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Rizzoni 6e SM CH04 - 전기및 전자공학개론 CH04 solution
전기및전자공학개론 (Chonnam National University)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Chapter 4: AC Network Analysis – Instructor Notes
The chapter starts by developing the dynamic equations for energy storage elements in Section 4.1. The
analogy between electrical and hydraulic circuits (Make The Connection: Fluid (hydraulic) Capacitance, p. 150,
Make The Connection: Fluid (hydraulic) inertance, p. 162, Table 4.2, p.162) 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 box: Capacitive displacement transducer and microphones, pp. 159-160, 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 in Section 4.2, with special emphasis on sinusoids. The material in this
section can also accompany a laboratory experiment on signal sources. Section 4.3 introduces the formulation (and
solution) of circuits described by differential equations, focusing on sinusoidal excitations. The emphasis placed on
sinusoidal signals is motivated by the desire to justify the concepts of phasors and impedance, which are introduced
next, in Section 4.4. This section covers phasor notation, impedance and admittance. It is followed by Section 4.5,
which extends the circuit analysis methods developed in Chapter 3 to AC circuits. The material in Sections 4.3 to
4.5 is of a mathematical nature,however, a Focus on Measurements box: Capacitive displacement transducer, pp.
187-189, extends the concept of bridge circuits to the case of sinusoidal excitation. 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 author has found that presenting the impedance concept early, is an efficient way of using the (invariably too
short) semester or quarter. Chapter 4 is specifically 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.
The homework problems in this chapter are mostly mathematical exercises aimed at mastery of the techniques. The
5th Edition of this book includes 18 new problems; some of the 4th Edition problems were removed, increasing the
end-of-chapter problem count from 69 to 85.
Learning Objectives for Chapter 4
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.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Section 4.1: Energy Storage Elements
Problem 4.1
The current through a 0.8H inductor is given by iL = sin(100t+π/4). Write the expression for the voltage across the inductor.
Solution:
Known quantities:
Inductance value, L = 0.8 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:
i L (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 )

 
π
π

= 0.8 L = 0.8 ⋅  cos100t +  ⋅ 100  = 80 cos100t + V
dt
dt
4
4

 

Problem 4.2
For each case shown below, derive the expression for
the current through a 200 μF capacitor. vC (t) is the
voltage across the capacitor.
a) v c (t ) = 22 cos(20t − π / 3)V
b) v c (t ) = −40 sin(90t + π / 2)V
c) v c (t ) = 28 cos(15t + π / 8)V
d) v c (t ) = 45 sin(120t + π / 4)V .
Solution:
Known quantities:
Capacitance value C = 200 µF ; capacitor terminal voltage as a function of time.
Find:
The current through the capacitor as a function of time for each case:
a) v c (t ) = 22 cos(20t − π / 3)V
b) v c (t ) = −40 sin(90t + π / 2)V
c) v c (t ) = 28 cos(15t + π / 8)V
d) v c (t ) = 45 sin(120t + π / 4)V .
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:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
a)
iC (t ) = C
dvC (t )
dv (t )
dv (t )
π
= 200 × 10 −6 C = 2 ⋅ 10 − 4 C = 0.088 cos(20t + ) A
dt
dt
dt
6
b)
iC (t ) = −40 ⋅ 200 ⋅ 10 −6 ⋅ 90 ⋅ cos(90t + π / 2) A = −0.72 cos(90t + π / 2 − π ) A =
= 0.72 cos(90t − π / 2) A
c)

π 
π


iC (t ) = 200 ⋅ 10 −6 ⋅ 28 ⋅ 15− sin 15t +  = −8.4 ⋅ 10 − 2 sin 15t + A
8 
8



d)
 
π 
π

iC (t ) = 200 ⋅ 10 −6 ⋅ 45 ⋅ 120 cos120t +  = 1.08 ⋅ cos120t +  A
4 
4

 
Problem 4.3
Derive the expression for the voltage across a
200mH inductor when its current is:
i L (t ) = −2 sin(10)tA
b) i L (t ) = 2 cos(3t ) A
c) i L (t ) = −10 sin(50t − π / 4) A
d) i L (t ) = 7 cos(10t + π / 4) A .
a)
Solution:
Known quantities:
Inductance value, L = 200 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) i L (t ) = −2 sin(10)tA
i L (t ) = 2 cos(3t ) A
c) i L (t ) = −10 sin(50t − π / 4) A
d) i L (t ) = 7 cos(10t + π / 4) A .
b)
Assumptions:
i L (t ≤ 0) = 0
Analysis:
Using the differential relationship for the inductor, (Eq. 4.9), we may obtain the voltage by differentiating the
current:s
v L (t ) = L
di L (t )
di (t )
di (t )
= 200 × 10 −3 L = 0.2 L
dt
dt
dt
a)
v L (t ) = 0.2 ⋅ (−2) ⋅ (10)[cos10t ] = 4 cos(10t − π ) V
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
b)
v L (t ) = 0.2 ⋅ (2) ⋅ (3) ⋅ (− sin 3t ) = 1.2 sin(3t + π ) V
c)

π 
π


v L (t ) = 0.2 − 10 × 50 sin  50t −  = −100 cos 50t +  = 100 cos(50t − 5π / 4)V
4 
4



d)

