G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Chapter 4 Instructor Notes
The chapter starts by developing the dynamic equations for energy storage elements. The analogy
between electrical and hydraulic circuits (Make The Connection: Fluid (hydraulic) Capacitance, p. 138,
Make The Connection: Fluid (hydraulic)Iinertance, p. 150, Table 4.2) is introduced early to permit a
connection with ideas that may already be familiar to the student from a course in fluid mechanics, such as
mechanical, civil, chemical and aerospace engineers are likely to have already encountered. A Focus on
Measurements boxes: Capacitive displacement transducer and microphones, pp. 147-148, permits
approaching the subject of capacitance in a pragmatic fashion, if so desired. The instructor wishing to gain
a more in-depth understanding of such transducers will find a detailed analysis in1.
Next, signal sources are introduced, with special emphasis on sinusoids. The material in this
section can also accompany a laboratory experiment on signal sources. The emphasis placed on sinusoidal
signals is motivated by the desire to justify the concepts of phasors and impedance, which are introduced
next. The author has found that presenting the impedance concept early on is an efficient way of using the
(invariably too short) semester or quarter. The chapter is designed to permit a straightforward extension of
the resistive circuit analysis concepts developed in Chapter 3 to the case of dynamic circuits excited by
sinusoids. The ideas of nodal and mesh analysis, and of equivalent circuits, can thus be reinforced at this
stage. The treatment of AC circuit analysis methods is reinforced by the usual examples and drill exercises,
designed to avoid unnecessarily complicated circuits. Two Focus on Methodology boxes (pp. 165 and 180)
provide the student with a systematic approach to the solution of basic AC analysis problems using phasor
and impedance concepts.
The capacitive displacement transducer example is picked up again in Focus on Measurements:
Capacitive displacement transducer (pp.175-177) to illustrate the use of impedances in a bridge circuit.
This type of circuit is very common in mechanical measurements, and is likely to be encountered at some
later time by some of the students.
The homework problems in this chapter are mostly exercises aimed at mastery of the techniques.
Learning Objectives
1. Compute currents, voltages and energy stored in capacitors and inductors.
2. Calculate the average and root-mean-square value of an arbitrary (periodic) signal.
3. Write the differential equation(s) for circuits containing inductors and capacitors.
4. Convert time-domain sinusoidal voltages and currents to phasor notation, and vice-versa, and
represent circuits using impedances.
5. Apply the circuit analysis methods of Chapter 3 to AC circuits in phasor form.
1
E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York,
1990.
4.1
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Section 4.1: Energy Storage Elements
Problem 4.1
Solution:
Known quantities:
Inductance value, L
= 0.5 H ; the current through the inductor as a function of time.
Find:
The voltage across the inductor, (Eq. 4.9), as a function of time.
Assumptions:
iL (t ≤ 0) = 0
Analysis:
Using the differential relationship for the inductor, we may obtain the voltage by differentiating the current:
v L (t ) = L
di L (t )
di (t )
ª
π ·º
§
= 0.5 L = 0.5 × «− 377 × 2 sin¨ 377t + ¸»
dt
dt
6 ¹¼
©
¬
π
5π ·
·
§
§
= 377 sin¨ 377t + − π ¸ = 377 sin¨ 377t −
¸ V
6
6 ¹
©
¹
©
______________________________________________________________________________________
Problem 4.2
Solution:
Known quantities:
Capacitance value C
= 100 µF ; capacitor terminal voltage as a function of time.
Find:
The current through the capacitor as a function of time for each case:
a) vc (t ) = 40 cos(20t − π / 2)V
b)
vc (t ) = 20 sin(100t )V
vc (t ) = −60 sin(80t + π / 6)V
d) vc (t ) = 30 cos(100t + π / 4)V .
c)
Assumptions:
The capacitor is initially discharged:
vC (t = 0) = 0
Analysis:
Using the defining differential relationship for the capacitor, (Eq. 4.4), we may obtain the current by
differentiating the voltage:
iC (t ) = C
dvC (t )
dv (t )
dv (t )
= 100 × 10 −6 C = 10 − 4 C
dt
dt
dt
a)
4.2
G. Rizzoni, Principles and Applications of Electrical Engineering
b)
Problem solutions, Chapter 4
ª
π ·º
π·
§
§
iC (t ) = 10 − 4 «− 20 × 40 sin¨ 20t − ¸ » = −0.08 sin¨ 20t − ¸
2 ¹¼
2¹
©
©
¬
π
π·
§
·
§
= 0.08 sin¨ 20t − + π ¸ = 0.08 sin¨ 20t + ¸ A
2
2¹
©
¹
©
iC (t ) = 10−4 [100 × 20 cos100t ] = 0.2 cos100t A
c)
d)
ª
π ·º
π·
§
§
iC (t ) = 10 − 4 « − 80 × 60 cos¨ 80t + ¸» = −0.48 cos¨ 80t + ¸
6 ¹¼
6¹
©
©
¬
π
5π ·
§
·
§
= 0.48 cos¨ 80t + − π ¸ = 0.48 cos¨ 80t −
¸ A
6
6 ¹
©
¹
©
ª
π ·º
π·
§
§
iC (t ) = 10 − 4 «− 100 × 30 sin¨100t + ¸» = −0.3 sin¨100t + ¸
4 ¹¼
4¹
©
©
¬
π
3π ·
§
·
§
= 0.3 sin¨100t + − π ¸ = 0.3 sin¨100t −
¸ A
4
4 ¹
©
¹
©
______________________________________________________________________________________
Problem 4.3
Solution:
Known quantities:
Inductance value, L
= 250 mH ; the current through the inductor, as a function of time.
Find:
The voltage across the inductor as a function of time for each case
a) iL (t ) = 5 sin 25tA
iL (t ) = −10 cos 50tA
c) iL (t ) = 25 cos(100t + π / 3) A
d) iL (t ) = 20 sin(10t − π / 12) A .
b)
Assumptions:
iL (t ≤ 0) = 0
Analysis:
Using the differential relationship for the inductor, (Eq. 4.9), we may obtain the voltage by differentiating
the current:
v L (t ) = L
a)
b)
di L (t )
di (t )
di (t )
= 250 × 10 −3 L = 0.25 L
dt
dt
dt
v L (t ) = 0.25[25 × 5 cos 25t ] = 31.25 cos 25t V
v L (t ) = 0.25[− 50 × (− 10 sin 50t )] = 125 sin 50t V
c)
4.3
G. Rizzoni, Principles and Applications of Electrical Engineering
d)
Problem solutions, Chapter 4
ª
π ·º
π·
§
§
v L (t ) = 0.25«− 100 × 25 sin¨100t + ¸» = −625 sin¨100t + ¸
3 ¹¼
3¹
©
©
¬
2π ·
π
§
·
§
= 625 sin¨100t + − π ¸ = 625 sin¨100t −
¸ V
3
3 ¹
©
¹
©
ª
π ·º
π ·
§
§
v L (t ) = 0.25«10 × 20 cos¨10t − ¸» = 50 cos¨10t − ¸ V
12 ¹¼
12 ¹
©
©
¬
______________________________________________________________________________________
Problem 4.4
Solution:
Known quantities:
Inductance value; resistance value; the current through the circuit shown in Figure P4.4 as a function of
time.
Find:
The energy stored in the inductor as a function of time.
Analysis:
The magnetic energy stored in an inductor may be
found from, (Eq. 4.16):
wL (t ) =
For − ∞ < t < 0 ,
1
1
2
Li (t ) = (2 ) i 2 (t ) = i 2 (t )
2
2
wL (t ) = 0
For 0 ≤ t < 10 s
wL (t ) = t 2 J
For 10 s ≤ t < +∞
wL (t ) = 100 J
______________________________________________________________________________________
Problem 4.5
Solution:
Known quantities:
Inductance value; resistance value; the current through the circuit in Figure P4.4 as a function of time.
Find:
The energy delivered by the source as a function of time.
Analysis:
The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy
stored in the inductor:
wS (t ) = wR (t ) + wL (t ) = Ri 2 (t ) +
1 2
Li (t )
2
4.4
G. Rizzoni, Principles and Applications of Electrical Engineering
= (1) i 2 (t ) +
For − ∞ < t < 0 ,
Problem solutions, Chapter 4
1
(2)i 2 (t ) = 2i 2 (t )
2
wS (t ) = 0
For 0 ≤ t < 10 s
wS (t ) = 2t 2 J
For 10 s ≤ t < +∞
wS (t ) = 200 J
______________________________________________________________________________________
Problem 4.6
Solution:
Known quantities:
Inductance value; resistance value; the current through the circuit shown in Figure P4.4 as a function of
time.
Find:
The energy stored in the inductor and the energy delivered by the source as a function of time.
Analysis:
The magnetic energy stored in an inductor may be found from, (Eq. 4.16):
wL (t ) =
1
1
2
Li (t ) = (2 ) i 2 (t ) = i 2 (t )
2
2
−∞ < t < 0,
wL (t ) = 0
For 0 ≤ t < 10 s
For
wL (t ) = t 2 J
For 10 ≤ t < 20 s
wL (t ) = (20 − t ) = 400 − 40t + t 2 J
For 20 s ≤ t < +∞
wL (t ) = 0 J
2
The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy
stored in the inductor:
wS (t ) = wR (t ) + wL (t ) = Ri 2 (t ) +
= (1) i 2 (t ) +
−∞ < t < 0,
wL (t ) = 0
For 0 ≤ t < 10 s
1 2
Li (t )
2
1
(2)i 2 (t ) = 2i 2 (t )
2
For
wL (t ) = 2t 2 J
4.5
G. Rizzoni, Principles and Applications of Electrical Engineering
For
Problem solutions, Chapter 4
10 ≤ t < 20 s
wL (t ) = 2 (20 − t ) = 800 − 80t + 2t 2 J
For 20 s ≤ t < +∞
wL (t ) = 0 J
2
______________________________________________________________________________________
Problem 4.7
Solution:
Known quantities:
Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of
time.
Find:
The energy stored in the capacitor as a function of time.
Analysis:
The energy stored in a capacitor may be found from:
1
1
2
2
2
Cv (t ) = (0.1)v(t ) = 0.05v(t )
2
2
wC (t ) =
−∞ <t < 0
wC (t ) = 0
For 0 ≤ t < 10 s
wC (t ) = 0.05 t 2 J
For 10 s ≤ t < +∞
2
wC (t ) = 0.05(10 ) = 5 J
For
______________________________________________________________________________________
Problem 4.8
Solution:
Known quantities:
Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of
time.
