Chapter 4:
4.1, 4.2, 4.4, 4.7, 4.10, 4.12, 4.14
4.29, 4.32, 4.33, 4.37, 4.38
4.43, 4.45, 4.47, 4.49, 4.51, 4.54, 4.55
4.68, 4.69, 4.72, 4.73
Q4.1.
Solution
Using the differential relationship for the inductor, we may obtain the voltage by differentiating the current:
di L (t )
di L (t )
⎡
⎛
π ⎞⎤
= 0.5× ⎢−377 × 2sin⎜ 377t + ⎟⎥
⎝
dt
dt
6 ⎠⎦
⎣
⎞
⎛
⎛
5π ⎞
π
= 377sin⎜ 377t + − π ⎟ = 377sin⎜ 377t −
⎟ V
⎠
⎝
⎝
6
6 ⎠
v L (t ) = L
= 0.5
Q4.2. Solution.
Using the defining differential relationship for the capacitor, (Eq. 4.4), we may obtain the current by
differentiating the voltage:
iC (t ) = C
a)
dvC (t )
dt
= 100 × 10−6
dvC (t )
dt
= 10−4
dvC (t )
dt
⎡
⎛
⎛
π ⎞⎤
π⎞
iC (t ) = 10−4 ⎢−20 × 40 sin⎜ 20t − ⎟⎥ = −0.08 sin⎜ 20t − ⎟
⎝
⎠
⎝
2 ⎦
2⎠
⎣
⎛
⎞
⎛
π
π⎞
= 0.08 sin⎜ 20t − + π ⎟ = 0.08sin⎜ 20t + ⎟ A
⎝
⎠
⎝
2
2⎠
b)
iC (t ) = 10−4 [100× 20cos100t ] = 0.2 cos100t A
c)
⎡
⎛
⎛
π ⎞⎤
π⎞
iC (t ) = 10−4 ⎢−80 × 60 cos⎜80t + ⎟⎥ = −0.48 cos⎜80t + ⎟
⎝
⎝
6 ⎠⎦
6⎠
⎣
⎛
⎞
⎛
π
5π ⎞
= 0.48 cos⎜80t + − π ⎟ = 0.48 cos⎜80t −
⎟ A
⎝
⎠
⎝
6
6 ⎠
d)
⎡
⎛
⎛
π ⎞⎤
π⎞
iC (t ) = 10−4 ⎢−100 × 30sin⎜100t + ⎟⎥ = −0.3sin⎜100t + ⎟
⎝
⎝
4 ⎠⎦
4⎠
⎣
⎛
⎞
⎛
π
3π ⎞
= 0.3sin⎜100t + − π ⎟ = 0.3sin⎜100t −
⎟ A
⎝
⎠
⎝
4
4 ⎠
1
Q4.4, Solution.
The magnetic energy stored in an inductor may be found
from, (Eq. 4.16):
1
1
Li(t )2 = (2)i 2 (t ) = i 2 (t )
2
2
For −∞ < t < 0 ,
wL (t ) = 0
wL (t ) =
For 0 ≤ t < 10 s
wL (t ) = t 2 J
For 10 s ≤ t < +∞
wL (t ) = 100 J
Q 4.7, Solution .
The energy stored in a capacitor may be found from:
1
1
Cv(t )2 = (0.1)v(t )2 = 0.05v(t )2
2
2
For −∞ < t < 0
wC (t ) = 0
wC (t ) =
For 0 ≤ t < 10s
wC (t ) = 0.05 t 2 J
For 10s ≤ t < +∞
wC (t ) = 0.05(10 )2 = 5 J
Q4.10, Solution.
Under steady-state conditions, all the currents are constant, no current can flow through the capacitors, and
the voltage across any inductor is equal to zero.
6 − v 4Ω v 4 Ω v 4Ω
1
1
=
+
⇒ v 4Ω = 3.43 V
⇒ w2 F = C 2 F v 22F = (2 F )(3.43 V )2 = 11.76 J
2
4
8
2
2
1
1
2
2
v1F = v 2H = 0
⇒ w1F = C1F v1F = (1 F )(0) = 0
2
2
v4Ω
1
1
2
2
i 2H =
= 0.43 A ⇒ w 2H = L2H i 2H = (2 H )(0.43 A) = 0.18 J
8
2
2
1
1
2
2
v 3F = v 4 Ω = 3.43 V ⇒ w 3F = C 3F v 3F = (3 F)(3.43 V) = 17.65 J
2
2
v 2 F = v 4Ω ⇒
2
Q4.12, Solution.
Since the voltage waveform is piecewise continuous, the derivative must be evaluated over each continuous
segment.