π 
π


v L (t ) = 0.2 7 × 10 cos10t −  = 14 cos10t −  V
4 
4



Problem 4.4
In the circuit shown in Figure P4.4, assume R = 1ohm and L = 2H. Also, let:
I(t)=
0 for−∞< t < 0
t for 0 ≤ t < 10 s
10 for 10 s ≤ t < ∞
Find the energy stored in the inductor for all time.
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):
1
1
Li (t )2 = (2 )i 2 (t ) = i 2 (t )
2
2
For −∞ < t < 0 ,
wL (t ) = 0
For 0 ≤ t < 10 s
wL (t ) =
wL (t ) = t 2 J
For 10 s ≤ t < +∞
wL (t ) = 100 J
Problem 4.5
Refer to Problem 4.4 and find the energy delivered by the source for all time.
Solution:
Known quantities:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
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 energy stored in
inductor and the energy dissipated in resistor.
t
t
t
0
0
0
The energy dissipated in resistor is: wR (t ) = ∫ p(t )dt = R ∫ i 2 (t )dt = ∫ i 2 (t )dt
For −∞ < t < 0 , wR (t ) = 0 . During this time scope,
wS (t ) = wR (t ) + wL (t ) = 0
t
For 0 ≤ t < 10 s , wR (t ) = ∫ t 2 dt =
0
wS (t ) = wR (t ) + wL (t ) = t 2 +
t3
. During this time scope,
3
t3
103
= 333.3
. At t = 10 s , wR (10 ) =
2
3
t
For 10 ≤ t < +∞ , wR (t ) = 333.3 + ∫ 10 2 dt = 333.3 + 100(t − 10) = 100t − 666.7 J . During this time scope,
10
wS (t ) = w R (t ) + w L (t ) = −566.7 + 100t .
Problem 4.6
In the circuit shown in Figure P4.4, assume R = 2ohm and L = 4H. Also, let:
0
f or −∞< t < 0
i (t) = 2t
f or 0 ≤ t < 5s
10-4t
f or 5 ≤ t < 12s
F2
or 12 ≤ t < ∞
Find:
a. The energy stored in the inductor for all time;
b. b. The energy delivered by the source for all time.
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:
a) The magnetic energy stored in an inductor may be found from, (Eq. 4.18):
wL (t ) =
1
1
2
Li (t ) = (4 )i 2 (t ) = 2i 2 (t )
2
2
For −∞ < t < 0 , w L (t ) = 0
For
0 ≤ t < 5s , wL (t ) = 8t 2 J
For 5 ≤ t < 12 s , wL (t ) = 2(10 − 4t ) = 200 − 160t + 32t J
2
For 12 s ≤ t < +∞ ,
2
wL (t ) = 8 J
b) The energy delivered by the source is the sum of energy stored in inductor and the energy dissipated in resistor.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
The energy dissipated in resistor is:
t
t
t
0
0
0
wR (t ) = ∫ p(t )dt = R ∫ i 2 (t )dt =2 ∫ i 2 (t )dt
For −∞ < t < 0 , wR (t ) = 0 . During this time scope, wS (t ) = wR (t ) + wL (t ) = 0
t
For 0 ≤ t < 5 s ,
wR (t ) = 2 ∫ 4t 2 dt =
0
8t 3
8t 3
2
. During this time scope, wS (t ) = wR (t ) + wL (t ) = 8t +
. At
3
2
3
10
= 333.3
3
t = 5 s , w R (10) =
For 5 ≤ t < 12 s ,
t

t2 
wR (t ) = 333.3 + 2 ∫ (10 − 4t ) dt = 333.3 + 2 100t + 16 / 3t 3 − 80  = 10.66t 3 − 40t 2 + 200t + 0.3 J
2 5

5
3
2
. During this time scope, wS (t ) = wR (t ) + wL (t ) = 10.66t − 48t + 40t + 200.3 J . At t = 12 s ,
t
2
wR (20 ) = 9300 J
t
For 12 s ≤ t < +∞ wR (t ) = 9300 + 2 4dt = 8t + 9204 J . During this time scope,
∫
12
wS (t ) = wR (t ) + wL (t ) = 8t + 9212 J .
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.7
In the circuit shown in Figure P4.7, assume R = 2ohm
and C = 0.1F. Also, let:
0 for−∞< t < 0
v(t) =
t for 0 ≤ t < 10 s
10 for 10 s ≤ t < ∞
Find the energy stored in the capacitor for all time.
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
Cv(t )2 = (0.1)v(t )2 = 0.05v(t )2
2
2
For −∞ < t < 0
wC (t ) = 0
For 0 ≤ t < 10s
wC (t ) =
wC (t ) = 0.05 t 2 J
For 10s ≤ t < +∞
wC (t ) = 0.05(10 )2 = 5 J
Problem 4.8
Refer to Problem 4.7 and find the energy delivered
by the source for all time.
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 energy stored in capacitor and the energy dissipated in resistor.
The energy dissipated in resistor is:
t
t
1t
w R (t ) = ∫ p(t )dt = ∫ v 2 (t )dt = 0.5 ∫ v 2 (t )dt
R0
0
0
For −∞ < t < 0 , wR (t ) = 0 . During this time scope,
wS (t ) = w R (t ) + wC (t ) = 0
t
For 0 ≤ t < 10 s , w R (t ) = 0.5 ∫ t 2 dt = 0.1667t 3 . During this time
0
scope, wS (t ) = wR (t ) + wL (t ) = 0.05t 2 + 0.1667t 3 . At t = 10 s ,
wR (10 ) = 0.1667t 3 = 166.7
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
t
For 10 ≤ t < +∞ , wR (t ) = 166.7 + 0.5 ∫ 10 2 dt = 166.7 + 50(t − 10) = 50t − 333.3 J . During
10
this time scope, wS (t ) = w R (t ) + w L (t ) = 50t − 228.3 J .
Problem 4.9
In the circuit shown in Figure P4.7, assume R = 4ohm and C = 0.2F. Also, let:
0
f or −∞< t < 0
v(t) =
4t
f or 0 ≤ t < 4s
2-0.5t
f or 4 ≤ t < 10s
0
f or t > 10s
Find:
a. The energy stored in the inductor for all time
b. The energy delivered by the source for all time.
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:
a) The energy stored in a capacitor may be found from: wC (t ) =
1 2
1
Cv (t ) = (0.2 )v 2 (t ) = 0.1v 2 (t )
2
2
For −∞ < t < 0 wC (t ) = 0
For
For
0 ≤ t < 4 s wC (t ) = 1.6 t 2 J
4 ≤ t < 10 s wC (t ) = 0.1 (2 − 0.5t ) = 0.4 − 0.2t + 0.025t 2 J
2
For 10s ≤ t < +∞ wC (t ) = 0
b) The energy delivered by the source is the sum of the energy dissipated by the resistance and the energy stored in
the capacitor.
t
t
t
1 2
2
The energy dissipated in resistor is: wR (t ) = ∫ p (t )dt = ∫ v (t )dt =0.25∫ v (t )dt
R0
0
0
For −∞ < t < 0 , wR (t ) = 0 . During this time scope, wS (t ) = wR (t ) + wL (t ) = 0
t
2
3
For 0 ≤ t < 4 s , wR (t ) = 0.25∫ 16t dt = 1.333t . During this time scope,
0
wS (t ) = wR (t ) + wL (t ) = 1.6t + 1.333t 3 . At t = 4 s , wR (10 ) = 1.333(4 ) = 85.31
3
2
For 4 ≤ t < 10 s , wR (t ) = 85.31 + 0.25
t
∫ (2 − 0.5t )
2
dt = t − 0.25t 2 + 0.0208t 3 + 85.31 . During this time
4
t = 10 s , wR (20 ) = 91.1433 J
For 10 s ≤ t < +∞ wR (t ) = 91.1433 J . During this time scope, wS (t ) = wR (t ) + wL (t ) = 91.1433 J .
scope, wS (t ) = wR (t ) + wL (t ) = 85.71 + 0.8t − 0.225t + 0.0208t . At
2
3
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.10
Solution:
Known quantities:
Inductor value L = 20 mH ; the voltage applied to the inductor as
a function of time as shown in Figure P4.12.
Find:
The current through 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 ) = mv L t + qv L
where:
mv L =
[40 V ] − [− 20 V ] = 12 V
[5ms] − [0]
ms
q vL = −20 V
(
t
)
1 t
1 t
1 1

