Chapter 6, Solution 1.
i=C
dv
dt
( )
= 5 2e −3t − 6 + e −3 t = 10(1 - 3t)e-3t A
p = vi = 10(1-3t)e-3t ⋅ 2t e-3t = 20t(1 - 3t)e-6t W
Chapter 6, Solution 2.
1 2 1
w1 = Cv1 = (40)(120) 2
2 2
1 2 1
w2 = Cv1 = (40)(80) 2
2 2
∆w = w 1 − w 2 = 20(120 2 − 80 2 ) = 160 kW
Chapter 6, Solution 3.
dv 280 − 160
i=C = 40x10 −3 = 480 mA
dt 5
Chapter 6, Solution 4.
1 t
C ∫o
v= idt + v(0)
1
2∫
= 6 sin 4 tdt + 1
= 1 - 0.75 cos 4t
Chapter 6, Solution 5.
1 t
C ∫o
v= idt + v(0)
For 0 < t < 1, i = 4t,
1 t
− 6 ∫o
v= 4t dt + 0 = 100t2 kV
20x10
v(1) = 100 kV
For 1 < t < 2, i = 8 - 4t,
1 t
20x10 −6 ∫1
v= (8 − 4t )dt + v(1)
= 100 (4t - t2 - 3) + 100 kV
100t 2 kV, 0 < t Chapter 6, Solution 8.
dv
(a) i = C = −100 ACe −100t − 600 BCe −600t (1)
dt
i (0) = 2 = −100 AC − 600 BC
→ 5 = − A − 6B (2)
v (0 + ) = v (0 − )
→ 50 = A + B (3)
Solving (2) and (3) leads to
A=61, B=-11
1 2 1
(b) Energy = Cv (0) = x 4 x10 −3 x 2500 = 5 J
2 2
(c ) From (1),
i = −100 x61x 4 x10 −3 e −100t − 600 x11x 4 x10 −3 e −600t = − 24.4e −100t − 26.4e −600t A
Chapter 6, Solution 9.
v(t) =
1 t
12
( ) (
∫o 6 1 − e dt + 0 = 12 t + e V
−t −t
)
v(2) = 12(2 + e-2) = 25.62 V
p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t)
p(2) = 72(2-e-4) = 142.68 W
Chapter 6, Solution 10
dv dv
i=C = 2 x10 −3
dt dt
16t , 0 < t < 1µs
v = 16, 1 < t < 3 µs
64 - 16t, 3 < t < 4 µs
16 x10 6 , 0 < t < 1µs
dv
= 0, 1 < t < 3 µs
dt
- 16x10 , 3 < t < 4 µs
6
32 kA, 0 < t < 1µs
i (t ) = 0, 1 < t < 3 µs
- 32 kA, 3 < t < 4µs
Chapter 6, Solution 11.
1 t
C ∫o
v= idt + v(0)
For 0 < t < 1,
1 t
− 6 ∫o
v= 40 x10 −3 dt = 10t kV
4x10
v(1) = 10 kV
For 1 < t < 2,
1 t
C ∫1
v= vdt + v(1) = 10kV
For 2 < t < 3,
1 t
∫ (−40x10
−3
v= )dt + v(2)
4x10 −6 2
= -10t + 30kV
Thus
10 t ⋅ kV, 0 < t Chapter 6, Solution 13.
Under dc conditions, the circuit becomes that shown below:
i1 10 Ω i2 50 Ω
+ 20 Ω +
30 Ω v1 v2
− +
−
60V −
i2 = 0, i1 = 60/(30+10+20) = 1A
v1 = 30i2 = 30V, v2 = 60-20i1 = 40V
Thus, v1 = 30V, v2 = 40V
Chapter 6, Solution 14.
(a) Ceq = 4C = 120 mF
1 4 4
(b) = = Ceq = 7.5 mF
C eq C 30
Chapter 6, Solution 15.