For 0 < t < 5 ms
vC (t ) = m vC t + qvC
where:
mvC =
[−10 V] − [+20 V] = −6
[5 ms] − [0]
V
ms
qvC = +20 V
⎛
dvC (t )
d
V⎞
=C
m vC t + qvC = CmvC = (80 μF )⎜−6
⎟ = −480 mA
⎝ ms ⎠
dt
dt
For 5 ms < t < 10 ms
[
iC = C
]
vC(t)=−10 V
dvC (t )
d
iC = C
= C [−10 V] = 0
dt
dt
For t > 10 ms
vC (t ) = 0
dv (t )
d
iC = C C
= C [0] = 0
dt
dt
A capacitor is fabricated from two conducting plates
separated by a dielectric constant. Dielectrics are also
insulators; therefore, current cannot really flow through
a capacitor. Positive charge, however, entering one plate exerts a repulsive force on and forces positive
carriers to exit the other plate. Current then appears to flow through the capacitor. Such currents are called
electric displacement currents.
Q4.14, Solution.
Since the voltage waveform is a piecewise continuous function of time, integration must be performed over
each continuous segment. Where not indicated, t is expressed in seconds.
For 0 < t ≤ 5 μs
v L (t ) = mvL t + q vL
where: mv L =
qv L = 0 V
[3.5 V] − [0 V] = 0.7 V
μs
[5 μs] − [0]
For t > 5 μs
v L (t) = c v L = −1.9 V
Therefore:
i L (t = 15 ?s) =
(
)
( )
1 15 μs
1
1
v L (τ )dτ = i L (0) + ∫ 05 μs m v L τ dτ + ∫ 515μμs s c v L dτ =
∫
L −∞
L
L
V
5 μs
0.7
15 μs
1 ⎡1
1
1
2⎤
ms
+ c vL τ
= 0+
⋅
= i L (0) + ⎢ m v L τ ⎥
5 μs
⎦0
L⎣2
L
0.75 mH
2
[ ]
+
1
⋅ (−1.9 V) ⋅ (15 μs − 5 μs) = −13.67 mA
0.75 mH
3
2
⎛
⎞
⋅ ⎜(5 μs) − 0⎟ +
⎝
⎠
Q4.29, Solution.
The average value is:
⎡ 2π
⎤
2π
⎡
2π
2π
⎤
⎥ 1
ω
ω ⎢ω
⎢− sin(ωt) ω + 2.5t ω ⎥
v (t) =
⎢ ∫ 2cos (ωt)dt + ∫ (2.5)dt ⎥ =
0
0 ⎥
⎢
2π ⎢ 0
π
2
0
⎥
⎣
⎦
1
1
=
[sin(0) − sin(2π )] + 2.5 = [0 − 0] + 2.5 = 2.5
2π
2π
⎣
⎦
The rms value is:
2π
2π
ω ω
ω ω⎡
2
⎤
x rms =
∫ (2cos (ωt) + 2.5) 2 dt =
∫ ⎢ 4 ⋅ [cos(ωt )] + 10 ⋅ cos(ωt ) + 6.25⎥dt =
⎣
⎦
2π 0
2π 0
⎡
=
2π
2π
2π
⎤
ω
ω ⎥
ω ⎢ ω
2
⋅ ⎢ 4 ⋅ ∫ [cos(ωt )] dt + 10 ⋅ ∫ sin(ωt )dt + 6.25 ⋅ ∫ dt ⎥ =
2π ⎢
0
0
0 ⎥
⎣
⎦
2π ⎤
ω ⎡ 1 2π
⋅ ⎢4 ⋅ ⋅
+ 10 ⋅ 0 + 6.25⋅
= 8.25 = 2.87
ω ⎥⎦
2π ⎣ 2 ω
Note: The integral of a sinusoid over an integer number of period is identically zero.
=
Q4.32, Solution.
The average value is: v =
The rms value is: v rms =
Therefore,
v
v rms
=−
[
]
1
0.004
∫ 0.002 (−9)dt + ∫ 0.002
(1)dt = 250(−0.018 + 0.002) = −4 V
0.004 0
[
]
1
2
0.004 2
∫ 0.002 (−9) dt + ∫ 0.002
(1) dt = 250 ⋅ [81⋅ 0.002 + 0.004 − 0.002] = 6.40 V
0.004 0
4
= −0.625
6.40
Q4.33, Solution.
The rms value is: i rms =
(
)
2
1 p
∫ 0 10 ⋅sin 2 (t ) dt =
p
1 p
∫ 100 ⋅sin 4 (t )dt =
p 0
1
3p
⋅100 ⋅
= 6.12 A
p
8
2
2
Therefore, the power dissipated by a 1-Ω resistor is: P1Ω = Ri rms
= (1)(6.12) W = 37.5 W
Q 4.37, Solution.