i L = i L (0 ) + ∫ v L (τ )dτ = 0.05 A + ∫ m vL τ + q vL dτ = 0.05 A +  m vL τ 2 + q vL τ  =
0
0
L
L
L 2
0
1
1

 V 2
m vL t 2 + 2q vL t = 0.05 A +
= 0.05 A +
⋅ t − 40 V ⋅ t  = 0.05 A + 300 ⋅ 10 3 t 2 − 1000t A
12
2L
2 ⋅ 20mH  ms

(
)
(
)
i L (t = 5ms = 0.005s ) = 2.505A
For 5 ms < t ≤ 10 ms
mv L =
[20 V ] − [40 V ] = −4 V
[5ms] − [0]
ms
q vL = 60 V
i L = i L (0.005) +
(
)
t
1 t
1 t
1 1

v L (τ )dτ = 2.505 A + ∫
mvL τ + q vL dτ = 2.505 A +  mvL τ 2 + q vL τ  =
∫
0
.
005
0
.
005
L
L
L 2
0
= 1.1A + 1500t − 10 5 t 2
i L (t = 10 ms = 0.01s ) = 6.1A
For t > 10 ms
v L (t) = c v L = 0
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
1 t
1 t
(
)
(
)
(0)dτ = i L (0.01) + 1 [0]t0.005 =
τ
τ
v
d
i
=
+
0
.
01
L
L
∫
∫
0
.
01
0
.
01
L
L
L
= i L (0.01) = 6.1A
i L = i L (0.01) +
Problem 4.11
Solution:
Known quantities:
Capacitor value, C = 100 µF ; the voltage applied to the inductor as shown in Figure P4.12; the initial condition
for the current ic (0 ) = 0 .
Find:
The current across 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 ) = mcl t + q cL
where:
mvC =
[40 V ] − [− 20 V ] = 12 V
[5ms] − [0]
ms
q vC = −20 V
iC = C
dvC (t )
d
 V 
=C
mvC t + q vC = CmvC = (100 µF )12
 = 1.2 A
dt
dt
 ms 
[
]
For 5 ms < t < 10 ms
V
ms
dvC (t )
d
V 

=C
mvC t + q vC = CmvC = (100 µF ) − 4
iC = C
 = −0.4 A
dt
dt
 ms 
mvC = −4
[
]
For t > 10 ms
vC (t ) = 20
iC = C
dvC (t )
d
= C [20] = 0 A
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. Some dielectrics however do allow very
minute amounts of current to flow directly through them. These currents are known as leakage currents.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.12
Solution:
Known quantities:
Inductance value L = 0.5mH ; the voltage applied to the
inductor as a function of time as shown in Figure P4.12.
Find:
The current through the inductor at the time
t = 6 ms .
Assumptions:
i L (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 ≤ 2 ms
v L (t ) = mvL t + q vL where: mvL =
[6V ] − [0 V ] = 3 V
[2 µs] − [0] ms
qv L = 0 V
For 2 < t ≤ 4 ms
v L (t ) = mvL t + q vL where: mvL =
[− 6V ] − [3V ] = −4.5 V
[2 ms] − [0]
ms
q vL = 15V
For t > 4ms
v L (t ) = cvL = −3 V
Therefore:
i L (t = 6ms ) =
(
)
1 6 ms
1 2 ms
1 4ms
1 6 ms
v L (τ )dτ = i L (0 ) + ∫ mvL τ dτ + ∫ (mvL + cvL )dτ + ∫ (mvL + cvL )dτ =
∫
L 0
L 2 ms
L 4ms
L −∞
2 ms
1 1