In parallel, as in Fig. (a),
v1 = v2 = 100
+
v1 −
+ + C1 +
+ C1 C2 + C2
100V − v1 v2 100V − v2
− − −
(a) (b)
1 2 1
w20 = Cv = x 20x10 −6 x100 2 = 0.1J
2 2
1
w30 = x30x10 −6 x100 2 = 0.15J
2
(b) When they are connected in series as in Fig. (b):
C2 30
v1 = V= x100 = 60, v2 = 40
C1 + C 2 50
1
w20 = x30x10 −6 x 60 2 = 36 mJ
2
1
w30 = x30x10 −6 x 40 2 = 24 mJ
2
Chapter 6, Solution 16
Cx80
C eq = 14 + = 30
→ C = 20 µF
C + 80
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) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F
(c) 3F in series with 6F = (3 x 6)/9 = 6F
1 1 1 1
= + + =1
C eq 2 6 3
Ceq = 1F
Chapter 6, Solution 18.
For the capacitors in parallel
C1 = 15 + 5 + 40 = 60 µF
eq
1 1 1 1 1
Hence = + + =
C eq 20 30 60 10
Ceq = 10 µF
Chapter 6, Solution 19.
We combine 10-, 20-, and 30- µ F capacitors in parallel to get 60 µ F. The 60 - µ F
capacitor in series with another 60- µ F capacitor gives 30 µ F.
30 + 50 = 80 µ F, 80 + 40 = 120 µ F
The circuit is reduced to that shown below.
12 120
12 80
120- µ F capacitor in series with 80 µ F gives (80x120)/200 = 48
48 + 12 = 60
60- µ F capacitor in series with 12 µ F gives (60x12)/72 = 10 µ F
Chapter 6, Solution 20.
3 in series with 6 = 6x3/(9) = 2
2 in parallel with 2 = 4
4 in series with 4 = (4x4)/8 = 2
The circuit is reduced to that shown below:
20
1 6 2
8
6 in parallel with 2 = 8
8 in series with 8 = 4
4 in parallel with 1 = 5
5 in series with 20 = (5x20)/25 = 4
Thus Ceq = 4 mF
Chapter 6, Solution 21.
4µF in series with 12µF = (4x12)/16 = 3µF
3µF in parallel with 3µF = 6µF
6µF in series with 6µF = 3µF
3µF in parallel with 2µF = 5µF
5µF in series with 5µF = 2.5µF
Hence Ceq = 2.5µF
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 C1 , where
eq
1 1 1 1 1
1
= + + = C1 = 10µF
eq
C eq 60 20 30 10
Thus
Ceq = 10 + 40 = 50 µF
Chapter 6, Solution 23.
(a) 3µF is in series with 6µF 3x6/(9) = 2µF
v4µF = 1/2 x 120 = 60V
v2µF = 60V
3
v6µF = (60) = 20V
6+3
v3µF = 60 - 20 = 40V
(b) Hence w = 1/2 Cv2
w4µF = 1/2 x 4 x 10-6 x 3600 = 7.2mJ
w2µF = 1/2 x 2 x 10-6 x 3600 = 3.6mJ
w6µF = 1/2 x 6 x 10-6 x 400 = 1.2mJ
w3µF = 1/2 x 3 x 10-6 x 1600 = 2.4mJ
Chapter 6, Solution 24.
20µF is series with 80µF = 20x80/(100) = 16µF
14µF is parallel with 16µF = 30µF
(a) v30µF = 90V
v60µF = 30V
v14µF = 60V
80
v20µF = x 60 = 48V
20 + 80
v80µF = 60 - 48 = 12V
1 2
(b) Since w = Cv
2
w30µF = 1/2 x 30 x 10-6 x 8100 = 121.5mJ
w60µF = 1/2 x 60 x 10-6 x 900 = 27mJ
w14µF = 1/2 x 14 x 10-6 x 3600 = 25.2mJ
w20µF = 1/2 x 20 x 10-6 x (48)2 = 23.04mJ
w80µF = 1/2 x 80 x 10-6 x 144 = 5.76mJ
Chapter 6, Solution 25.
(a) For the capacitors in series,
v1 C 2
Q1 = Q2 C1v1 = C2v2 =
v 2 C1
C2 C + C2 C1
vs = v1 + v2 = v2 + v2 = 1 v2 v2 = vs
C1 C1 C1 + C 2
C2
Similarly, v1 = vs
C1 + C 2
(b) For capacitors in parallel
Q1 Q 2
v1 = v2 = =
C1 C 2
C C + C2
Qs = Q1 + Q2 = 1 Q 2 + Q 2 = 1 Q2
C2 C2
or
C2
Q2 =
C1 + C 2
C1
Q1 = Qs
C1 + C 2
dQ C1 C2
i= i1 = is , i2 = is
dt C1 + C 2 C1 + C 2
Chapter 6, Solution 26.