In phasor form:
a) V(jw) = 155∠-25° V
b) V(jw) = 5∠-130° V
c) I(jw) = 10∠63° + 15∠-42° = (4.54 + j8.91) + (11.15 − j 10.04 ) = 15.69 − j1.13 = 15.73∠-4.12° A
d) I(jw) = 460∠-25° - 220∠75° = (416.90 - j194.40) − (56.94-j 212.50) = 359.96 + j18.10 = 360.4∠2.88° A
4
Q4.38, Solution.
a) 4 + j 4=4 2∠45°=5.66∠45°
b) -3 + j 4 = 5 ∠126.9°
c) j + 2 − j 4 − 3 = -1 -j 3 = 3.16∠-108.4°
Q 4.43, Solution.
a) The current and the voltage can be expressed in phasor form:
I = 17∠ − 15o mA , V = 3.5∠75o V
Z=
V
3.5∠75o V
=
= 205.9∠90 o Ω = 0 + j ⋅ 205.9 Ω
I 17∠ − 15o mA
The impedance has a positive imaginary or reactive component and a positive angle of 90 degree indicating
that this is an inductor (see Fig. 4.39).
b)
Z L = j ⋅ X L = j ⋅ ωL = j ⋅ 205.9 Ω ⇒ L =
205.9 Ω
Vs
= 327.7 m
= 327.7 mH
rad
A
628.3
s
Solution 45.
From graph:
3π 180 o
2π
rad
= −135o , I 0 = 8 mA, ω = 2πf =
= 1571
s
4 π
T
⎛
⎞
rad
i (t ) = I 0 cos(ωt + φ ) = 8 cos⎜1571
⋅ t − 135o ⎟ mA
⎝
⎠
s
φ =−
Phasor form:
V = V0∠φ = 8∠(−135) V = 8 V ⋅ e− j 135
o
o
Solution 47.
⎛
rad ⎞
X L = ωL = ⎜ 3 k
⎟(190 mH ) = 0.57 kΩ ⇒ Z L = + j ⋅ X L = + j ⋅ 0.57 kΩ
⎝
s ⎠
XC =
1
1
=
= 6.061 kΩ ⇒ ZC = − j ⋅ XC = − j ⋅ 6.061 kΩ
⎛
⎞
rad
ωC
⎜3 k
⎟(55 nF)
⎝
s ⎠
Z eq1 = Z R1 + Z L = R1 + jX L = 2.3+ j ⋅ 0.57 kΩ = 2.37∠13.92 o kΩ
Z eq2 = Z R1 + ZC = R1 − jXC = 1.1− j ⋅ 6.061 kΩ = 6.16∠ − 79.71o kΩ
Z eq =
=
Z eq1 ⋅ Z eq 2
Z eq1 + Z eq2
=
(2.37∠13.92
o
)(
kΩ ⋅ 6.16∠ − 79.71o kΩ
)=
(2.3+ j ⋅ 0.57 kΩ) + (1.1− j ⋅ 6.061 kΩ)
14.60∠ − 65.79 o kΩ 2 14.60∠ − 65.79 o kΩ 2
=
= 2.261∠ − 7.56 o kΩ
3.4 − j ⋅ 5.491 kΩ
6.458∠ − 58.23o kΩ
5
Solution 49.
a) Phasor notation:
I s = I 0 ∠φ = 13∠30 o mA
b) ZC = − jXC − j
1
1
=−j
= 0 − j2 kΩ = 2∠ − 90 o kΩ
⎛
rad ⎞
ωC
⎜1000
⎟(0.5 ΩF)
⎝
s ⎠
c)
(
)(
vC (t ) = 26 cos(1000t − 60 o ) V
)
VC = I s ⋅ ZC = 13∠30 o mA ⋅ 2∠ − 90 o kΩ = 26∠ − 60 o V
Note that conversion from phasor notation to time notation or vice versa can be done at any time.
Solution 51.
KVL:
Vs1 = Vs2 = 170∠0 o V = (170 + j0) V
−Vs1 − Vs2 + I 3Z 3 = 0
Vs1 + Vs2 170∠0 o V + 170∠0 o V 340∠0 o V
=
=
= 20∠ − 11o A
Z3
17∠11o Ω
17∠11o Ω
⎛
⎞
rad
i 3(t ) = 20 cos⎜ 377
⋅ t − 11o ⎟ A
⎝
⎠
s
I3 =
Note also:
KVL:
−Vs1 + I1Z1 = 0 ⇒ I1 =
Vs1
V
, −Vs2 − I 2 Z 2 = 0 ⇒ I 2 = − s2
Z1
Z2
Solution 54.