= i L (0 ) +  mvL τ 2 
L 2
0
4 ms
[ ]
1
1 1
1

+  mv L τ 2 + q v L τ  + c v L τ
2
L 2
 2 ms L
6 ms
4 ms
= 12 A + 4 A − 12 A = 4 A
Problem 4.13
Solution:
Known quantities:
Vpk = 20V, T = 40μs, C = 680nF the voltage applied across the
capacitor as a function of time as shown in Figure P4.13.
Find:
Plot the current through the capacitor.
Analysis:
Current through a capacitor is equal to the capacitance times the derivative of the voltage across it:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
끫뢬끫뢠 = 끫롬
끫뢢끫뢢
끫뢢끫뢢
The derivative of a line is the slope of the line. Therefore, the derivative can be found by breaking up the equation
into 4 parts and multiplying each part by the capacitance. This also means that since the 4 sections are made up of
straight lines, the derivative will consist of horizontal lines.
time = 0-26.666μs:
20
끫뢬1 = 680 ∗ 10−9 �
� = 0.51끫롨
26.66 ∗ 10−6
time = 26.666μs-40μs:
−20
끫뢬2 = 680 ∗ 10−9 �
� = −1.02끫롨
13.33 ∗ 10−6
time = 40μs-66.66μs:
20
끫뢬3 = 680 ∗ 10−9 �
� = 0.51끫롨
26.66 ∗ 10−6
time = 66.666μs-80μs:
−20
끫뢬4 = 680 ∗ 10−9 �
� = −1.02끫롨
13.33 ∗ 10−6
It is noticeable that the discontinuities result in a discontinuity in the current graph. Also when the slope of the
voltage is negative, the current becomes negative.
Problem 4.14
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 .
Find:
The current through the inductor at 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 30 µs
1
∫ −∞ v L (τ )dτ = i L (0) + ∫ 020 µs v L (τ )dτ + ∫ 20
µs v L (τ )dτ =
L
L
L
= i L (0) +
+
20 µs
1 3 3 V 
V
1
1
30 µs
⋅1 2
+ [1.2τ nV]20 µs = 0 +
 τ

2
L 3
16 µH
L
s 0
s
3


⋅ (20 µs) − 0 +


1
⋅ (1.2 nV) ⋅ (30 µs − 20 µs) = 1.250 nA
16 µH
Problem 4.15
Solution:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Known quantities:
Resistance value R = 7 Ω ; inductance value L = 7 mH ;
capacitance value C = 0.5 µF ; the voltage across the
components as shown in Figure P4.17.
Find:
The current through each component.
Assumptions:
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.
For
t ≤ 0:
i R (t ) = i L (t ) = iC (t ) = 0
For 0 < t < 5 ms :
v (t ) = m v t + qv
where:
[15 V] − [0] = 3 V
mv =
ms
[5 ms] − [0]
qvC = 0 V
v (t )
m ⋅t
= v =
i R (t ) =
R
R
i L (t ) =
3
V
⋅t
ms = 428.6 ⋅ t A
7Ω
t
1 t
1
1 1
1
mv ⋅ t 2 =
∫ 0 v (τ )dτ = ∫ 0t mv ⋅ τ dτ =  mv ⋅ τ 2  =
0 2L
L
L
L 2
V 2
1
⋅3
⋅ t = 214.3 ⋅10 3 ⋅ t 2 A
ms
2 ⋅ 7 mH

dv (t )
d
µAs  V 
= C [mv t ] = Cmv =  0.5
iC (t ) = C C
 3
 = 1.5 mA

dt
dt
V  ms 
For t = 5 ms :
=
( )
(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
iL
iC
−3
3
(
⋅ t 2 A = 214.3 ⋅10 3 ⋅ 5⋅10−3
) A = 5.357 A
2
−3
For 5 ms < t < 10 ms :
v (t ) = c v = 15 V
v (t ) c v 15 V
=
=
= 2.143 A
i R (t ) =
R
R
7Ω
1
1
i L (t ) = i L 5⋅10−3 + ∫ t −3 v (τ )dτ = i L 5⋅10−3 + ∫ t −3 c v dτ =
L 5⋅10
L 5⋅10
1
1
t
= i L 5⋅10−3 + [c v ⋅ τ ]5⋅10 −3 = i L 5⋅10−3 + ⋅c v ⋅ t − 5⋅10−3 =
L
L
1
−3
= 5.357 A +
⋅15 V ⋅ t − 5⋅10 s = −5.357 + 2.143⋅10 3 ⋅ t A
7 mH
(
(
)
)
(
( )
( )
)
(
)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
dvC (t )
d
= C [c v ] = 0
dt
dt
For t = 10 ms :
i R (0.01) = 2.143 A
iC (t ) = C
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
= =0
i R (t ) =
R
R
1
1
i L (t ) = i L 10 ⋅10−3 + ∫ t −3 v (τ )dτ = i L 10 ⋅10−3 + ∫ t −3 0 dτ =
10⋅10
L 10⋅10
L
1
t
= i L 10 ⋅10−3 + [0]10⋅10 −3 = i L 10 ⋅10−3 = 16.07 A
L
dv (t )
d
iC (t ) = C C = C [0] = 0
dt
dt
(
(
)
)
(
(
)
)
Problem 4.16
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.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Analysis:
Considering the period: −2.5 µs < t < +2.5 µs :
ic = C
dv c
∆v
= C c , since the voltage has a linear waveform.
dt
∆t
Substituting:
12 A = C
[+10 V] − [−10 V]
5 µs
⇒ C = 12 A ⋅
5 µs
= 3 µF
20 V
Problem 4.17
Solution:
Known quantities:
The voltage across and the current through an ideal inductor as shown in Figure P4.19.
Find:
The inductance of the inductor.
Analysis:
vL = L
∆i
di L
= L L , since the current has a linear waveform.
dt
∆t
Substituting:
− 3V = L
[− 2 A] − [− 1A] ⇒ L = −3V ⋅
10 ms − 5ms
4 ms
= 12 mH
−1 A
Problem 4.18
Solution:
Known quantities:
The voltage across and the current through an ideal capacitor as shown in Figure P4.20.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Find:
The capacitance of the capacitor.
Analysis:
Considering the period: 0 < t < 5 ms :
ic = C
∆v
dv c
= C c , since the voltage has a linear waveform. Substituting:
dt
∆t
3.5mA = C
[40 V ] − [0] ⇒ C = 3.5mA ⋅
3ms
3 ms
= 0.2625 µF
40 V
Problem 4.19
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
dv c
∆v
= C c , since the voltage has a linear waveform.
dt
∆t
Substituting:
3 mA = C
[7 V] − [0]
5 ms
⇒ C = 3 mA ⋅
5 ms
= 2.14 µF
7 V
Problem 4.20
Solution:
Known quantities:
The voltage across the inductor as shown in Figure P4.22.
Find:
The current in the inductor.
Analysis:
1t
∫ v L (t )dt
L0
The current in the inductor is shown in the figure:
i L (t ) =
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.21
Solution:
Known quantities:
The current through the inductor as shown in Figure P4.23.
Find:
The voltage across the inductor.
Analysis:
4t − 15 mA 0 < t < 5ms
i(t ) = 
5 mA t > 5ms
For a 2 H inductor, since v L = L
di L
,
dt
8 V 0 < t < 5ms
v L (t ) = 
0 t > 5ms
The voltage waveform is sketched in the figure.
Problem 4.22
Solution:
Known quantities:
The voltage across the inductor as shown in Figure
P4.24.
Find:
The current through the inductor and the capacitor.
Analysis:
 15
 0.004 t V 0 < t < 4ms
v(t ) = 
30 − 15 t V 4ms<t<6ms