Find:
The energy delivered by the source as a function of time.
Analysis:
The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy
stored in the capacitor:
wS (t ) = wR (t ) + wC (t ) =
=
For − ∞ < t < 0 ,
v 2 (t ) 1 2
+ Cv (t )
R
2
1
1 2
v (t ) + (0.1)v 2 (t ) = 0.55i 2 (t )
2
2
wS (t ) = 0
For 0 ≤ t < 10 s
4.6
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
wS (t ) = 0.55t 2 J
For 10 s ≤ t < +∞
wS (t ) = 55 J
______________________________________________________________________________________
Problem 4.9
Solution:
Solution:
Known quantities:
Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of
time.
Find:
The energy stored in the capacitor and the energy delivered by the source as a function of time.
Analysis:
The energy stored in a capacitor may be found from:
wC (t ) =
For − ∞ < t < 0
1
1
2
2
2
Cv (t ) = (0.1)v(t ) = 0.05v(t )
2
2
wC (t ) = 0
For 0 ≤ t < 10 s
wC (t ) = 0.05 t 2 J
For 10 ≤ t < 20 s
2
wC (t ) = 0.05 (20 − t ) = 20 − 2t + 0.05t 2 J
For 10 s ≤ t < +∞
wC (t ) = 0
The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy
stored in the capacitor:
wS (t ) = wR (t ) + wC (t ) =
=
For − ∞ < t < 0
v 2 (t ) 1 2
+ Cv (t )
R
2
1
1 2
v (t ) + (0.1)v 2 (t ) = 0.55i 2 (t )
2
2
wC (t ) = 0
For 0 ≤ t < 10 s
wC (t ) = 0.55 t 2 J
For 10 ≤ t < 20 s
2
wC (t ) = 0.55 (20 − t ) = 220 − 22t + 0.55t 2 J
For 10 s ≤ t < +∞
wC (t ) = 0
______________________________________________________________________________________
4.7
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.10
Solution:
Known quantities:
Capacitance, resistance and inductance values; the source voltage vS
= 6 V applied to the circuit shown in
Figure P4.10.
Find:
The energy stored in each capacitor and inductor.
Analysis:
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.
v2 F = v4 Ω Ÿ
6 − v4 Ω v4 Ω v4Ω
=
+
Ÿ v4Ω = 3.43 V
2
4
8
1
1
2
Ÿ w2 F = C2 F v22F = (2 F )(3.43 V ) = 11.76 J
2
2
v1F = v 2 H = 0
i2 H =
v4 Ω
= 0.43 A
8
Ÿ w1F =
1
1
2
C1F v12F = (1 F )(0 ) = 0
2
2
1
1
2
L2 H i22H = (2 H )(0.43 A ) = 0.18 J
2
2
1
1
2
= C3 F v32F = (3 F )(3.43 V ) = 17.65 J
2
2
Ÿ w2 H =
v3 F = v4Ω = 3.43 V Ÿ w3 F
______________________________________________________________________________________
Problem 4.11
Solution:
Known quantities:
Capacitance, resistance and inductance values; the voltage v A = 12 V applied to the circuit shown in
Figure P4.11.
Find:
The energy stored in each capacitor and inductor.
Analysis:
Under steady-state conditions, all the currents are constant, no current can flow across the capacitors, and
the voltage across any inductor is equal to zero.
The voltage for the 1-F capacitor is equal to the
12-Volt input.
Since the voltage is the same on either end of the
3-Ω resistor in parallel with the 2-F capacitor,
there is no voltage drop through either
component.
Finally, since there is no voltage drop through the 3-Ω resistor in parallel with the 2-F capacitor, there is no
current flow through the resistor, and the current through the 1-H inductor is equal to the current through
the 2-H inductor.
Therefore,
4.8
G. Rizzoni, Principles and Applications of Electrical Engineering
v1F = v A = 12 V Ÿ w1F =
12 V
=2A
6Ω
Problem solutions, Chapter 4
1
1
2
C1F v12F = (1 F )(12 V ) = 72 J
2
2
1
1
2
L1H i12H = (1 H )(2 A ) = 2 J
2
2
1
1
2
i2 H = i1H = 2 A Ÿ w2 H = L2 H i22H = (2 H )(2 A ) = 4 J
2
2
1
1
2
v2 F = 0 V Ÿ w2 F = C2 F v22F = (2 F )(0 V ) = 0 J
2
2
i1H =
Ÿ w1H =
______________________________________________________________________________________
Problem 4.12
Solution:
Known quantities:
Capacitance value C = 80 µF ; the voltage applied to the capacitor as a function of time as shown in
Figure P4.12.
Find:
The current through the capacitor as a function of time.
Analysis:
Since the voltage waveform is piecewise continuous, the derivative must be evaluated over each continuous
segment.
For 0 < t < 5 ms
vC (t ) = mvC t + q vC
where:
m vC =
[− 10 V] − [+ 20 V] = −6 V
ms
[5 ms] − [0]
q vC = +20 V
dvC (t )
d
V·
§
=C
mvC t + q vC = CmvC = (80 µF )¨ − 6
¸ = −480 mA
dt
dt
ms ¹
©
For 5 ms < t < 10 ms
vC (t ) = −10 V
[
iC = C
iC = C
For
]
dvC (t )
d
= C [− 10 V ] = 0
dt
dt
t > 10 ms
vC (t ) = 0
iC = C
dvC (t )
d
= C [0] = 0
dt
dt
A capacitor is fabricated from two conducting plates
separated by a dielectric constant. Dielectrics are also
isolators; 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
4.9
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
flow through the capacitor. Such currents are called electric displacement currents.
______________________________________________________________________________________
Problem 4.13
Solution:
Known quantities:
Inductance value, L
= 35 mH ; the voltage applied to the inductor as shown in Figure P4.12; the initial
condition for the current i L (0 ) = 0 .
Find:
The current across the inductor as a function of time.
Analysis:
Since the voltage waveform is piecewise continuous, integration must be performed over each continuous
segment. Where not indicated t is supposed to be expressed in seconds.
For
0 < t ≤ 5 ms
v L (t ) = mvL t + q vL
where:
mvL =
[− 10 V] − [+ 20 V] = −6 V
ms
[5 ms] − [0]
q vL = +20 V
(
t
)
1 t
1 t
1 ª1
º
i L = i L (0) + ³ v L (τ )dτ = 0 + ³ mvL τ + qv L dτ = « mvL τ 2 + qvL τ » =
L 0
L 0
L ¬2
¼0
1
1
V 2
§
·
=
mvL t 2 + q vL t =
⋅ t + 20 V ⋅ t ¸ = − 85.71 ⋅ 10 3 t 2 + 571.4t A
¨− 6
2L
2 ⋅ 35 mH ©
ms
¹
(
)
(
(
)
)
i L (t = 5 ms = 0.005 s ) = − 85.71 ⋅ 10 3 ⋅ (0.005) + 571.4 ⋅ 0.005 A = 714.3 mA
For 5 ms < t ≤ 10 ms
v L ( t ) = cvL = −10 V
2
( )
[ ]
1 t
1 t
1
v L (τ )dτ = i L (0.005) + ³
cvL dτ = i L (0.005) + cvL τ
³
0
005
0
005
.
.
L
L
L
1
= 714.3 mA −
⋅ (− 10 V ) ⋅ (t − 0.005 s ) = (2.143 − 285.7t ) A
35 mH
i L = i L (0.005) +
i L (t = 10 ms = 0.01 s ) = (2.143 − 285.7(0.01)) A = −713.5 mA
For t > 10 ms
v L ( t ) = cvL = 0
1 t
1 t
1 t
v L (τ )dτ = i L (0.01) + ³ (0 )dτ = i L (0.01) + [0]0.005 =
³
L 0.01
L 0.01
L
= i L (0.01) = −713.5 mA
i L = i L (0.01) +
4.10
t
0.005
=
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
______________________________________________________________________________________
Problem 4.14
Solution:
Known quantities:
Inductance value L = 0.75 mH ; the voltage applied to the inductor as a function of time as shown in
Figure P4.14.
Find:
The current through the inductor at the time t = 15 µs .
Assumptions:
iL (t ≤ 0) = 0
Analysis:
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:
mvL =
[3.5 V] − [0 V] = 0.7 V
µs
[5 µs ] − [0]
q vL = 0 V
For t > 5 µs
v L ( t ) = cvL = −1.9 V
Therefore:
(
)
( )
1 15 µs
1 5 µs
1 15 µs
v L (τ )dτ = iL (0 ) + ³ mvLτ dτ + ³
c v dτ =
³
0
−
∞
L
L
L 5 µs L
V
5 µs
0 .7
1 ª1
1
1
º
15 µs
ms ⋅ (5 µs )2 − 0 +
= iL (0 ) + « mvLτ 2 » + cvLτ 5 µs = 0 +
⋅
L ¬2
L
0.75 mH
2
¼0
iL (t = 15 µs ) =
[ ]
+
(
1
⋅ (− 1.9 V ) ⋅ (15 µs − 5 µs ) = −13.67 mA
0.75 mH
______________________________________________________________________________________
4.11
)
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.15
Solution:
Known quantities:
Capacitance value C
v Peak
= 680 nF ; the periodic voltage applied to the capacitor as shown in Figure P4.15:
= 20 V , T = 40 µs .
Find:
The waveform and the plot for the current through the capacitor as a function of time.
Analysis:
Since the voltage waveform is not a continuous function of time, differentiation can be performed only over
each continuous segment. In the discontinuity points the derivative of the voltage will assume an infinite
value, the sign depending on the sign of the step. Where not indicated, t is expressed in seconds.
For each period 0 < t < T , T
consider only the first period:
For 0 < t < T
< t < 2T , the behavior of the capacitor will be the same; thus, we
vC (t ) = mvC t + q vC
where:
m vC =
[v Peak ] − [0] = 0.5 V
[T ] − [0]
µs
q vC = 0 V
dvC (t )
d
nAs ·§
V·
§
=C
mvC t = CmvC = ¨ 480
¸ ¨ 0 .5
¸ = 340 mA
dt
dt
V ¹©
µs ¹
©
For t = T , 2T , dv (t )
iC = C C = C [− ∞]
dt
iC = C
[
]
Figure P 4.15 shows the current waveform.
Note: For the voltage across the capacitor to decrease
instantaneously to zero at t = T , 2T , , the charge
on the plates of the capacitor should be instantaneously
discharged. This requires an infinite current which is
not physically possible.
