sc Finally:
0.3 A, Rt i (t )
40 , (0.1 0.3
Homework #6 Solutions Ex. 7.3-2 a)
W t
W 0
t 0
vi dt 0
t 0
First, W 0 Next, v t
0 since v 0 v 0
t 0
1 C
t 0
i dt
104
2 dt
2 104 t
W t
W 1s
b) Ex 7.4-2
W 100s
2 104 t 2 dt
2 104 t 2
2 104 J = 20 kJ
2 104 100
2
2 108 J = 200 MJ
Ex. 7.
Homework #5 Solutions
P6.3-1
P6.3-3 The voltages at the input nodes of an ideal op amp 2V. are equal so va Apply KCL at node a:
vo
2 8000
12 2 4000
0
vo
resistor
30 V
Apply Ohm's law to the 8 k
io 2 vo 8000
3.5 mA
P6.3-5 The voltages at th
P4.4-6 Pick a reference node and label the unknown node voltages:
Express the controlling current of the dependent source in terms of the node voltages: va va . Then v b 2 i 4 . i4 3 6 Apply KCL at node a: v a 12 v a v a v b 0 3 6 4 So: va va v a 12
Homework #3 Solutions P4.3-2
Express the branch voltage of each voltage source in terms of its node voltages to get:
va 12 V, vb vc vd 8
KCL at node b: vb va 4000
0.002 i
vb
12 4000
0.002 i
vb 12 8 4000 i
KCL at the supernode corresponding to
Homework #2 Solutions Ex. 4.3-2
v 8 b 10
12
v b 40
3
v b
8 V and v a
16 V
Ex. 4.4-1 Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb 4 ia and solve for vb .
6 vb 8 12 i a v b 4i a 4 9 v b 12
HW#1 Solutions P1.2-1
i t d 41 e dt
5t
20 e
5t
A
P1.2-2
q t
P1.2-6
t 0
i
d
q 0
t 0
41 e
5
d
0
t 0
4d
t 0
4e
5
d
4t
4 e 5
5t
4 C 5
i = 600 A = 600
C s C s mg 20 min 60 1.118 = 8.05 105 mg=805 g s min C
Silver deposited = 600
Homework #8 Solutions P8.3-1 Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open c
Homework #9 Solutions
Ex. 14.6-1
Taking Laplace Transform of the differential equation:
s2 F s sf 0 f' 0 5 sF s f 0 6F s 10 s 3
Using the given initial conditions
(s 2 5s 6) F (s)
10 2s 10 s 3
A s 3
2
2s 2 16s 40 s 3
B s 3 C s 2
F s
2s 2 16s 4