0.004
3750t V 0 < t < 4ms
=
30 − 3750 t V 4ms < t < 6ms
The capacitor current is
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
dv
dv
= 500 × 10 −6
dt
dt
1.875 A 0 < t < 4ms
=
− 1.875 A 4ms < t < 6ms
iC = C
The inductor current is
iL =
1 t
1 t
t
∫ v(τ )dτ = iL (t0 ) + ∫t0 v(τ )dτ = iL (t0 ) + 10 ∫t0 v(τ )dτ
L −∞
L
Assume i L (0 ) = 0 .
For 0 < t < 4ms, we have
i L = 0 + 10∫0t 3750τdτ = 37500
t2
2
t
= 18750t 2 A
0
For 4ms < t < 6ms, we have
i L = 18750(0.004)2 + 10∫0t .004 (30 − 3750τ )dτ
[
]
t
= 0.3 + 300τ − 18750τ 2 0.004
= 0.3 + 300t − 18750t 2 − 300(0.004) + 18750(0.004)2
= 300t − 18750t 2 − 0.6 A
The two functions are sketched in the figures:
Problem 4.23
Solution:
Known quantities:
Circuit as shown in Figure P4.25 and the value of current
source.
Find:
The energy stored in the inductor.
Analysis:
1 2 1
Li = (2 )i 2 = i 2
2
2
For −∞ < t < 0 ,
wL (t ) = 0
For 0 ≤ t < 1s
wL (t ) =
wL (t ) = t 2 J
For 1s ≤ t < 2s
wL (t ) = [− (t − 2 )]2 = t 2 − 4t + 4 J
For 2s ≤ t < ∞
wL (t ) = 0
Problem 4.24
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Solution:
Known quantities:
Circuit as shown in Figure P4.26 and the value of voltage
source.
Find:
The energy stored in the capacitor.
Analysis:
1 2 1
Cv = (0.1)v 2 = 0.05v 2
2
2
For −∞ < t < 0
wC (t ) = 0
For 0 ≤ t < 1s
wC (t ) =
wC (t ) = 0.05(2t )2 = 0.2t 2 J
For 1s ≤ t < 2s
wC (t ) = 0.05[− (2t − 4 )]2 = 0.2t 2 − 0.8t + 0.8 J
For 2s ≤ t < ∞
wC (t ) = 0
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.25
Solution:
Known quantities:
Circuit as shown in Figure P4.27 and the value of voltage
source.
Find:
The current through the capacitor.
Analysis:
In an ideal capacitor,
dV
ic = C C
dt
For 0 ≤ t ≤ 0.5 s, the voltage is:
Vc = 30 t
∴ i c = 0.01 × 30 = 0.3 A
For 0.5 ≤ t ≤ 1 s,
Vc = - 30t + 30
∴ ic = -0.01 × 30 = -0.3 A
For 1 ≤ t ≤ 1.5 s,
Vc = 30t − 30
∴ ic = 0.01 × 30 = 0.3 A
For 1.5 ≤ t ≤ 2 s,
Vc = -30t + 60
∴ ic = -0.01 × 30 = -0.3 A
The waveform of ic(t) is shown in the right hand side.
Problem 4.26
Solution:
Known quantities:
Circuit as shown in Figure P4.28 and the value of voltage source.
Find:
The current through the inductor.
Analysis:
vL
=
L
di L
dt
1
∫ v L dt
L
For 0 < t < 0.1 s
iL = ∫ 0.5 − 0.5e −t/ 0.02
iL
=
= 0.5t + (0.5 × 0.02 )e -t/ 0.02 − 0.01 = 0.5t + 0.01e -t/ 0.02 − 0.01 A
For 0.1 < t < 0.2 s
iL = iL (0.1) + ∫ 0.4866dτ = 0.0501 + 0.4966(t − 0.1) = 0.4966t + 0.00044 A
For 0.2 < t < 0.3
iL = iL (0.2 ) + ∫ (0.5 − 0.5e (τ − 0.3) / 0.02 )dτ
= 0.99364 + 0.5(t − 0.2) − (0.5 × 0.02)(e (t − 0.3)/ 0.02 − e − 5 ) = 0.006283 + 0.5t − 0.01e (t − 0.3)/ 0.02 A
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
For 0.3 < t < 0.4
iL = iL (0.3) = 0.146283 A
The resulting waveform is shown in the right hand side.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.27
Solution:
Known quantities:
Find:
The average and rms voltage.
끫룊(끫뢢) = 3 cos(7끫븨끫뢢) + 4
Analysis:
The average value of a sinusoid is 0. x(t) is a normal sinusoid raised by 4 on the y-axis(Voltage).
Therefore the average of x(t) is equal to 4.
The rms voltage is found using the following equation:
Plug in the equation:
1 끫뢎
끫뢖끫뢾끫뢾끫뢾 = � � 끫룊 2 (끫뢢)끫뢢끫뢢
끫뢎 0
끫뾞 끫룶
끫룾끫뤦끫뤦끫뤦 = � � (끫뾢 끫렸끫렸끫렸(끫뾪끫뾪끫뾪) + 끫뾤)끫뾠 끫뤊끫뾪
끫룶 끫뾜
Problem 4.28
Solution:
Known quantities:
The sinusoidal voltage v (t ) of 110 V rms shown in Figure
P4.30.
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:
2π