If this were a practical waveform, the slope at
t = T , 2T , , would be finite, not infinite. A large
negative spike of current over a finite period of time
would result instead of the infinite spike over zero
time. These large spike of current (or voltage) degrade
the performance of many circuits.
______________________________________________________________________________________
Problem 4.16
Solution:
Known quantities:
Inductance value, L
= 16 µH ; the voltage applied to the inductor as a function of time as shown in
Figure P4.16; the initial condition for the current i L (0 ) = 0 .
4.12
G. Rizzoni, Principles and Applications of Electrical Engineering
Find:
The current through the inductor at
Problem solutions, Chapter 4
t = 30 µs .
Analysis:
Since the voltage waveform is piecewise continuous, the integration can be performed over each
continuous segment. Where not indicated t is supposed to be expressed in seconds.
i L (t = 30 µs ) =
1 30 µs
1 20 µ s
1 30 µ s
v L (τ )dτ = i L (0) + ³
v L (τ )dτ + ³ v L (τ )dτ =
³
L −∞
L 0
L 20 µ s
(
20 µ s
)
1 ª3
Vº
1
1
V
30 µ s
3
= i L (0) + « τ 3 2 »
+ [1.2τ nV ]20 µ s = 0 +
⋅ 1 2 ⋅ (20 µs ) − 0 +
L ¬3
L
16 µH
s ¼0
s
1
+
⋅ (1.2 nV ) ⋅ (30 µs − 20 µs ) = 1.250 nA
16 µH
______________________________________________________________________________________
Problem 4.17
Solution:
Known quantities:
Resistance value R = 7 Ω ; inductance value L =
across the components as shown in Figure P4.17.
Find:
The current through each component.
Assumptions:
7 mH ; capacitance value C = 0.5 µF ; the voltage
i R (t ≤ 0) = i L (t ≤ 0) = iC (t ≤ 0 ) = 0
Analysis:
Since the voltage waveform is piecewise continuous, integration and differentiation can only be performed
over each continuous segment. Where not indicated, t is expressed in seconds.
t ≤ 0:
i R (t ) = i L (t ) = iC (t ) = 0
For 0 < t < 5 ms :
v(t ) = mv t + qv
For
where:
mv =
[15 V] − [0] = 3 V
[5 ms] − [0] ms
q vC = 0 V
v(t ) mv ⋅ t
i R (t ) =
=
=
R
R
3
V
⋅t
ms = 428.6 ⋅ t A
7Ω
t
1 t
1 t
1 ª1
1
º
i L (t ) = ³ v(τ )dτ = ³ mv ⋅ τ dτ = « mv ⋅ τ 2 » =
mv ⋅ t 2 =
0
0
L
L
L ¬2
¼ 0 2L
=
1
V 2
⋅3
⋅ t = 214.3 ⋅ 10 3 ⋅ t 2 A
2 ⋅ 7 mH ms
4.13
G. Rizzoni, Principles and Applications of Electrical Engineering
iC (t ) = C
For
t = 5 ms :
Problem solutions, Chapter 4
dvC (t )
d
µAs ·§ V ·
§
= C [mv t ] = Cmv = ¨ 0.5
¸¨ 3
¸ = 1.5 mA
dt
dt
V ¹© ms ¹
©
(
)
(5 ⋅ 10 ) = 214.3 ⋅10
(5 ⋅ 10 ) = 1.5 mA
i R 5 ⋅ 10 −3 = 428.6 ⋅ t A = 428.6 ⋅ 5 ⋅ 10 −3 A = 2.143 A
(
)
2
−3
3
iL
⋅ t 2 A = 214.3 ⋅ 10 3 ⋅ 5 ⋅ 10 −3 A = 5.357 A
−3
iC
For 5 ms < t < 10 ms :
v(t ) = c v = 15 V
v(t ) cv 15 V
i R (t ) =
=
=
= 2.143 A
R
R
7Ω
1 t
1 t
i L (t ) = i L 5 ⋅ 10 −3 + ³ −3 v(τ )dτ = i L 5 ⋅ 10 −3 + ³ −3 cv dτ =
L 5⋅10
L 5⋅10
1
1
t
= i L 5 ⋅ 10 −3 + [cv ⋅ τ ]5⋅10 −3 = i L 5 ⋅ 10 −3 + ⋅ cv ⋅ t − 5 ⋅ 10 −3 =
L
L
1
= 5.357 A +
⋅ 15 V ⋅ t − 5 ⋅ 10 −3 s = −5.357 + 2.143 ⋅ 10 3 ⋅ t A
7 mH
dv (t )
d
iC (t ) = C C = C [cv ] = 0
dt
dt
For t = 10 ms :
i R (0.01) = 2.143 A
i L (0.01) = −5.357 + 2.143 ⋅ 10 3 ⋅ t A = −5.357 + 2.143 ⋅ 10 3 ⋅ 0.01 A = 16.07 A
iC (0.01) = 0
For t > 10 ms :
v(t ) = 0
v(t ) 0
i R (t ) =
= =0
R
R
1 t
1 t
i L (t ) = i L (10 ⋅ 10 −3 ) + ³ − 3 v(τ )dτ = i L (10 ⋅ 10 −3 ) + ³ − 3 0 dτ =
L 10⋅10
L 10⋅10
1 t
= i L (10 ⋅ 10 −3 ) + [0]10⋅10− 3 = i L (10 ⋅ 10 −3 ) = 16.07 A
L
dvC (t )
d
iC (t ) = C
= C [0] = 0
dt
dt
(
)
(
)
(
(
)
)
(
)
4.14
(
)
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
______________________________________________________________________________________
Problem 4.18
Solution:
Known quantities:
The voltage across and the current through an ideal capacitor as shown in Figure P4.18.
Find:
The capacitance of the capacitor.
Analysis:
Considering the period: − 2.5 µs < t < +2.5 µs :
ic = C
dvc
∆v
= C c , since the voltage has a linear waveform.
dt
∆t
Substituting:
12 A = C
[+ 10 V] − [− 10 V]
5 µs
Ÿ C = 12 A ⋅
5 µs
= 3 µF
20 V
______________________________________________________________________________________
Problem 4.19
Solution:
Known quantities:
The voltage across and the current through an ideal inductor as shown in Figure P4.19.
4.15
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Find:
The inductance of the inductor.
Analysis:
vL = L
di L
∆i
= L L , since the current has a linear waveform.
dt
∆t
Substituting:
2V=L
[2 A] − [1 A ]
10 ms − 5 ms
Ÿ L = 2 V⋅
5 ms
= 10 mH
1 A
______________________________________________________________________________________
Problem 4.20
Solution:
Known quantities:
The voltage across and the current through an ideal capacitor as shown in Figure P4.20.
Find:
The capacitance of the capacitor.
Analysis:
Considering the period: 0 < t < 5 ms :
ic = C
dvc
∆v
= C c , since the voltage has a linear waveform.
dt
∆t
Substituting:
1.5 mA = C
[15 V] − [0]
5 ms
Ÿ C = 1.5 mA ⋅
5 ms
= 0.5 µF
15 V
______________________________________________________________________________________
Problem 4.21
Solution:
Known quantities:
The voltage across and the current through an ideal capacitor as shown in Figure P4.21.
Find:
The capacitance of the capacitor.
Analysis:
Considering the period: 0 < t < 5 ms :
ic = C
dvc
∆v
= C c , since the voltage has a linear waveform.
dt
∆t
Substituting:
3 mA = C
[7 V] − [0]
5 ms
Ÿ C = 3 mA ⋅
5 ms
= 2.14 µF
7 V
______________________________________________________________________________________
4.16
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Section 4.2: Time-Dependent Signals
Problem 4.22
Solution:
Known quantities:
The signal x(t ) = 2 cos(ωt ) + 2.5 .
Find:
The average and rms value of the signal.
Analysis:
The average value is:
v(t ) =
ϖ
2π
=
1
2π
2π
ª 2ωπ
º
ω
2π
2π
º
«
» 1 ª
ω + 2.5t ω
(
)
2cos
(
t)dt
+
2
.
5
dt
sin
(
)
ω
ϖ
t
=
−
³
0
0 »
«³
» 2𠫬
¼
0
«¬ 0
»¼
[sin(0) − sin(2π )] + 2.5 = 1 [0 − 0] + 2.5 = 2.5
2π
The rms value is:
x rms
ω
=
2π
2π
ω
ω
2
³0 (2cos (ωt) + 2.5) dt = 2π
ª
ω «
=
⋅ 4⋅
2π «
«¬
=
ω
2π
2π
ω
³ [cos(ωt )]
0
2
2π
ω
³ [4 ⋅ [cos(ωt )]
2
]
+ 10 ⋅ cos(ωt ) + 6.25 dt =
0
2π
ω
2π
ω
º
»
dt + 10 ⋅ ³ sin (ωt )dt + 6.25 ⋅ ³ dt » =
0
0
»¼
2π º
ª 1 2π
⋅ «4 ⋅ ⋅
+ 10 ⋅ 0 + 6.25 ⋅ » = 8.25 = 2.87
ω¼
¬ 2 ω
Note: The integral of a sinusoid over an integer number of period is identically zero. This is a useful and
important result.
______________________________________________________________________________________
Problem 4.23
Solution:
Known quantities:
The sinusoidal voltage
v(t ) of 110 V rms shown in Figure P4.23.
Find:
The average and rms voltage.
Analysis:
The rms value of a sinusoidal is equal to 0.707 times the peak value:
V peak = 110 2
The average value is:
4.17
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
θ
2π
º
1 ª
(
)
+
t
dt
110
2
sin
110 2 sin (t )dt » =
«³
³
2π ¬ 0
2π −θ
¼
θ
2π
1 ª
º=
=
− 110 2 cos(t ) − 110 2 cos(t )
«
»¼
0
2
π
θ
−
2π ¬
v(t ) =
=−
The rms value is:
vrms
50 2
[cos(θ ) − 1 + cos(2π ) − cos(2π − θ )] = 0
π
­° 1
=®
°̄ 2π
(
)
(
)
12
2π
§θ
·½
2
2
¨ ³ 110 2 sin (t ) dt + ³ 110 2 sin (t ) dt ¸ °¾
¨
¸°
2π −θ
©0
¹¿
=
12
2π
θ
­°12100 § 1
· ½°
¸¾
¨ (− cos(t )sin (t ) + t ) + 1 (− cos(t )sin (t ) + t )
=®
¸
2
°̄ 𠨩 2
0
2π −θ ¹ °
¿
=
12
­ 6050
(− cos(θ )sin (θ ) + θ + 2π − (− cos(2π − θ )sin (2π − θ ) + 2π − θ ))½¾
=®
¯ π
¿
12
­ 6050
=®
(2θ )½¾
¯ π
¿
= 110
=
θ
π
______________________________________________________________________________________
Problem 4.24
Solution:
Known quantities:
The sinusoidal voltage
v(t ) of 110 V rms shown in Figure P4.23.
