Download ECE 2006 Homework #05 Solution

ECE 2006 Homework #05 Solution
Problem (1):
In the following circuit, V=10V, R1=8kΩ, R2=20kΩ, R3=1kΩ, R4=5kΩ. Please solve
for ix and iy..
Problem
Obtain ix and iy in the op amp circuit in Fig. 5.55.
iR3
R3
iR1
R1
ix
va
vb
iy
+
vo
iR4
+
iR2
vs
-
Solution :
Based on the KCL at node Va,
v − v a v a − vo
i R1 − i R3 = 0 ⇒ s
−
= 0 (Eq. 1)
R1
R3
Based on the KCL at node Vb,
R2
R4
0 − v a v a − vo
−
= 0 (Eq. 2)
R2
R4
Eq. 1 and Eq. 2 are the two linear equations about the two unknowns v a and vo .
Solving Eq. 1 and Eq. 2, v a and vo can be solved as
R2 R3
va =
vs ,
R2 R3 − R1 R4
R ( R + R4 )
vo = 3 2
vs .
R2 R3 − R1 R4
i R2 − i R4 = 0 ⇒
Then i x and i y can be obtained as
ix =
vs − va
R4
vs
=
R1
R1 R4 − R2 R3
 v − v a vo − v a 
R3 + R4
 =
+
vs
i y = − i R3 + i R4 = − o
R4  R2 R3 − R1 R4
 R3
With the values of provided for the resisters and input voltage,
ix=2.5 mA, iy=3 mA
(
)
Problems in the textbook: 5.59, 6.9, 6.13, 6.17, 6.34, 6.48, 6.53
Chapter 5, Problem 59.
In the op amp circuit of Fig. 5.86, determine the voltage gain vo/vs.
3R
R
4R
R
–
+
vs
+
_
Chapter 6, Problem 9.
The current through a 0.5-F capacitor is 6(1-e-t)A.
Determine the voltage and power at t=2 s. Assume v(0) = 0.
Chapter 6, Solution 9.
–
+
+
vo
–
v(t) =
(
)
(
)
1 t
−t
−t t
6
1
−
e
dt
+
0
=
12
t
+
e
V = 12(t + e-t) – 12
∫
o
0
12
v(2) = 12(2 + e-2) – 12 = 13.624 V
p = iv = [12 (t + e-t) – 12]6(1-e-t)
p(2) = [12 (2 + e-2) – 12]6(1-e-2) = 70.66 W
Chapter 6, Problem 13.
Find the voltage across the capacitors in the circuit of Fig. 6.49 under dc
conditions.
30 Ω
Figure 6.49
Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
+
v1
+
+
−
i2 = 0, i1 = 60/(30+10+20) = 1A
v2
v1 = 30i1 = 30V, v2 = 60–20i1 = 40V
Thus, v1 = 30V, v2 = 40V
Chapter 6, Problem 17.
Determine the equivalent capacitance for each of the circuits in
Fig. 6.51.
Figure 6.51
Chapter 6, Solution 17.
(a)
4F in series with 12F = 4 x 12/(16) = 3F
3F in parallel with 6F and 3F = 3+6+3 = 12F
4F in series with 12F = 3F
i.e. Ceq = 3F
(b)
(c)
Ceq = 5 + [6x(4 + 2)/(6+4+2)] = 5 + (36/12) = 5 + 3 = 8F
3F in series with 6F = (3 x 6)/9 = 2F
1
1 1 1
= + + =1
C eq 2 6 3
Ceq = 1F
Chapter 6, Problem 34.
The current through a 10-mH inductor is 6e-t/2 A.
the power at t = 3 s.
Find the voltage and
Chapter 6, Solution 34.
i = 6e-t/2
di
1
v = L = 10 x10 −3 (6) e − t / 2
dt
2
-t/2
= -30e mV
v(3) = -30e-3/2 mV = –6.694 mV
p = vi = -180e-t mW
p(3) = -180e-3 mW = –8.962 mW
Chapter 6, Problem 48.
Under steady-state dc conditions, find i and v in the circuit in Fig. 6.71.
i
2 mH
+
10 mA
30kΩ
Figure 6.71
v
-
6 µF
20 kΩ
For Prob. 6.48.
Chapter 6, Solution 48.
Under steady-state, the inductor acts like a short-circuit, while the capacitor acts like
an open circuit as shown below.
i
+
10 mA
30kΩ
v
20 kΩ
–
Using current division,
i = [30k/(30k+20k)]10 mA = 6 mA
v = 20ki = 120 V
Chapter 6, Problem 53.
Find Leq at the terminals of the circuit in Fig. 6.75.
Figure 6.75
Chapter 6, Solution 53.
L eq = 6 + 10 + 8 [5 (8 + 12) + 6 (8 + 4)]
= 16 + 8 (4 + 4) = 16 + 4
Leq = 20 mH