1 θ
v(t) =
 ∫ 110 2 sin(t )dt + ∫ 110 2 sin(t )dt  =
2π  0

2π −θ
=
θ
2π 
1 
−110 2 cos(t) 0 − 110 2 cos(t) 2π −θ  =
2π 

=−
50 2
π
[cos(θ ) − 1+ cos(2π ) − cos(2π − θ )] = 0
The rms value is:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
v rms
 1
=
 2π
12
2π
θ

2
2
 ∫ 110 2 sin (t ) dt + ∫ 110 2 sin (t ) dt 


2π −θ
0

(
)
(
)
=
2π
θ
12100  1
 (− cos(t )sin (t ) + t ) + 1 (− cos(t )sin (t ) + t )
=
2
 π  2
0
2π −θ
12




=
12
 6050
(− cos(θ )sin (θ ) + θ + 2π − (− cos(2π − θ )sin (2π − θ ) + 2π − θ ))
=

 π
12
 6050
(sin(2θ ) − 2θ )
=

 π
= 55
2
π
=
sin( 2θ ) − 2θ
Problem 4.29
Solution:
Known quantities:
The sinusoidal voltage v (t ) of 110 V rms shown in Figure P4.30.
Find:
The angle θ that correspond to delivering exactly one-half of the
total available power in the waveform to a resistive load.
Analysis:
θ
, we obtain:
π
θ 110 2
π
2
v rms
= 110 2 =
⇒ θ=
π
2
2
From v rms = 110
Problem 4.30
Solution:
Known quantities:
The signal v (t ) shown in Figure P4.32.
Find:
The ratio between average and rms value of the signal.
Analysis:
The average value is: V =
The rms value is:
v rms =
0.006
1  0.003
(
6 )dt + ∫ (− 2 )dt  = 2V
∫

0.003
0.006  0
0.006
1  0.003 2
(
)
(− 2)2 dt  = 20 = 4.47 V
6
dt
+
∫
∫

0
0
.
003

0.006 
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Therefore,
V
v rms
=
2
= 0.45
4.47
Problem 4.31
Solution:
Known quantities:
The signal i (t ) shown in Figure P4.33.
Find:
The power dissipated by a 1-Ω resistor.
Analysis:
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
= 6.12 A
⋅100 ⋅
p
8
2
= (1)(6.12) W = 37.5 W
Therefore, the power dissipated by a 1-Ω resistor is: P1Ω = Ri rms
2
Problem 4.32
Solution:
Known quantities:
The signal x (t ) shown in Figure P4.34.
Find:
The average and rms value of the signal.
Analysis:
The average value is: V =
The rms value is: Vrms =
t
1 t0 +τ
∫ t Vm dt = Vm where t0 is the left-hand side of the pulse.
0
T
T
1 t0 +τ 2
t
Vm dt =
Vm
∫
T
T t0
Therefore,
V
t
=
Vrms
T
Problem 4.33
Solution:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Known quantities:
The signal i (t ) shown in Figure P4.35.
Find:
The rms value of the signal.
Analysis:
Think of the problem graphically first. The current must be squared. This squares the maximum amplitude and cuts
the period in half. If you redraw the graph after it is squared it from 0 to T, it will consist of 4 identical triangles with
a base width of T/4 and a height of 4. The integral of these triangles is the area under the curve. Add up the area of
the triangles:
Plug in the area into the RMS equation:
끫롨끫롨끫롨끫롨 =
1
끫뢎
끫룀 ∗ 4끫롨 ∗ ∗ 4 = 2끫뢎끫롨 ∗ 끫룀
2
4
1 끫뢎
1
끫뢬끫뢾끫뢾끫뢾 = � � 끫뢬 2 (끫뢢)끫뢢끫뢢 = � ∗ 2끫뢎 = √2 = 끫뾞. 끫뾤끫뾞끫뾤
끫뢎 0
끫뢎
Problem 4.34
Solution:
Known quantities:
The signal v (t ).
Find:
The rms value of the signal.
Analysis:
The rms value is: v rms =
2
1 2π
∫ (v (t )) d (ωt )
2π 0
2
1 2π 2
1 2π
∫ (VDC + V0 cos(ωt )) d (ωt ) =
∫ V DC + 2VDC V0 cos(ωt ) + V02 cos 2 (ωt ) d (ωt ) =
2π 0
2π 0

V2 V2
1 2π  2
=
∫ VDC + 2VDC V0 cos(ωt ) + 0 + 0 cos 2 (ωt )d (ωt ) =
2
2
2π 0 

 1 

V2
1  2
V2
2
VDC [ωt ]20π + 0 + 0 [ωt ]20π + 0 =
VDC
=
(2π − 0) + 0 + 0 (2π − 0) + 0
2
2
2π 
 2π 