Find:
The angle θ that correspond to delivering exactly one-half of the total available power in the waveform to
a resistive load.
Analysis:
From
vrms = 110
θ
π
2
vrms
= 110 2
, we obtain:
θ 110 2
=
π
2
Ÿ
θ=
π
2
______________________________________________________________________________________
Problem 4.25
Solution:
Known quantities:
The signal v(t ) shown in Figure P4.25.
Find:
The ratio between average and rms value of the signal.
4.18
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Analysis:
The average value is:
v =
0.004
1 ª 0.002
(
− 9 )dt + ³ (1)dt º = 250(− 0.018 + 0.002 ) = −4 V
³
»¼
0.002
0.004 «¬ 0
The rms value is:
8
0.004
1 ª 0.002
2
2
(
− 9) dt + ³ (1) dt º = 250 ⋅ [81 ⋅ 0.002 + 0.004 − 0.002] = 6.40 V
³
»¼
0.002
0.004 «¬ 0
v rms =
Therefor,
v
vrms
=−
4
= −0.625
6.40
______________________________________________________________________________________
Problem 4.26
Solution:
Known quantities:
The signal i (t ) shown in Figure P4.26.
Find:
The power dissipated by a 1-Ω resistor.
Analysis:
The rms value is:
(
)
2
1 p
2
10
⋅
sin
(
t
)
dt =
p ³0
irms =
1 p
100 ⋅ sin 4 (t )dt =
³
0
p
1
3p
⋅ 100 ⋅
= 6.12 A
p
8
Therefore, the power dissipated by a 1-Ω resistor is:
2
P1Ω = Rirms
= (1)(6.12) W = 37.5 W
2
______________________________________________________________________________________
Problem 4.27
Solution:
Known quantities:
The signal x(t ) shown in Figure P4.27.
Find:
The average and rms value of the signal.
Analysis:
The average value is:
V =
1 t0 +τ
t
Vm dt = Vm
³
T t0
T
where t0 is the left-hand side of the pulse.
The rms value is:
4.19
G. Rizzoni, Principles and Applications of Electrical Engineering
Vrms =
Problem solutions, Chapter 4
t
1 t0 +τ 2
Vm dt =
Vm
³
T t0
T
Therefore,
V
Vrms
=
t
T
______________________________________________________________________________________
Problem 4.28
Solution:
Known quantities:
The signal i (t ) shown in Figure P4.28.
Find:
The rms value of the signal.
Analysis:
The rms value is:
irms
3T
º
ª T4
2
2
2
T
4
1« §8 ·
§ 8
·
§8
· »
=
¨ t ¸ dt + ³ ¨ − t + 4 ¸ dt + ³ ¨ t − 8 ¸ dt » =
T « ³0 © T ¹
T
¹
¹ »
3T © T
T©
«
4
4
¼
¬
3T
º
ª T4
2
2
T
4
1 « § 64 2 ·
§ 64 2 64
·
§ 64 2 128
· »
t ¸ dt + ³ ¨ 2 t − t + 16 ¸ dt + ³ ¨ 2 t −
t + 64 ¸ dt » =
=
¨
T « ³0 © T 2 ¹
T
T
¹
¹ »
3T © T
T ©T
«
4
4
¼
¬
=
1 ª1
1
64
º
T + 9T − 18T + 12T − T + 2T − 4T + T − 64T + 64T − 9T + 36T − 48T » =
«
T ¬3
3
3
¼
=
1
T
2
ª4 º
T
=
= 1.15 A
«3 »
3
¬ ¼
______________________________________________________________________________________
Problem 4.29
Solution:
Known quantities:
The signal v(t ) .
Find:
The rms value of the signal.
Analysis:
The rms value is:
vrms
1
=
2π
2π
³ (v(t )) d (ωt )
2
0
4.20
G. Rizzoni, Principles and Applications of Electrical Engineering
v
2
rms
1
=
2π
2π
=
1
2π
2π
=
1
2π
§ 2
· 1 § 2
·
V2
V2
¨¨ VDC [ωt ]02π + 0 + 0 [ωt ]02π + 0 ¸¸ =
¨¨VDC (2π − 0 ) + 0 + 0 (2π − 0 ) + 0 ¸¸
2
2
©
¹ 2π ©
¹
³ (V
+ V0 cos(ωt ))
2
DC
0
1
d (ωt ) =
2π
2π
Problem solutions, Chapter 4
³ (V
2
DC
)
+ 2VDCV0 cos(ωt ) + V02 cos 2 (ωt ) d (ωt ) =
0
§ 2
·
V02 V02
¨
(
)
+
+
+
V
V
V
cos
t
cos 2 (ωt )¸¸ d (ωt ) =
2
ω
³0 ¨© DC DC 0
2
2
¹
2
v rms = VDC
+
V02
=
2
(50 V )2 + 1 (70.7 V )2
2
= 70.7 V
Note:
1. T = period in units of time and ωt = period in angular units, i.e., 2π radians. Considering ωt as a
single variable is useful when dealing with sinusoids.
2. A sinusoid integrated over one or more whole periods gives 0 which is very useful.
______________________________________________________________________________________
4.21
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Section 4.4: Phasor Solution of Circuits with Sinusoidal
Excitation
Focus on Methodology: Phasors
1.
Any sinusoidal signal may be mathematically represented in one of two ways: a time
domain form: v(t) A cos( t
) , and a frequency domain form:
2.
3.
Ae j
j t
. Note the jω in the notation V( j ) , indicating the e
dependence of the phasor. In the remainder of this chapter, bold uppercase quantities
indicate phasor voltages and currents.
A phasor is a complex number, expressed in polar form, consisting of a magnitude equal
to the peak amplitude of the sinusoidal signal and a phase angle equalto the phase shift
of the sinusoidal signal referenced to a cosine signal.
when one is using phaso notation, it is important tonote the specific frequency w of the
sinusoidal signal,since this is not explicitly apparent in the phasor expression.
V( j )
A
Problem 4.30
Solution:
Known quantities:
The current through and the voltage across a component.
Find:
a) Whether the component is a resistor, capacitor, inductor
b) The value of the component in ohms, farads, or henrys.
Analysis:
a) The current and the voltage can be expressed in phasor form:
I = 17∠ − 15 o mA , V = 3.5∠75 o V
Z=
3.5∠75 o V
V
=
= 205.9∠90 o Ω = 0 + j ⋅ 205.9 Ω
o
I 17∠ − 15 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 Ÿ L =
205.9 Ω
Vs
= 327.7 m
= 327.7 mH
rad
A
628.3
s
______________________________________________________________________________________
Problem 4.31
Solution:
Known quantities:
The waveform of a signal shown in Figure P4.31.
Find:
The sinusoidal description of the signal.
Analysis:
From the graph of Figure P4.31:
4.22
G. Rizzoni, Principles and Applications of Electrical Engineering
φ =+
π 180 o
2π
= 60 o , V0 = 170 V , ω = 2πf =
3 π
T
(
Problem solutions, Chapter 4
… not given.
)
v r (t ) = V0 cos(ωt + φ ) = 170 cos ωt + 60 o V
Phasor form:
V = V0 ∠φ = 170∠60 o V = 170 V ⋅ e j 60
o
______________________________________________________________________________________
Problem 4.32
Solution:
Known quantities:
The waveform of a signal shown in Figure P4.32.
Find:
The sinusoidal description of the signal.
Analysis:
From graph:
3π 180 o
2π
rad
= −135 o , I 0 = 8 mA , ω = 2πf =
= 1571
4 π
T
s
rad
§
·
⋅ t − 135 o ¸ mA
i (t ) = I 0 cos(ωt + φ ) = 8 cos¨1571
s
©
¹
φ =−
Phasor form:
I = I 0 ∠φ = 8∠ − 135 o mA = 8 mA ⋅ e − j 135
o
______________________________________________________________________________________
Problem 4.33
Solution:
Known quantities:
The waveform of a signal shown in Figure P4.33.
Find:
The sinusoidal description of the signal.
Analysis:
From graph:
3π 180 o
2π
rad
= −135 o , I 0 = 8 mA , ω = 2πf =
= 1571
4 π
T
s
rad
§
·
i (t ) = I 0 cos(ωt + φ ) = 8 cos¨1571
⋅ t − 135 o ¸ mA
s
©
¹
φ =−
Phasor form:
I = I 0 ∠φ = 8∠ − 135 o mA = 8 mA ⋅ e − j 135
o
______________________________________________________________________________________
4.23
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.34
Solution:
Known quantities:
The current through
(
)
i (t ) = I 0 cos ωt + 45o , I 0 = 3 mA, ω = 6.283
v(t ) = V0 cos(ωt ), V0 = 700 mV , ω = 6.283
rad
, and the voltage across
s
rad
an electrical component.
s
Find:
a) Whether the component is inductive or capacitive.
b) The waveform of the instantaneous power p (t ) as a function of ωt over the range 0 < ωt < 2π .
c) The average power dissipated as heat in the component.
d) The same as b. and c. with the phase of the current equal to zero.