Assume clockwise currents:
rad
, VS = 12∠0 o V
s
1
ZC =
= − j Ω , Z L = jωL = j 9 Ω ⇒ Z total = 3 + j 9 − j = 3 + j8 Ω
jωC
12
I=
= 0.4932 − j1.3151 A = 1.4045∠ − 69.44 o A , i (t ) = 1.4 cos ωt − 69.4 o
3 + j8
ω=3
(
)
A
Solution 55.
Assume clockwise currents:
rad
1
, I S = 10∠0 o A , Z L = jωL = j 6 Ω , ZC =
= − j 1.5Ω
s
jωC
1
1
1
Z eq =
=
=
= 0.9231 − j1.3846 Ω
1
1
1
1
1
2 0.33+ j 0.5
+
+
−j +j
R Z L ZC
3
6
3
ω=2
V = I S Z eq = 10 A ⋅ (0.9231 − j1.3846) Ω = 9.231 − j13.846 a * 10 V = 16.641∠ − 56.31o V
6
Solution 68.
rad
, Vs = 36∠ − 60° V
s
ω=3
Z L2 = jωL2 = j3⋅ 3 = j9 Ω
1
1
=
= − j6 Ω
jωC j 3 ⋅ (1 18)
ZC =
Z L3 = jωL3 = j 3⋅ 3 = j9 Ω
Z eq =
1
(
Z L3 Z L2 + ZC
)
=
1
1
j9
=
=
= 2.25∠90 o Ω
1
1
1
1
4
+
+
Z L3
j 9 j3
Z L2 + ZC
(
)
ZT = Z R + Z L1 + Z eq = 9 + j3⋅ 3+ j2.25 = 9 + j11.25 = 14.407∠51.34 o Ω
V
36∠ − 60 o V
I= S =
= 2.499∠ −111.34 o A
ZT 14.407∠51.34 0 Ω
Veq = IZ eq = (2.499∠ −111.34 o )(2.25∠90 o ) = 5.623∠ − 21.34 o V
− j6
ZC
V=
Veq =
5.623∠ − 21.34 o = 11.25∠158.66 o V
j3
Z L + ZC
(
2
(
)
)
v = 11.25cos 3t −158.66 o V
Solution 69.
1
rad
, Z L2 = jωL2 = j2Ω , ZC =
= − j Ω , Z L3 = jωL3 = j 20 Ω
jωC
s
Z L3 + Z C
j 20 − j
j19
I=
IS =
6 ∠0 o =
6∠0 o = 5.28∠13.4 o A
( j 20 − j ) + (5 + j 2)
Z L3 + Z C + R + Z L2
5 + j 21
ω=2
(
(
) (
)
i = 5.28 cos 2t + 13.4 o
)A
Solution 72.
VS (t ) = 2∠0 o V
ZC =
1
1
= − j2Ω
1
j2
4
Z L = jω L = j 2 ⋅ 2 = j 4 Ω
jω C
=
Applying the voltage divider rule:
VL =
(Z L || (ZC + Z 2 )) V = 4∠36.8° 2∠0 o = 1.05∠18.4° V
S
4∠0° + 4∠36.8°
Z1 + (Z L || (ZC + Z 2 ))
Therefore, the current is:
IL =
VL 1.05∠18.4°
=
= 0.2635∠ − 71.6 o A
ZL
4∠90°
(
i L (t ) = 0.2635 cos 2t − 71.6 o
)A
7
Solution 73.
In the phasor domain:
ZC =
-j
-6
1500(1 × 10 )
=−j
2000
= − j 666.7 Ω , Z L = j (1500)(0.5) = j750 Ω
3
By applying KVL in the first loop, we have
VS = R1I 1 + ZC (I 1 − I 2 )
By applying KVL in the second loop, we have
0 = (ZC )(I 2 − I 1 ) + (Z L + R2 )I 2
That is:
⎧
2000 ⎞
2000
o ⎛
I2
⎟I 1 + j
⎪15∠0 = ⎜ 75 − j
⎪
⎝
⎠
3
3
⎨
⎪0 = j 2000 I + ⎛100 + j 250 ⎞I
⎟ 2
1 ⎜
⎪⎩
⎝
3
3 ⎠
By solving above equations, we have
I 1 = 3.8 ⋅10−3 ∠46.6 o A
I 2 = 19.6 ⋅10−3∠ − 83.2 o A
i1 (t ) = 3.8 cos (1,500t + 46.6o ) mA
i2 (t ) = 19.6 cos (1,500t − 83.2o ) mA
8