Chapter 6, Problem 2.
A 20-μF capacitor has energy w(t) = 10 cos2 377t J. Determine the current through
the capacitor.
Chapter 6, Solution 2.
w=
1
Cv2
2
⎯⎯
→ v2 =
2W 20 cos2 377t
=
= 106 cos2 377t
−6
C
20 x10
v = ±103cos(377t) V, let us assume the v = +cos(377t) mV, this then leads to,
i = C(dv/dt) = 20x10–6(–377sin(377t)10–3) = –7.54sin(377t) A.
Please note that if we had chosen the negative value for v,
then i would have been positive.
Chapter 6, Problem 8.
A 4-mF capacitor has the terminal voltage
50 V,
t≤0
⎧
v = ⎨ -100t
+ Be -600t V,
t≥0
⎩Ae
If the capacitor has initial current of 2A, find:
(a) the constants A and B,
(b) the energy stored in the capacitor at t = 0,
(c) the capacitor current for t > 0.
Chapter 6, Solution 8.
(a) i = C
dv
= −100 ACe −100t − 600 BCe −600t
dt
i (0) = 2 = −100 AC − 600 BC
⎯
⎯→
(1)
5 = − A − 6B
v (0 + ) = v (0 − )
⎯
⎯→ 50 = A + B
Solving (2) and (3) leads to
A=61, B=-11
(b) Energy =
(2)
(3)
1 2
1
Cv (0) = x 4 x10 −3 x 2500 = 5 J
2
2
(c ) From (1),
i = −100 x61x 4 x10 −3 e −100t − 600 x(−11) x 4 x10 −3 e −600t = − 24.4e −100t + 26.4e −600t A
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.
v(t) =
(
)
(
)
t
1 t
6 1 − e − t dt + 0 = 12 t + e − t V = 12(t + e-t) – 12
∫
0
12 o
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:
50 Ohms
10 Ohms
20 Ohms
30 Ohms
+
v1
+
60V
+
−
i2 = 0, i1 = 60/(30+10+20) = 1A
v2
v1 = 30i1 = 30V, v2 = 60–20i1 = 40V
Thus, v1 = 30V, v2 = 40V
Chapter 6, Problem 37.
The current through a 12-mH inductor is 4 sin 100t A. Find
the voltage, and also the energy stored in the inductor for
0 < t < π/200 s.
Chapter 6, Solution 37.
di
= 12 x10 −3 x 4(100) cos100t
dt
= 4.8 cos 100t V
v=L
p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
w=
t
11 / 200
o
o
∫ pdt = ∫
9.6 sin 200 t
9.6
/ 200
cos 200t 11
J
o
200
= −48(cos π − 1)mJ = 96 mJ
=−
Please note that this problem could have also been done by using (½)Li2.
Chapter 6, Problem 39.
The voltage across a 200-mH inductor is given by
v(t) = 3t2 + 2t + 4 V for t > 0.
Determine the current i(t) through the inductor. Assume that i(0) = 1 A.
Chapter 6, Solution 39
v=L
i=
di
1
⎯
⎯→ i = ∫ 0t idt + i(0)
dt
L
1
200x10
t
(3t 2
−3 ∫ 0
= 5( t 3 + t 2 + 4t )
t
0
+ 2t + 4)dt + 1
+1
i(t) = 5t3 + 5t2 + 20t + 1 A
Chapter 6, Problem 47.
For the circuit in Fig. 6.70, calculate the value of R that will
make the energy stored in the capacitor the same as that stored in the
inductor under dc conditions.
Figure 6.70
Chapter 6, Solution 47.
Under dc conditions, the circuit is equivalent to that shown below:
R
+
5A
iL =
2 Ohms
2
10
10R
(5) =
, v c = Ri L =
R+2
R+2
R+2
1 2
100R 2
Cv c = 80x10 −6 x
2
(R + 2) 2
1
100
w L = Li12 = 2x10 −3 x
2
(R + 2) 2
If wc = wL,
wc =
vC
80x10
−6
x
100R 2
(R + 2)
=
2
2x10 −3 x100
(R + 2)
80 x 10-3R2 = 2
2
R = 5Ω
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
Chapter 6, Solution 48.
v
-
6 μF
20 kΩ
For Prob. 6.48.
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