Analysis:
a) Phasor notation:
I = 3∠45o mA , V = 700∠0 o mV
Z=
V 700∠0o mV
=
= 233.3∠ − 45o Ω = 165.0 − j165.0 Ω
I
3∠90o mA
The component is inductive because it is lagging.
b)
(
)
( (
)
( ))
1
p(t ) = v(t )i (t ) = V0 I 0 cos ωt + 45o cos(ωt ) = V0 I 0 cos 2ωt + 45o + cos 45o =
2
1
= (700 mV )(3 mA ) cos 2ωt + 45o + 0.707 = 1050 cos 2ωt + 45o + 742.4 µW
2
( (
)
) (
(
c)
P=
1
ωT
ωT
³ p(t )dωt =
0
1
ωT
ωT
³ (1050 µW cos(2ωt + 45 ) + 742.4)dωt =
0
[ (
)]
1
(1050 µW ) 1 sin 2ωt + 45o 02π + 1 742.4(ωt ) 02π =
ωT
ωT
2
1
=
(1050 µW ) 1 sin 765o − sin (45°) + 742.4 =
ωT
2
1
1
(1050 µW ) [0 − 0] + 742.4 = 742.4µW
=
2π
2
=
[ (
d)
o
]
)
(
( ))
1
1
p(t ) = v(t )i (t ) = V0 I 0 cos(ωt ) cos(ωt ) = V0 I 0 cos(2ωt ) + cos 0o =
2
2
1
= (700 mV )(3 mA )(cos(2ωt ) + 1) = 1050(cos(2ωt ) + 1) µW
2
4.24
)
)
G. Rizzoni, Principles and Applications of Electrical Engineering
1
P=
ωT
ωt
³
0
1
p(t )dωt =
ωT
Problem solutions, Chapter 4
ωt
³ (1050 µW(cos(2ωt ) + 1))dωt =
0
=
1
(1050 µW )§¨ 1 [sin (2ωt )]02π + [ωt ]02π ·¸ =
ωT
©2
¹
=
1
(1050 µW )§¨ 1 (0 − 0) + (2π − 0)·¸ = 1050 µW
2π
©2
¹
______________________________________________________________________________________
Problem 4.35
Solution:
Known quantities:
The values of the impedance,
R1 = 2.3 kΩ , R2 = 1.1 kΩ , L = 190 mH , C = 55 nF and the
o
voltage applied to the circuit shown in Figure P4.35, v s (t ) = 7 cos (3000t + 30 ) V .
Find:
The equivalent impedance of the circuit.
Analysis:
rad ·
§
X L = ωL = ¨ 3 k
¸(190 mH ) = 0.57 kΩ Ÿ Z L = + j ⋅ X L = + j ⋅ 0.57 kΩ
s ¹
©
1
1
= 6.061 kΩ Ÿ Z C = − j ⋅ X C = − j ⋅ 6.061 kΩ
=
rad ·
ωC §
¨3 k
¸(55 nF)
s ¹
©
= Z R1 + Z L = R1 + jX L = 2.3 + j ⋅ 0.57 kΩ = 2.37∠13.92 o kΩ
XC =
Z eq1
Z eq 2 = Z R1 + Z C = R1 − jX C = 1.1 − j ⋅ 6.061 kΩ = 6.16∠ − 79.71o kΩ
Z eq =
Z eq1 ⋅ Z eq 2
Z eq1 + Z eq 2
=
(2.37∠13.92
)(
)
kΩ ⋅ 6.16∠ − 79.71o kΩ
=
(2.3 + j ⋅ 0.57 kΩ ) + (1.1 − j ⋅ 6.061 kΩ )
o
14.60∠ − 65.79o kΩ 2 14.60∠ − 65.79o kΩ 2
=
=
= 2.261∠ − 7.56o kΩ
o
3.4 − j ⋅ 5.491 kΩ
6.458∠ − 58.23 kΩ
______________________________________________________________________________________
Problem 4.36
Solution:
Known quantities:
The values of the impedance,
R1 = 3.3 kΩ , R2 = 22 kΩ , L = 1.90 H , C = 6.8 nF and the voltage
o
applied to the circuit shown in Figure P4.35, v s (t ) = 636 cos (3000t + 15 ) V .
Find:
The equivalent impedance of the circuit.
4.25
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Analysis:
rad ·
§
X L = ωL = ¨ 3 k
¸(1.90 H ) = 5.7 kΩ Ÿ Z L = + j ⋅ X L = + j ⋅ 5.7 kΩ
s ¹
©
1
1
= 49.02 kΩ Ÿ Z C = − j ⋅ X C = − j ⋅ 49.02 kΩ
=
rad ·
ωC §
¸(6.8 nF)
¨3 k
s ¹
©
= Z R1 + Z L = R1 + jX L = 3.3 + j ⋅ 5.7 kΩ = 6.59∠59.93o kΩ
XC =
Z eq1
Z eq 2 = Z R1 + Z C = R1 − jX C = 22 − j ⋅ 49.02 kΩ = 53.73∠ − 65.83o kΩ
Z eq =
Z eq1 ⋅ Z eq 2
Z eq1 + Z eq 2
=
(6.59∠59.93
)(
)
kΩ ⋅ 53.73∠ − 65.83o kΩ
=
(3.3 + j ⋅ 5.7 kΩ ) + (22 − j ⋅ 49.02 kΩ )
o
354.08∠ − 5.9o kΩ 2 354.08∠ − 5.9o kΩ 2
=
=
= 7.05∠53.81o kΩ
o
25.3 − j ⋅ 43.32 kΩ
50.17∠ − 59.71 kΩ
______________________________________________________________________________________
Problem 4.37
Solution:
Known quantities:
The current in the circuit,
(
)
i s (t ) = I 0 cos ωt + 30 o , I 0 = 13 mA , ω = 1000
the capacitance present in the circuit shown in Figure P4.37 C
= 0.5 µF .
rad
, and the value of
s
Find:
a) The phasor notation for the source current.
b) The impedance of the capacitor.
c) The voltage across the capacitor, showing all the passages and using phasor notation only.
Analysis:
a) Phasor notation:
I s = I 0 ∠φ = 13∠30 o mA
b)
Z C = − jX C − j
c)
(
1
1
= 0 − j 2 kΩ = 2∠ − 90 o kΩ
=−j
rad ·
ωC
§
¨1000
¸(0.5 µF )
s ¹
©
)(
)
VC = I s ⋅ Z C = 13∠30 o mA ⋅ 2∠ − 90 o kΩ = 26∠ − 60 o V
(
)
vC (t ) = 26 cos 1000t − 60 o V
Note that conversion from phasor notation to time notation or vice versa can be done at any time.
______________________________________________________________________________________
4.26
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.38
Solution:
Known quantities:
The values of two currents in the circuit shown in Figure P4.38:
(
)
i2 (t ) = 50 cos ωt + 53.13o mA , ω = 377
Find:
The current
(
)
i1 (t ) = 141.4 cos ωt + 135 o mA ,
rad
.
s
i3 (t ) .
Analysis:
A solution using trigonometric identities is possible but inefficient, cumbersome, and takes a lot of time.
Phasors are better! Note that one current is described with a sine and the other with a cosine function.
When using phasors, all currents and voltages must be described with either sine functions or cosine
functions. Which does not matter, but it is a good idea to adopt one and use it consistently. Therefore, first
converts to cosines.
KCL:
− i1 (t ) + i2 (t ) + i3 (t ) = 0 Ÿ + i3 (t ) = i1 (t ) − i 2 (t )
(
)
(
)
= 141.4 cos(ωt + 135 ) mA − 50 cos(ωt − 53.13 − 90 ) mA
i3 (t ) = 141.4 cos ωt + 135 o mA − 50 sin ωt − 53.13o mA =
o
o
o
I 3 = 141.4 mA ∠135 o − 50 mA ∠ − 143.13o =
= (− 99.98 + j ⋅ 99.98) mA − (− 40.00 − j ⋅ 30.00) mA =
= (− 59.98 + j ⋅ 129.98) mA = 143.2 mA ∠114.8 o
(
)
i3 (t ) = 143.2 cos ωt + 114.8 o mA
If sine functions were used, the result in phasor notation would differ in phase by 90 degrees.
______________________________________________________________________________________
Problem 4.39
Solution:
Known quantities:
Z 1 = 5.9∠7 o kΩ , Z 2 = 2.3∠0 o Ω , Z 3 = 17∠11o Ω and the
voltages applied to the circuit shown in Figure P4.39, v s1 (t ) = v s 2 (t ) = 170 cos (377t ) V .
The values of the impedance,
Find:
The current through
Z3 .
Analysis:
Vs1 = Vs 2 = 170∠0 o V = (170 + j 0 ) V
KVL:
− Vs1 − Vs 2 + I 3 Z 3 = 0
I3 =
Vs1 + Vs 2 170∠0 o V + 170∠0 o V 340∠0 o V
=
=
= 20∠ − 11o A
Z3
17∠11o Ω
17∠11o Ω
4.27
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
rad
§
·
i3 (t ) = 20 cos¨ 377
⋅ t − 11o ¸ A
s
©
¹
Note also:
KVL:
− Vs1 + I 1 Z 1 = 0 Ÿ I 1 =
Vs1
Vs 2
, − Vs 2 − I 2 Z 2 = 0 Ÿ I 2 = −
Z1
Z2
______________________________________________________________________________________
Problem 4.40
Solution:
Known quantities:
The values of the impedance in the circuit shown in Figure P4.40,
Z s = (13000 + jω 3) Ω ,
R = 120 Ω , L = 19 mH , C = 220 pF .
Find:
The frequency such that the current
I i and the voltage V0 are in phase.
Analysis:
Z s is not a factor in this solution. Only R, L, and C will determine if the voltage across this combination
is in phase with the current through it. If the voltage and current are in phase, then, the equivalent
impedance must have an "imaginary" or reactive part which is zero!
V0
= Z eq ∠0 o = Req + jX eq , X eq (ω ) = 0
Ii
(Z + Z L ) ⋅ Z C = (R + jX L ) ⋅ (− jX C ) = X L X C − jRX C R − j ( X L − X C ) =
Z eq = R
Z R + Z L + ZC
R + jX L − jX C
R + j( X L − X C ) R − j( X L − X C )
Z eq =
X L X C R − RX C ( X L − X C )] − j [R 2 X C + X L X C ( X L − X C )]
[
=
2
R 2 + (X L − X C )
At the resonant frequency the reactive component of this impedance must equal zero:
R2 X C + X L X C (X L − X C )
X eq (ω ) =
= 0 Ÿ R 2 + X L (X L − X C ) = 0
2
2
R + (X L − X C )
1 ·
L
§
2 2
2
R 2 + ωL¨ ωL −
¸=0 Ÿ ω L = −R
ωC ¹
C
©
ω=
1
R2
− 2 =
LC L
= 293.3 G
(120 Ω)2
(19 mH )(220 pF) (19 mH )2
1
−
=
rad
rad
rad
− 39.89 M
= 489.1 k
s
s
s
Notes:
1. To separate the equivalent impedance into real (resistive) and "imaginary" (reactive) components, the
denominator had to be "rationalized". This was done by multiplying numerator and denominator by the
complex conjugate of the denominator, and multiplying term by term. Remember that j = −1 , etc.
2. The term with R had a negligible effect on the resonant frequency in this case. If R is sufficiently
large, however, it will significantly affect the answer.
______________________________________________________________________________________
2
4.28
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.41
Solution:
Known quantities:
The circuit shown in Figure P4.40
Find:
The frequency such that the current
I i and the voltage V0 are in phase.
