HOMEWORK VII
Chapter 6, Problem 6.
The voltage waveform in Fig. 6.46 is applied across a 30-µF
capacitor. Draw the current waveform through it.
Figure 6.46
Chapter 6, Solution 6.
dv
= 30x10 −6 x slope of the waveform.
dt
For example, for 0 < t < 2,
i=C
dv
10
=
dt 2x10 −3
dv
10
i= C
= 30x10 −6 x
= 150mA
dt
2x10 −3
Thus the current i is sketched below.
i(t) (mA)
150
4
2
‐150
8
6
t (msec)
10
12
Chapter 6, Problem 12.
A voltage of 60 cos 4πt V appears across the terminals of
a 3-mF capacitor. Calculate the current through the capacitor and
the energy stored in it from t = 0 to t = 0.125 s.
Chapter 6, Solution 12.
dv
= 3x10 −3 x 60(4π)(− sin 4π t)
dt
= –0.72 π sin 4πt A
i=C
P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W
W=
∫
t
o
1
pdt = − ∫ 8 21.6π sin 8π t dt
o
21.6π
=
cos 8π
8π
Chapter 6, Problem 22.
1/ 8
o
= –5.4J
Obtain the equivalent capacitance of the circuit in Fig. 6.56.
Figure 6.56
Chapter 6, Solution 22.
Combining the capacitors in parallel, we obtain the equivalent circuit shown below:
a
b
40 µF
60 µF
30 µF
20 µF
Combining the capacitors in series gives C1eq , where
1
1
1
1
1
=
+
+
=
1
C eq 60 20 30 10
C1eq = 10µF
Thus
Ceq = 10 + 40 = 50 µF
Chapter 6, Problem 31.
If v(0)=0, find v(t), i1(t), and i2(t) in the circuit in Fig. 6.63.
Figure 6.63
Chapter 6, Solution 31.
⎡20 tmA,
i s ( t ) = ⎢⎢20mA,
⎢⎣− 50 + 10 t ,
0 < t <1
1< t < 3
3< t <5
Ceq = 4 + 6 = 10µF
1 t
v=
idt + v(0)
C eq ∫o
For 0 < t < 1,
v=
10 −3
10x10 −6
t
∫ 20t dt + 0 = t
2
o
kV
For 1 < t < 3,
10 3 t
v=
20dt + v(1) = 2( t − 1) + 1kV
10 ∫1
= 2 t − 1kV
For 3 < t < 5,
10 3
v=
10
=
t
∫ 10(t − 5)dt + v(3)
3
t2
t2
− 5t 3t +5kV =
− 5t + 15.5kV
2
2
⎡
⎢ t 2 kV,
0 < t <1
⎢
v( t ) = ⎢2t − 1kV,
1< t < 3
⎢ 2
⎢t
⎢⎣ 2 − 5t + 15.5kV, 3 < t < 5
dv
dv
i 1 = C1
= 6x10 −6
dt
dt
0 < t <1
⎡12tmA,
⎢
= ⎢12mA,
1< t < 3
⎢⎣6 t − 30mA, 3 < t < 5
0 < t <1
⎡8tmA,
dv
⎢
−6 dv
i2 = C2
= 4x10
= ⎢8mA,
1< t < 3
dt
dt
⎢⎣4 t − 20mA, 3 < t < 5
Chapter 6, Problem 32.
In the circuit in Fig. 6.64, let is = 30e-2t mA and v1(0) = 50 V, v2(0) = 20 V. Determine: (a)
v1(t) and v2(t), (b) the energy in each capacitor at t = 0.5 s.
Figure 6.64
Chapter 6, Solution 32.
(a) Ceq = (12x60)/72 = 10 µ F
v1 =
v2 =
10 − 3
12 x10
t
−6
− 2t
dt + v1 (0) = − 1250e − 2 t 0 + 50 = − 1250e − 2 t + 1300V
t
0
10 − 3
60 x10
∫ 30e
−6
t
∫ 30e
− 2t
dt + v 2 (0) = 250e − 2 t 0 + 20 = − 250e − 2 t + 270V
t
0
(b) At t=0.5s,
v1 = −1250e −1 + 1300 = 840.2,
w12 µF =
v 2 = −250e −1 + 270 = 178.03
1
x12 x10 −6 x(840.15) 2 = 4.235 J
2
1
x 20x10 − 6 x (178.03) 2 = 0.3169 J
2
1
w 40µF = x 40x10 − 6 x (178.03) 2 = 0.6339 J
2
w 20µF =