(a) Ceq = C1 + C2 + C3 = 35µF
(b) Q1 = C1v = 5 x 150µC = 0.75mC
Q2 = C2v = 10 x 150µC = 1.5mC
Q3 = C3v = 20 x 150 = 3mC
1 1
(c) w= C eq v 2 = x35x150 2 µJ = 393.8mJ
2 2
Chapter 6, Solution 27.
1 1 1 1 1 1 1 7
(a) = + + = + + =
C eq C1 C 2 C 3 5 10 20 20
20
Ceq = µF = 2.857µF
7
(b) Since the capacitors are in series,
20
Q1 = Q2 = Q3 = Q = Ceqv = x 200µV = 0.5714mV
7
1 1 20
(c) w = C eq v 2 = x x 200 2 µJ = 57.143mJ
2 2 7
Chapter 6, Solution 28.
We may treat this like a resistive circuit and apply delta-wye transformation, except that
R is replaced by 1/C.
Cb 50 µF
Ca
Cc 20 µF
1 1 1 1 1 1
+ +
1 10 40 10 30 30 40
=
Ca 1
30
3 1 1 2
= + + =
40 10 40 10
Ca = 5µF
1 1 1
+ +
1 2
= 400 300 1200 =
C6 1 30
10
Cb = 15µF
1 1 1
+ +
1 4
= 400 300 1200 =
Cc 1 15
40
Cc = 3.75µF
Cb in parallel with 50µF = 50 + 15 = 65µF
Cc in series with 20µF = 23.75µF
65x 23.75
65µF in series with 23.75µF = = 17.39µF
88.75
17.39µF in parallel with Ca = 17.39 + 5 = 22.39µF
Hence Ceq = 22.39µF
Chapter 6, Solution 29.
(a) C in series with C = C/(2)
C/2 in parallel with C = 3C/2
3C
Cx
3C 2 = 3C
in series with C =
2 C 5
5
2
C C
3 in parallel with C = C + 3 = 1.6 C
5 5
(b)
2C
Ceq
2C
1 1 1 1
= + =
C eq 2C 2C C
Ceq = C
Chapter 6, Solution 30.
1 t
C ∫o
vo = idt + i(0)
For 0 < t < 1, i = 60t mA,
10 −3 t
3x10 −6 ∫o
vo = 60tdt + 0 = 10 t 2 kV
vo(1) = 10kV
For 1< t < 2, i = 120 - 60t mA,
10 −3 t
3x10t −6 ∫1
vo = (120 − 60t )dt + v o (1)
= [40t – 10t2 ] 1 + 10kV
= 40t – 10t2 - 20
10t 2 kV, 0 < t = t 2 − 5 + 3 +5kV = t 2 − 5t + 11kV
t
t 2 kV, 0 < t Chapter 6, Solution 33
Because this is a totally capacitive circuit, we can combine all the capacitors using
the property that capacitors in parallel can be combined by just adding their
values and we combine capacitors in series by adding their reciprocals.
3F + 2F = 5F
1/5 +1/5 = 2/5 or 2.5F
The voltage will divide equally across the two 5F capacitors. Therefore, we get:
VTh = 7.5 V, CTh = 2.5 F
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) = -300e-3/2 mV = -0.9487 mV
p = vi = -180e-t mW
p(3) = -180e-3 mW = -0.8 mW
Chapter 6, Solution 35.
di V 60 x10 −3
v=L L= = = 200 mH
dt ∆i / ∆t 0.6 /(2)
Chapter 6, Solution 36.
di 1
v=L = x10 −3 (12)(2)(− sin 2 t )V
dt 4
= - 6 sin 2t mV
p = vi = -72 sin 2t cos 2t mW
But 2 sin A cos A = sin 2A
p = -36 sin 4t mW
Chapter 6, Solution 37.
di
v=L = 12 x10 −3 x 4(100) cos100t
dt
= 4.8 cos 100t V
p = vi = 4.8 x 4 sin 100t cos 100t = 9.6 sin 200t
t 11 / 200
w= ∫ pdt = ∫
o o
9.6 sin 200 t
9.6
=− cos 200t 11 / 200 J
o
200
= −48(cos π − 1)mJ = 96 mJ
Chapter 6, Solution 38.