Analysis:
If the voltage and current are in phase, then, the equivalent impedance must have an "imaginary" or
reactive part which is zero!
V0
= Z eq ∠0 o = Req + jX eq , X eq (ω ) = 0
Ii
(Z + Z L ) ⋅ Z C = (R + jX L ) ⋅ (− jX C ) = X L X C − jRX C R − j ( X L − X C ) =
Z eq = R
Z R + Z L + ZC
R + jX L − jX C
R + j( X L − X C ) R − j( X L − X C )
Z eq =
=
[X L X C R − RX C ( X L − X C )] − j[R 2 X C + X L X C ( X L − X C )]
2
R 2 + (X L − X C )
At the resonant frequency the reactive component of this impedance must equal zero:
X eq (ω ) =
R2 X C + X L X C (X L − X C )
= 0 Ÿ R 2 + X L (X L − X C ) = 0
2
2
R + (X L − X C )
L
1 ·
§
2 2
2
R 2 + ωL¨ ωL −
¸=0 Ÿ ω L = −R
C
ω
C
©
¹
ω=
1
R2
− 2
LC L
______________________________________________________________________________________
Problem 4.42
Solution:
Known quantities:
The values of the impedance,
Rs = 50 Ω , Rc = 40 Ω , L = 20 µH , C = 1.25 nF , and the voltage
rad
o
applied to the circuit shown in Figure P4.42, v s (t ) = V0 cos (ωt + 0 ), V0 = 10 V , ω = 6 M
.
s
Find:
The current supplied by the source.
Analysis:
Assume clockwise currents:
rad ·
§
o
X L = ωL = ¨ 6 M
¸(20 µH ) = 1203 Ω Ÿ Z L = 0 + j120 Ω = 120∠90 Ω
s ¹
©
XC =
1
1
=
= 133.3 Ω Ÿ Z C = 0 − j133.3 Ω = 133.3∠ − 90 o Ω
rad ·
ωC §
¨6 M
¸(1.25 nF)
s ¹
©
4.29
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Z Rc = 40 − j Ω = 40∠0 o Ω , Z Rs = 50 − j Ω = 50∠0 o Ω
Equivalent impedances:
Z eq1 = Z Rc + Z L = 40 + j120 Ω = 126.5∠71.56 o Ω
Z eq = Z Rs +
Z C ⋅ Z eq1
Z C + Z eq1
= 50 + j 0 Ω +
OL:
Is =
(133.3∠ − 90 Ω)⋅ (126.5∠71.56 Ω) =
= 50 + j 0 Ω +
o
o
133.3∠ − 90 o Ω + 126.5∠71.56 o Ω
16.87∠ − 18.44 o kΩ2
= 50∠0 o Ω + 400∠0 o Ω = 450∠0 o Ω
o
42.161∠ − 18.44 Ω
Vs
10∠0 o V
=
= 22.22∠0 o mA Ÿ i s (t ) = 22.22 cos ωt + 0 o mA
Z eq 450∠0 o Ω
(
)
Note:
The equivalent impedance of the parallel combination is purely resistive; therefore, the frequency given is
the resonant frequency of this network.
______________________________________________________________________________________
Problem 4.43
Solution:
Known quantities:
The values of the impedance and the voltage applied to the circuit shown in Figure P4.43.
Find:
The current in the circuit.
Analysis:
Assume clockwise currents:
rad
o
, VS = 12∠0 V
s
1
ZC =
= − j Ω , Z L = jωL = j 9 Ω Ÿ Z total = 3 + j 9 − j = 3 + j8 Ω
jω C
12
I=
= 0.4932 − j1.3151 A = 1.4045∠ − 69.44 o A ,
3 + j8
i(t ) = 1.4 cos ωt − 69.4 o A
ω =3
(
)
______________________________________________________________________________________
Problem 4.44
Solution:
Known quantities:
The values of the impedance and the current source shown in Figure P4.44.
Find:
The voltage.
Analysis:
Assume clockwise currents:
ω=2
rad
1
o
, I S = 10∠0 A , Z L = jωL = j 6 Ω , Z C =
= − j 1.5 Ω
jωC
s
4.30
G. Rizzoni, Principles and Applications of Electrical Engineering
Z eq =
Problem solutions, Chapter 4
1
1
1
=
=
= 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
V = I S Z eq = 10 A ⋅ (0.9231 − j1.3846) Ω = 9.231 − j13.846 a *10 V = 16.641∠ − 56.31o V
______________________________________________________________________________________
Problem 4.45
Solution:
Known quantities:
The values of the impedance and the current source for the circuit shown in Figure P4.45.
Find:
The current I1.
Analysis:
Specifying the positive directions of the currents as in figure P4.45:
Z eq =
1
1 § 1 ·
¸
+¨
2 ¨© − j 4 ¸¹
= 1.79∠26.56 o Ω
(
) (
)
VS = I S Z eq = 10∠ − 22.5o A ⋅ 1.79∠26.56 o Ω = 17.9∠4.06 o V
I1 =
VS
= 8.95∠4.06 o A
R
______________________________________________________________________________________
Problem 4.46
Solution:
Known quantities:
The values of the impedance and the voltage source for circuit shown in Figure P4.46.
Find:
The voltage V2.
Analysis:
Specifying the positive directions as in figure P4.46:
Z L = jωL = j12 Ω
V2 =
R12Ω
R6Ω
6Ω
150∠0 o Ω
V = 6.93∠ − 33.7 o V
V=
25∠0 o V =
(12 + j12 + 6) Ω
+ Z L + R6Ω
18 + j12 Ω
______________________________________________________________________________________
Problem 4.47
Solution:
Known quantities:
The values of the impedance and the current source of circuit shown in Figure P4.44.
Find:
The value of ω for which the current through the resistor is maximum.
4.31
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Analysis:
Assume clockwise currents:
I S = 10∠0 o A , Z L = jωL = j 3ω Ω , Z C =
1
3
=−j Ω
jωC
ω
1
1
10ω
3
R
IS =
10∠0 o =
IR =
ω
1 1
1
1
1
ω + j ω 2 −1
+
+
−j
+j
R Z L ZC
3
3ω
3
(
)
The maximum of I R is obtained for ω = 1. , therefore iR (t ) = 10 cos(t ).
______________________________________________________________________________________
Problem 4.48
Solution:
Known quantities:
The values of the impedance and the current source for circuit shown in Figure P4.48.
Find:
The current through the resistor.
Analysis:
Specifying the positive directions as in figure P4.48:
By current division:
IR = −
1
R
1 1
+
R ZC
⋅ Is = −
1
1
1 − jωRC
⋅ Is = −
⋅ Is = −
⋅ Is =
2
R
1+jωRC
(
)
1
+
RC
ω
1+
jωC
= −(0.0247 − j 0.1552) A ⋅ 1∠0 o A = 157 ⋅ 10 −3 ∠99.04 o A
i R (t ) = 157 cos (200πt + 99.04 o ) mA
______________________________________________________________________________________
4.32
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.49
Solution:
Known quantities:
The values of the reactance
X L = 1 kΩ , X C = 10 kΩ , and the current source I = 10∠45o mA for
circuit shown in Figure P4.49.
Find:
The voltage vout .
Analysis:
Specifying the positive directions of the currents as in figure P4.49:
Vout = Z eq I = (Z L + Z C )I = (0 + jX L + 0 − jX C )I = ( j1 kΩ − j10 kΩ ) ⋅10∠45o mA =
= (− j 9 kΩ ) ⋅10∠45o mA = 9∠ − 90o kΩ ⋅10∠45o mA = 90∠ − 45o V
vout = 90 cos(ωt − 45°) V
______________________________________________________________________________________
Problem 4.50
Solution:
Known quantities:
The circuit shown in Figure P4.50, the values of the resistance,
inductance,
L = 1 4 H , and the frequency ω = 4
Find:
The impedance
Analysis:
R = 2 Ω , capacitance, C = 1 8 F ,
rad
.
s
Z.
1
1
1
1
Z L = jω L = j 4 Ω = j Ω , Z C =
=−j
=−j
= − j2 Ω
jω C
ωC
4
4 ⋅ (1 8)
( j 2) (− 1 − j )
1
1
j2
= j+
= j+
= j+
Z = Z L + ZC R = Z L +
1
1
1
1
(− 1 + j ) (− 1 − j )
−1+ j
+
+
ZC R
− j2 2
= j+
j 2(− 1 − j )
= j − j +1 = 1Ω
1+1
______________________________________________________________________________________
Problem 4.51
Solution:
Known quantities:
Circuit shown in Figure P4.51, the values of the resistance,
inductance,
L = 4 5 H , and the frequency ω = 5
rad
.
s
4.33
R = 3 Ω , capacitance, C = 1 10 F ,
G. Rizzoni, Principles and Applications of Electrical Engineering
Find:
The admittance
Analysis:
Problem solutions, Chapter 4
Y.
1
1
4
=
= − j 2 Ω , Z L = j5 ⋅ = j 4 Ω
jωC j 5 ⋅ (1 10 )
5
1
1
1
1
1
1
1
=
=
+
=
+
Y= =
1
Z Z C (R + Z L )
ZC R + Z L − j2 3 + j4
1
1
+
ZC R + Z L
ZC =
(1) (3 − j 4) = j 1 + 3 − j 4
1
= j +
2 (3 + j 4 ) (3 − j 4 )
2 9 + 16
1 3 − j4
1 3
4
= j +
= j +
−j
= 0.12 + j 0.34 S
2
25
2 25
25
______________________________________________________________________________________
4.34
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Section 4.5: AC Circuit Analysis methods
Focus on Methodology: AC Circuit Analysis
1.
2.
3.
4.
5.
Identify the sinusoidal source(s) and note the excitation frequency.
Convert the source(s) tophasor form.
Represent each circuit element by ots impedance.
Splve the resulting phasor circuit, using appropriate circuit analysis tools.
Convert the (phasor-form) answer to its time-domain equivalent, using equation 4.50.
Problem 4.52
Solution:
Known quantities:
Circuit shown in Figure P4.52, the values of the resistance,
R = 9 Ω , capacitance, C = 1 18 F ,
π·
§
inductance, L1 = 3 H , L2 = 3 H , L3 = 3 H , and the voltage source v s (t ) = 36 cos¨ 3t − ¸ V .
3¹
©
Find:
The voltage across the capacitance
tehcniques.