(
v 2 rms =
2
v rms = V DC
+
V02
=
2
)
(35 V )2 + 1 (63V )2
2
= 56.65 V
Notes:
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. The integral of a sinusoid over one or more whole periods is equal to 0.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.35
Solution:
Known quantities:
Capacitance, resistance and inductance values; the voltage
v S = 6 V applied to the circuit of Fig. P4.35.
Find:
The energy stored in each capacitor and inductor.
Analysis:
The first step is to redraw the circuit and place a break where
there is a capacitor and a wire where there is an inductor.
The circuit can be simplified to find the Req and total current:
끫뢊끫뢤끫뢤 = 4.66 끫뢄ℎ끫뢴끫룀
6
끫롸끫룂 =
= 1.29끫롨
4.66
There is no current through R3, therefore, the 2F and the 3F capacitor
have the same voltage across them. This can be found by calculating the
voltage drop across R1:
6 − 끫뢒1
= 1.29끫롨
2
끫뢒1 = 3.43끫뢒
Plug this voltage into the equation for energy storage in a capacitor to get the energy stored in the 2F and the 3F
capacitor:
끫뾞
끫룼끫룔끫뾠끫룔 = ∗ 끫뾠끫룔 ∗ (끫뾢. 끫뾤끫뾢끫뾤)끫뾠 = 끫뾞끫뾞. 끫뾪끫뾪끫뾪
끫뾠
끫뾞
끫룼끫룔끫뾢끫룔 = ∗ 끫뾢끫룔 ∗ (끫뾢. 끫뾤끫뾢끫뾤)끫뾠 = 끫뾞끫뾪. 끫뾪끫뾢끫뾪
끫뾠
The 1F capacitor is connected to the same node with both of its leads. This means there is no voltage across it and
that its energy stored is:
끫룼끫룔끫뾞끫룔 = 끫뾜
The inductor requires the use of current division to find the current through the 8 Ohm resistor:
끫뢊2
끫롸8 =
∗ 끫롸 = 0.43끫롨
끫뢊2 + 끫뢊4 끫룂
This current is also the current through the inductor. Plug it into the energy storage of an inductor equation:
끫뾞
끫룼끫룦끫뾠끫룦 = ∗ 끫뾠끫룦 ∗ (끫뾜. 끫뾤끫뾢끫뾤)끫뾠 = 끫뾜. 끫뾞끫뾞
끫뾠
Problem 4.36
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.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
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,
1
1
2
2
= (1 F)(12 V) = 72 J
v1F = v A = 12 V ⇒ w1F = C1F v1F
2
2
1
1
2
2
L1H i1H
= (1 H )(2 A) = 2 J
2
2
1
1
2
2
i 2H = i1H = 2 A ⇒ w 2H = L2H i 2H
= (2 H )(2 A) = 4 J
2
2
1
1
2
2
= (2 F)(0 V) = 0 J
v 2F = 0 V ⇒ w 2F = C 2F v 2F
2
2
i1H =
12V
=2 A
6Ω
⇒ w1H =
Problem 4.37
Find the phasor form of the following functions:
a. v(t) = 155 cos (377t – 25°) V
b. v(t) = 5 sin (1,000t − 40◦) V
c. i (t) = 10 cos (10t + 63°) + 15 cos (10t − 42◦) A
d. i (t) = 460 cos (500πt − 25◦) − 220 sin (500πt + 15◦) A
Solution:
Known quantities:
Functions.
Find:
The phasor form.
Analysis:
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
Problem 4.38
Convert the following complex numbers to polar form:
a. 7 + j9
b. −2 + j7
c. J2/3+4-j1/3+3
Solution:
Known quantities:
Complex number.
Find:
The polar form.
Analysis:
a) 7 + j 9= 130∠52.12°=11.4∠52.12°
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
b) - 2 + j 7 = 7.28 ∠74.05°
c) j ⋅ 2 / 3 + 4 − j ⋅ 1 / 3 + 3 = 7 + j
1
= 7.008∠2.73°
3
Problem 4.39
Convert the rectangular factors to polar form and
compute the product. Also compute the product
directly using the rectangular factors. Compare the
results.
a. (50 + j10) (4 + j8)
b. ( j2 − 2) (4 + j5) (2 +j7)
Solution:
Known quantities:
Complex number.
Find:
The polar form.
Analysis:
a) (50 + j10 )(4 + j8) = (50.99∠11.30°)(8.94∠63.43°) = 456.1∠74.7°
(50 + j10)(4 + j8)= 200 + j 400 + j 40 + j 2 80 = 120 + j 440 = 456.1∠74.7°
b) ( j 2 − 2)(4 + j 5)(2 + j 7 ) = (2.82∠135°)(6.40∠51.34°)(7.28∠74.05°) = 131.8∠260.4° = 131.8∠-99.6°
( j 2 − 2)(4 + j5)(2 + j 7 ) = -36 - j126 - j 4 - j 2 14 = -22 - j130 = 131.8∠-99.6°
Problem 4.40
Complete the following exercises in complex
arithmetic.
a. Find the complex conjugate of (4 +j 4), (2 −j 8),
(−5 + j 2).
b. Multiply the numerator and denominator of each
ratio by the complex conjugate of the denominator.
Use the result to express each ratio in polar form.
(1+j7)/(4+j4)
j4/(2-j8) 1/(-5+j2)
c. Convert the numerator and denominator of each
ratio in part (b) to polar form. Use the result to
express each ratio in polar form.
Solution:
Known quantities:
Complex number.
Find:
a)
b)
c)
Complex conjugate
Polar form, by first multiplying numerator and denominator by the complex conjugate.
Polar form, by converting into polar coordinates.
Analysis:
A = 4 + j 4 , A* = 4 - j 4
a) B = 2- j 8, B* = 2 + j 8
C = -5 + j 2 , C* = -5 - j 2
b)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
1 + j 7 (1 + j 7 )(4 - j 4 ) 4 - j 4 + j 28 - j 2 28
=
=
=
4 + j 4 (4 + j 4 )(4 - j 4 )
16 + 16
j4
j 4(2 + j8)
- 32 + j8
32
8
=
=
=- +j
2 - j8 (2 − j8)(2 + j8) 4 + 64
68
68
- 5 - j2
1
1(− 5 − j 2 )
5
=
=
=−
−
- 5 + j 2 (− 5 + j 2 )(− 5 − j 2 ) 25 + 4
29
c) Repeat b) converting to polar form first:
1 + j7
7.071∠81.87°
= 1.25∠36.87°
=
4 + j4
4 2∠45°
j4
4∠90°
=
= 0.485∠165.96°
2 - j8
8.246∠75.96°
1∠0°
1
= 0.1857∠-158.2°
=
-5 + j 2 5.385∠158.2°
32 + j 24
= 1 + j 0.75 = 1.25∠36.87°
32
= 0.485∠165.96°
j
2
= 0.1857∠-158.2°
29
Problem 4.41
Convert the following expressions to rectangular form:
jj
e − jπ
e − j 2π
Solution:
Known quantities:
Complex number.
Find:
Real-imaginary form
Analysis:
j j = e jπ / 2 j = e −π / 2 = 0.2079
e − jπ = cos(− π ) + j sin (− π ) = −1 + j 0 = −1
e j 2 π = cos(2π ) + j sin (2π ) = 1 + j 0 = 1
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.42
Find v(t) = v1(t)+v2(t), where v1(t) = 10
cos(ωt + 30◦) and v2(t) = 20 cos(ωt + 60◦), using:
a. Trigonometric identities.
b. Phasors.
Solution:
Known quantities:
Sinusoidal functions.
Find:
Sum of them, by using trigonometric identity and phasors
Analysis:
(a) Using the trigonometric identity cos(‡ + ‰) = cos ‡ cos ‰− sin ‡ sin ‰, we expand the voltages:
 3
1
 cos 
10 cos(
t + 30°)= 10 cos 30° cos 
t − 10 sin 30° sin 
t = 10
t − 10  sin 
t = 8.66 cos 
t − 5 sin 
t
 2 
2