= 40x10 −3 (e − 2 t − 2te − 2 t )dt
di
v=L
dt
= 40(1 − 2t )e −2 t mV, t > 0
Chapter 6, Solution 39
di 1 t
v=L i = ∫ 0 idt + i(0)
→
dt L
1
i= t
(3t 2 + 2t + 4)dt + 1
−3 ∫ 0
200x10
t
= 5( t 3 + t 2 + 4t ) +1
0
i(t) = 5t3 + 5t2 + 20t + 1 A
Chapter 6, Solution 40
di di
v=L = 20 x10 −3
dt dt
10t , 0 < t < 1 ms
i = 20 - 10t, 1 < t < 3 ms
- 40 + 10t, 3 < t < 4 ms
10 x10 3 , 0 < t < 1 ms
di
= - 10x10 3 , 1 < t < 3 ms
dt
10x10 ,
3
3 < t < 4 ms
200 V, 0 < t < 1 ms
v = - 200 V, 1 < t < 3 ms
200 V, 3 < t < 4 ms
which is sketched below.
v(t) V
200
0 1 2 3 4 t(ms)
-200
Chapter 6, Solution 41.
i=
1 t
L
1 t
( )
∫0 vdt + i(0) = 2 ∫o 20 1 − 2 dt + 0.3
−2 t
1
= 10 t + e − 2t to +0. 3 = 10t + 5e − 2t − 4. 7 A
2
At t = ls, i = 10 - 4.7 + 5e-2 = 5.977 A
1 2
w= L i = 35.72J
2
Chapter 6, Solution 42.
1 t 1 t
L ∫o
i= vdt + i(0) = ∫ v( t )dt − 1
5 o
10 t
For 0 < t < 1, i = ∫ dt − 1 = 2t − 1 A
5 0
For 1 < t < 2, i = 0 + i(1) = 1A
1
∫ 10dt + i(2) = 2t t +1
2
For 2 < t < 3, i =
5
= 2t - 3 A
For 3 < t < 4, i = 0 + i(3) = 3 A
1 t
5 ∫4
For 4 < t < 5, i = 10dt + i(4) = 2 t 4 +3
t
= 2t - 5 A
2t − 1A, 0 < t Chapter 6, Solution 43.
2
t 1 1
w = L ∫ idt = Li( t ) − Li (−∞)
−∞ 2 2
1
(
= x80 x10 −3 x 60x10 −3 − 0
2
)
= 144 µJ
Chapter 6, Solution 44.
1 t 1 t
i=
L ∫to vdt + i(t o ) = 5 ∫o (4 + 10 cos 2t )dt − 1
= 0.8t + sin 2t -1
Chapter 6, Solution 45.
1 t
L ∫o
i(t) = v( t ) + i(0)
For 0 < t < 1, v = 5t
1 t
i=
10x10 −3 ∫ 5t dt + 0
o
= 0.25t2 kA
For 1 < t < 2, v = -10 + 5t
1 t
i=
10x10 −3 ∫ (−10 + 5t )dt + i(1)
1
t
= ∫ (0.5t − 1)dt + 0.25kA
1
= 1 - t + 0.25t2 kA
0.25t 2 kA, 0 < t Chapter 6, Solution 46.
Under dc conditions, the circuit is as shown below:
2Ω
iL
+
4Ω vC
3A −
By current division,
4
iL = (3) = 2A, vc = 0V
4+2
1 2 11 2
wL = L i L = (2) = 1J
2 22
1 1
wc = C v c = (2)( v) = 0J
2
2 2
Chapter 6, Solution 47.
Under dc conditions, the circuit is equivalent to that shown below:
R
+ vC − iL
2Ω
5A
2 10 10R
iL = (5) = , v c = Ri L =
R+2 R+2 R+2
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