Analysis:
v using phasor
rad
, Vs = 36∠ − 60° V
s
= jωL2 = j 3 ⋅ 3 = j 9 Ω
ω =3
Z L2
1
1
=
= − j6 Ω
jωC j 3 ⋅ (1 18)
Z L3 = jωL3 = j 3 ⋅ 3 = j 9 Ω
ZC =
Z eq =
(
1
Z L3 Z L2 + Z C
)
=
1
1
1
+
Z L3
Z L2 + Z C
(
=
)
1
1
1
+
j9 j3
=
j9
= 2.25∠90 o Ω
4
Z T = Z R + Z L1 + Z eq = 9 + j 3 ⋅ 3 + j 2.25 = 9 + j11.25 = 14.407∠51.34 o Ω
I=
VS
36∠ − 60o V
=
= 2.499∠ − 111.34o A
Z T 14.407∠51.340 Ω
Veq = IZ eq = (2.499∠ − 111.34o )(2.25∠90o ) = 5.623∠ − 21.34o V
V=
− j6
ZC
5.623∠ − 21.34o = 11.25∠158.66o V
Veq =
j3
Z L2 + Z C
(
(
)
)
v = 11.25 cos 3t − 158.66o V
______________________________________________________________________________________
4.35
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.53
Solution:
Known quantities:
Circuit shown in Figure P4.53, the values of the resistance,
R = 5 Ω , capacitance, C = 1 2 F ,
inductance, L1 = 0.5 H , L2 = 1 H , L2 = 10 H , and the current source i s (t ) = 6 cos(2t ) A .
Find:
The current through the inductance iL2 .
Analysis:
rad
, Z L2 = jωL2 = j 2 Ω
s
1
ZC =
= − j Ω , Z L3 = jωL3 = j 20 Ω
jω C
Z L3 + Z C
j 20 − j
I=
IS =
6∠0 o
Z L3 + Z C + R + Z L2
( j 20 − j ) + (5 + j 2)
ω=2
(
) (
)
j19
6∠0 o = 5.28∠13.4 o A
5 + j 21
i = 5.28 cos 2t + 13.4 o A
=
(
)
______________________________________________________________________________________
Problem 4.54
Solution:
Known quantities:
Circuit shown in Figure P4.54 the values of the resistance,
inductance,
RS = RL = 500 Ω , R = 1 kΩ and the
L = 10 mH .
Find:
a) The Thèvenin equivalent circuit if the voltage applied to the circuit is
vS (t ) = 10 cos(1,000t ) .
b) The Thèvenin equivalent circuit if the voltage applied to the circuit is vS (t ) = 10 cos(1,000 ,000t ) .
Analysis:
a)
Z L = jωL = j1000
rad
⋅ 10 mH = j10 Ω ,
s
The equivalent impedance is:
ZT =
ZL ⋅ R
( j10)1000 + 500 = 500 + j103 = 500.1 + j9.999 Ω
+ RS =
(Z L + R )
100 + j
j10 + 1000
The equivalent Thèvenin voltage is:
VT = VS = 10∠0 o V
6 rad
⋅ 10 mH = j10 4 Ω ,
b) Z L = jωL = j10
s
The equivalent impedance is:
4.36
G. Rizzoni, Principles and Applications of Electrical Engineering
ZT =
Problem solutions, Chapter 4
ZL ⋅ R
( j10 4 )1000 + 500 = 500 + j10 4 = 1490.1 + j99.01 Ω
+ RS =
(Z L + R )
1 + j10
j10 4 + 1000
The equivalent Thèvenin voltage is:
VT = VS = 10∠0 o V
______________________________________________________________________________________
Problem 4.55
Solution:
Circuit shown in Figure P4.55 the values of the impedance,
the voltage source
vin (t ) = 12 cos(10t ) V .
L = 0.1 H , capacitance, C = 100 µF , and
Find:
The Thèvenin equivalent of the circuit as seen by the load resistor
RL .
Analysis:
1
= − j1000 Ω
rad
⋅ 100 µF
j10
s
rad
Z L = jωL = j10
⋅ 0.1 H = j1 Ω
s
ZC =
1
=
jω C
The equivalent impedance is:
ZT = Z L Z C =
Z L ⋅ ZC
j (− j1000 ) 1000
=
=
= 1.001∠90 o Ω = j1.001 Ω
Z L + ZC
j − j1000 − j 999
The Thèvenin voltage is:
VT =
ZC
− j1000
1000
Vin =
⋅ 12∠0 o =
⋅ 12∠0 o = 12.012∠0 o V
Z L + ZC
j − j1000
999
______________________________________________________________________________________
Problem 4.56
Solution:
Known quantities:
Circuit shown in Figure P4.56 the values of the resistance,
R1 = 4 Ω , R2 = 4 Ω , capacitance,
C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V .
Find:
The current in the circuit
i L (t ) using phasor techniques.
Analysis:
VS (t ) = 2∠0 o V
1
1
=
= − j2 Ω
ZC =
1
jωC
j2
4
Z L = jω L = j 2 ⋅ 2 = j 4 Ω
Applying the voltage divider rule:
4.37
G. Rizzoni, Principles and Applications of Electrical Engineering
(Z L || (Z C + Z 2 )) V
S
Z1 + (Z L || (Z C + Z 2 ))
VL =
=
Problem solutions, Chapter 4
4∠36.8°
2∠0o = 1.05∠18.4° V
4∠0° + 4∠36.8°
Therefore, the current is:
IL =
VL 1.05∠18.4°
=
= 0.2635∠ − 71.6o A
ZL
4∠90°
(
)
iL (t ) = 0.2635 cos 2t − 71.6o A
______________________________________________________________________________________
Problem 4.57
Solution:
Known quantities:
Circuit shown in Figure P4.57, the values of the resistance,
R1 = 75 Ω , R2 = 100 Ω , capacitance,
C = 1 µF , inductance, L = 0.5 H , and the voltage source v s (t ) = 15 cos(1,500t ) V .
Find:
The currents in the circuit
i1 (t ) and i2 (t ) .
Analysis:
In the phasor domain:
ZC =
-j
2000
=−j
= − j 666.7 Ω , Z L = j (1500)(0.5) = j 750 Ω
-6
3
1500( 1 × 10 )
By applying KVL in the first loop, we have
VS = R1 I 1 + Z C (I1 − I 2 )
By applying KVL in the second loop, we have
0 = (Z C )(I 2 − I 1 ) + (Z L + R2 )I 2
That is:
­
2000 ·
2000
§
o
°15∠0 = ¨ 75 − j 3 ¸ I1 + j 3 I 2
°
©
¹
®
°0 = j 2000 I + §¨100 + j 250 ·¸ I
1
2
°¯
3
3 ¹
©
By solving above equations, we have
I1 = 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 + 46.6 o mA
i2 (t ) = 19.6 cos 1,500t − 83.2 o mA
(
(
)
)
______________________________________________________________________________________
Problem 4.58
Solution:
Known quantities:
Circuit shown in Figure P4.58, the values of the resistance,
R1 = 40 Ω , R2 = 10 Ω ,
capacitance, C = 500 µF , inductance, L = 0.2 H , and the current source is (t ) = 40 cos(100t ) A.
4.38
G. Rizzoni, Principles and Applications of Electrical Engineering
Find:
The voltages in the circuit
Problem solutions, Chapter 4
v1 (t ) and v 2 (t ) .
Analysis:
ZC =
1
-j
=
= -j20 Ω , Z L = jωL = j100 ⋅ 0.2 = j20 Ω
jωC 100 ⋅ 500 ⋅ 10 - 6
Applying KCL at node 1, we have:
IS =
§1
V1 V1 − V2
1 ·
1
j ·
j
§ 1
¸¸V1 −
V2 Ÿ 40∠0o = ¨ + ¸V1 − V2
Ÿ I S = ¨¨ +
+
R1
ZC
ZC
20
© 40 20 ¹
© R1 Z C ¹
Applying KCL at node 2, we have
§ 1
V1 − V2 V2 V2
V
1
1
=
+
Ÿ 1 = ¨¨ +
+
ZC
R2 Z L
Z C © R2 Z L Z C
Therefore:
·
V
1
1
1
¸¸V2 Ÿ j 1 = §¨ − j + j ·¸V2
20 © 10
20
20 ¹
¹
­
j ·
j
§ 1
o
­
j ·
j
§ 1
o
°40∠0 = ¨ 40 + 20 ¸V1 − 20 V2
©
¹
°40∠0 = ¨ + ¸(− j 2V2 ) − V2
°
Ÿ
Ÿ ®
20
© 40 20 ¹
®
°V = − j 2V
° j V1 = §¨ 1 ·¸V
2
¯ 1
°¯ 20 © 10 ¹ 2
­
j
1
j
j·
§1
o
°40∠0 = − V2 + V2 − V2 = ¨ − ¸V2
20
10
20
© 10 10 ¹
®
°V = − j 2V
¯ 1
2
40∠0o
= 282.84∠45o V, V1 = − j 2V2 = 565.68∠ − 45o V
j·
§1
¨ − ¸
© 10 10 ¹
v2 (t ) = 282.84 cos 100t + 45o V, v1 (t ) = 568.68 cos 100t − 45o V
V2 =
(
)
(
)
______________________________________________________________________________________
Problem 4.59
Solution:
Known quantities:
The circuit called Wheatstone bridge shown in Figure P4.59.
a) The balanced status for the bridge: v ab = 0 .
R1 = 100 Ω , R2 = 1 Ω , the capacitance, C 3 = 4.7 µF , the
inductance, L3 = 0.098 H , that are necessary to balance the bridge: v ab = 0 , and the voltage
applied to the bridge, v s = 24 sin(2 ,000t ) V .
b) The values of the resistance,
Find:
a) The unknown reactance
X 4 in terms of the circuit elements.
b) The value of the unknown reactance X 4 .
c) The source frequency that should be avoided in this circuit.
Analysis:
a) Assuming a balanced circuit, we have v ab = 0 , that is, v a
4.39
= vb
G. Rizzoni, Principles and Applications of Electrical Engineering
From the voltage divider:
jX L3
Problem solutions, Chapter 4
R2
jX 4
=
Ÿ
- jX C3 + R2 R1 + jX 4
R2
jX 4
=
j
R1 + jX 4
jωL3 + R2
ωC 3
Inverting both sides and equating imaginary parts:
b)
§
1 ·
R1R2
¸¸ X 4 Ÿ X 4 =
R1R2 = ¨¨ − ωL3 +
ωC3 ¹
§ 1
·
©
¨¨
− ωL3 ¸¸
© ωC3
¹
X4 =
100 ⋅ 1
1
§
·
- 2000 ⋅ 0.098 ¸
¨
−6
© 2000 ⋅ 4.7 ⋅ 10
¹
= −1.116 Ω
Negative reactance implies that the component is a capacitor.
