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 yi
P8.6-4 Determine vo(t) for t > 0 for the circuit shown in Figure P8.6-4.
Figure P8.6-4
Solution: Determine the initial condition, vo(o), by considering the circuit when t < 0 and the circuit i
P 5.2-6 Use source transformations to find
the value of the voltage va in Figure P 5.2-6.
Answer: va = 7 V
Figure P 5.2-6
Solution:
A source transformation on the right side of the circuit, followed b
P 6.3-1 Determine the value of voltage
measured by the voltmeter in Figure P 6.3-1.
Answer: 4 V
Figure P 6.3-1
Solution:
(checked using LNAP 8/16/02)
P 6.3-2
Find vo and io for the circuit of Figure P
P 4.2-1
The node voltages in the circuit of Figure P 4.2-1 are
v1 = 4 V and v2 = 2 V.
Determine i, the current of the current source.
Answer: i = 1.5 A
Figure P 4.2-1
Solution:
v
v v
4 4 2
1
1
P 4.2-2 Determine the node voltages for the
circuit of Figure P 4.2-2.
Answer: v1 = 2 V, v2 = 30 V, and v3 = 24 V
Figure P 4.2-2
Solution:
v v
v
1 2
1
KCL at node 1:
+ + 1 = 0 5 v v = 20
1 2
P 1.2-5
The total charge q(t), in coulombs, that enters the terminal of an element is
0
q(t ) = 2t
3 + e2t (t 2)
t<0
0t 2
t>2
Find the current i(t) and sketch its waveform for t 0.
Answer:
0
t <0
NAME: _x_:wl_-E_c _
BME 2H EXAM #1
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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
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
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 1
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
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 n
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
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
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 =
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
P7.4-5 Determine the value of the capacitance C in the circuit shown in Figure P 7.4-5, given
that Ceq = 8 F.
Answer: C = 20 F
Figure P 7.4-5
Solution: The 16 F capacitor is in series with a parallel