1
1
= 1.116 Ω Ÿ C =
= 448 µF
ωC
ω ⋅ 1.116
c)
If the reactances of L3 and C3 cancel, the bridge can not measure X4. Thus, the condition to be avoided is:
ωL 3
−
1
1
= 0 Ÿ L3C 3 = 2 Ÿ ω =
ωC 3
ω
1
L3 C 3
=
1
0.098 ⋅ 4.7 ⋅ 10 − 6
= 1473
rad
s
f = 234.5 Hz
______________________________________________________________________________________
Problem 4.60
Solution:
Known quantities:
Circuit shown in Figure P4.56, the values of the resistance,
R1 = 4 Ω , R2 = 4 Ω , capacitance,
C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V .
Find:
The Thévenin impedance seen by resistor
R2 .
Analysis:
Z T = (R1 Z L ) + (Z C ) = (4 j 4 ) + (-j 2 ) = j 2(1-j )-j 2 = 2 + j 2 + (-j 2) = 2 Ω
______________________________________________________________________________________
Problem 4.61
Solution:
Known quantities:
Circuit shown in Figure P4.58, the values of the resistance,
R1 = 10 Ω , R2 = 40 Ω , capacitance,
C = 500 µF , inductance, L = 0.2 H , and the current source i s (t ) = 40 cos(100t ) A .
Find:
The Thévenin voltage seen by inductance L .
Analysis:
The Thévenin equivalent voltage source is the open-circuit voltage at the load terminals:
4.40
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
VT = R2 I 2 = 40I 2
From the current division, we have
I2 =
R1
10
IS =
40∠0 o = 7.43∠21.8 o A
(R2 + Z C ) + R1
(40-j 20) + 10
VT = R2 I 2 = 40 ⋅ 7.43∠21.8 o = 297∠21.8 o V
vT (t ) = 297 cos 100t + 21.8 o V
(
)
______________________________________________________________________________________
Problem 4.62
Solution:
Known quantities:
Circuit shown in Figure P4.62, the values of the impedance,
the voltage source
R = 8 Ω , Z C = − j8 Ω , Z L = j8 Ω , and
Vs = 5∠ − 30 V .
o
Find:
The Thévenin equivalent circuit seen from the terminals a-b.
Analysis:
The Thévenin equivalent circuit is given by:
§ 8 + j8 ·
¸¸5∠ − 30° = (1 + j )5∠ − 30 o = 7.07∠15 V
VTH = ¨¨
+
−
8
j
8
j
8
©
¹
(8 + j8)(− j8) = (8 − j8) = 8 2∠ − 45o Ω
Z TH =
8 + j8 − j 8
______________________________________________________________________________________
Problem 4.63
Solution:
Known quantities:
Circuit shown in Figure P4.56, the values of the resistance,
R1 = 4 Ω , R2 = 4 Ω , capacitance,
C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V .
Find:
The Thévenin equivalent voltage seen by the resistor
R2 .
Analysis:
The Thévenin equivalent circuit is given by:
j4
2∠0 o = (1 + j ) = 2∠45 o = 1.414∠45 o V
4+j 4
vT (t ) = 1.414 cos 2t + 45 o V
VT =
(
)
______________________________________________________________________________________
4.41
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
Problem 4.64
Solution:
Known quantities:
Circuit shown in Figure P4.56, the values of the resistance,
R1 = 4 Ω , R2 = 4 Ω , capacitance,
C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V .
Find:
The Norton equivalent circuit seen by the resistor
Analysis:
From the result of Problem 4.60, we have
IN =
R2 .
Z T = 2 Ω . From the current divider:
j4
I = 2I
j 4-j 2
and
j4 − j2 =
(− j 2)( j 4) = − j 4
j2
The current is:
I=
2∠0 o
2
=
∠45 o = 0.353∠45 o A
4-j 4
4
Therefore:
I N = 2I = 0.707∠45 o A
______________________________________________________________________________________
Problem 4.65
Solution:
Known quantities:
Circuit shown in Figure P4.66.
Find:
The equations required to solve for the loop currents in the circuit in:
a. Integral-differential form;
b. Phasor form.
Analysis:
KVL:
KVL:
1 t
(i1 − i2 )dt + (i1 − i2 )R1 = 0
C ³0
t
(i2 − i1 )R1 − vc (0 ) + 1 ³0 (i2 − i1 )dt + L di2 + i2 R2 = 0
dt
C
− v S + i1 RS + vc (0 ) +
Note: The initial voltage across the capacitor must, in general, be considered. It is modeled as an ideal
voltage source in series with the capacitor.
KVL: − VS + I 1 RS + (I 1 − I 2 )Z C + (I 1 − I 2 )R1 = 0
KVL:
(I 2 − I 1 )R1 + (I 2 − I1 )Z C + +I 2 Z L + I 2 R2
=0
Note:
1. The i-v characteristics of the inductor and capacitor, i.e. the integral and derivative, have been replaced
here by the impedance.
2. This form of the equation is applicable only when the waveforms of the currents and voltages are
sinusoids!
4.42
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
______________________________________________________________________________________
Problem 4.66
Solution:
Known quantities:
Circuit shown in Figure P4.65.
Find:
The node equations required to solve for all currents and voltages in the circuit.
Analysis:
I1 = I 2 + I 3
VS − I1RS = I 2 (Z C + R1 ) + I 3 (Z L + R2 )
______________________________________________________________________________________
Problem 4.67
Solution:
Known quantities:
The voltages at the nodes of the circuit shown in Figure P4.67,
Va = 450∠0 o V , Vb = 440∠30 o V ,
Vc = 420∠ − 200 o V , Vbc = 779.5∠5.621o V , Vcd = 153.9∠68.93 o V ,
Vba = 230.6∠107.4 o V , and the voltage sources, v s1 = 450 cos(ωt ) V , v s 2 = 450 cos(ωt ) V .
Find:
The new values of
Vb and Vbc , if the ground is moved from Node e to Node d.
Analysis:
A node voltage is defined as the voltage between a node and the ground node. If the ground node is
changed, then all node voltages in the circuit will change. With the ground at Node d:
Vb = Vbd = Vbe + Ved = Vbe + Vs 2 = 440∠30 o V + 450∠0 o V =
= (381.1 + j 220.0 ) V + (450 + j 0) V = 831.1 + j 220.0 V = 859.6∠14.83o V
The voltages between any two nodes in a circuit do not depend on which is the ground node; therefore, the
voltage between Node b and Node c remains the same when the ground is moved from Node e to Node d:
Vbc = 779.5∠5.621o V
______________________________________________________________________________________
Problem 4.68
Solution:
Known quantities:
Circuit shown in Figure P4.68, the values of the resistance,
RL = 120 Ω , the capacitance,
C = 12.5 µF , and the inductance, L = 60 mH , and the voltage source,
(
)
vi = 4 cos 1,000t + 30 o V .
Find:
The new value of
V0 .
Analysis:
The circuit has 3 unknown mesh currents but only 1 unknown node voltage.
4.43
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
1
1
=−j
= − j80 Ω = 80∠ − 90 o Ω
rad ·
ωC
§
¨1,000 k
¸(12.5 µF )
s ¹
©
rad ·
§
o
Z L = jX L = jωL = j ¨1,000 k
¸(60 mH ) = j 60 Ω = 60∠90 Ω
s
©
¹
o
Reference phasor: Vi = 4∠30 V
Z C = − jX C = − j
KCL:
V0 − 0 V0 − 0 V0 − Vi
+
+
=0
Z RL
ZC
ZL
Vi
Vi
ZL
4∠30 o V
=
V0 =
=
=
ZL ZL
1
1
1
60∠90 o Ω 60∠90 o Ω
+
+
+
+1
+1
+
Z RL Z C Z L
Z RL Z C
120∠0 o Ω 80∠ − 90 o Ω
=
4∠30 o V
4∠30 o V
=
=
0.5∠90 o + 0.75∠180 o + 1 (0 + j 0.5) + (− 0.75 + j 0 ) + (1 + j 0 )
=
4∠30 o V
4∠30 o V
=
= 7.155∠ − 33.43 o V
o
0.25 + j 0.5 0.559∠63.43
(
)
v0 (t ) = 7.155 cos ωt − 33.43o V
______________________________________________________________________________________
Problem 4.69
Solution:
Known quantities:
Circuit shown in Figure P4.69, the mesh currents and node voltages,
(
)
(
)
(
(
)
)
i1 (t ) = 3.127 cos ωt − 47.28 o A , i2 (t ) = 3.914 cos ωt − 102.0 o A ,
i3 (t ) = 1.900 cos(ωt + 37.50 o ) A , v1 (t ) = 130.0 cos ωt + 10.08 o V ,
rad
v 2 (t ) = 130.0 cos ωt − 25.00 o V , where ω = 377.0
.
s
Find:
One of the following:
Analysis:
KCL:
− I1
L1 , C 2 , R3 , L3 .
+ I Z1 + I 3 = 0
I Z1 = I1 − I 3 = (2.121 − j 2.297 ) A − (1.507 + j1.157 ) A = 3.508∠ − 79.92 o A
V1
130∠10.08 o V
=
= 37.05∠90 o Ω = ωL1∠90 o
I Z1 3.508∠ − 79.92 o A
OL:
Z1 =
KCL:
37.05 Ω
= 98.29 mH
rad
377
s
I 2 + I Z2 − I 3 = 0
L1 =
4.44
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 4
I Z 2 = I 3 − I 2 = (1.507 − j1.157 ) A − (− 0.8138 − j 3.828) A = 5.499∠65.03o A
OL:
V2
130∠24.97 o V
1
Z2 =
=
= 23.64∠ − 90 o Ω =
∠ − 90 o
o
I Z 2 5.499∠65.03 A
ωC 2
1
= 112.2 µF
rad ·
§
¨ 377
¸(23.64 Ω)
s ¹
©
V2 − V1 + VZ3 = 0
C2 =
KVL:
VZ 3 = V1 − V2 = (128.0 + j 22.75) V − (117.8 − j 54.88) V = 78.29∠82.56 o V
OL:
VZ 3
78.29∠82.56 o V
Z3 =
=
= 4.21∠45.06 o Ω = 29.11 + j 29.17 Ω = R3 + jωL3
o
I3
1.9∠37.5 A
29.17 Ω
R3 = 29.11 Ω, R3 =
= 77.37 mH
rad
377
s
______________________________________________________________________________________
4.45