 3
1
 sin 
20 cos(
t + 60°)= 20 cos 60° cos 
t − 20 sin 60° sin 
t = 20  cos 
t − 20
t = 10 cos 
t − 17.32 sin 
t
 2 
2


Hence,
v(t) = (8.66 + 10 )cos 
t − (5 + 17.32)sin 
t = 18.66 cos 
t − 22.32 sin 
t
Now, v(t) is of the form
v(t) = A cos φ cos ωt − A sin φ sin ωt = A cos(ωt + φ )
18.66
A sin š
 22.32 
and since
= 29.09 ∴ v(t ) = 29.09 cos(ϖt + 50.1°)
= tan š , we have š = tan −1 
 = 50.1° and A =
18
.
66
A cos š
cos š


(b) Using phasors,
3
1
1
3
+ j10 = 8.66 + j 5
V2 () = 20∠60° = 20 + j 20
= 10 + j17.32
2
2
2
2
V (ω ) = (8.66 + 10 )+ j (5 + 17.32 ) = 18.66 + j 22.32 = 29.09∠50.1°
∴ v(t) = 29.09 cos(
t + 50.1°)
V1 () = 10∠30° = 10
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition 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(ωt + θ ) , and a frequency domain form:
jωt
2.
3.
V( jω ) = Ae jθ = A∠θ . Note the jω in the notation V( jω ) , 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 using phasor notation, it is important to note the specific frequency ω of the
sinusoidal input.
Problem 4.43
The current through and the voltage across a circuit
element are:
i (t) = 8 cos(ωt + π/4)A
v(t) = 2 cos( ωt − π/4 )V
where ω = 600 rad/s
Determine:
a. Whether the element is a resistor, capacitor, or inductor;
b. The value of the element in ohms, farads, or henrys.
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 = 8∠45 o mA , V = 2∠ − 45 o V
Z=
V 2∠ − 45 o V
=
= 0.25∠ − 90 o Ω = 0 − j ⋅ 0.25Ω
o
I 8∠45 mA
The component is a capacitor.
b)
ZC = − j ⋅ X C =
1
j ⋅ ωC
= − j 0.25 Ω ⇒ C =
1
= 0.0066 F
ω ⋅ 0.25
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.44
Express the sinusoidal waveform shown in
Figure P4.44 using time-dependent and phasor
notation.
Solution:
Known quantities:
The waveform of a signal shown in Figure P4.44.
Find:
The sinusoidal description of the signal.
Analysis:
From the graph of Figure P4.44:
φ =+
2π
π 180 o
= 60 o , V0 = 170 V , ω = 2πf =
T
3 π
(
)
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 4.45
Express the sinusoidal waveform shown in
Figure P4.45 using time-dependent and phasor
notation.
Solution:
Known quantities:
The waveform of a signal shown in Figure P4.45.
Find:
The sinusoidal description of the signal.
Analysis:
From graph:
3π 180 o
2π
rad
= −135o , I 0 = 8 mA, ω = 2πf =
= 1571
4 π
T
s


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
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.46
Convert the following pairs of voltage and current
waveforms to phasor form. Each pair of waveforms
corresponds to an unknown element. Determine
whether each element is a resistor, a capacitor, or an
inductor, and compute the value of the corresponding
parameter R, C or L.
a. v(t) = 20 cos(400t +1.2), i (t) = 4 sin(400t +1.2)
b. v(t) = 9 cos(900t –π/3), i (t) = 4 sin(900t+2/3π)
c. v(t) = 13 cos(250t +π/3), i (t) = 7 sin(250t+5/6π)
Solution:
Known quantities:
The current and the voltage values of the components.
Find:
The components is inductive, capacitive or resistive?.
Analysis:
a)
Phasor notation:
V = 20e
π
j1.2
, I = 4e
j (1.2 − )
2
∠Z = ∠V − ∠ I =
V = ZI
.
π
2
V
⇒ Z = j ωL ⇒ L =
Iω
=
1
H
80
The component is inductive because it is lagging.
b) Phasor notation:
V = 9e
−j
π
3
, I = 4e
j(
2π π
− )
3 2
= 4e
∠Z = ∠V − ∠ I = −
π
2
−j
π
6
.
V = ZI
⇒Z =−
I
j
4
⇒C =
=
F
ωC
V ω 8100
The component is capacitive.
c)
Phasor notation:
V = 13e
+j
π
3
, I = 7e
j(
5π π
− )
6 2
= 7e
j
π
3
.
V = ZI
∠Z = ∠V − ∠ I = 0 ⇒ Z = R ⇒ C =
V
I
=
13
Ω
7
The component is resistive.
Problem 4.47
Determine the equivalent impedance seen by the
source Vs in Figure P4.47 when:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Vs (t) = 10 cos(4000t + 60◦) V R1 = 800ohm, R2 = 500ohm, L = 200mH, C = 70nF
Solution:
Known quantities:
R1 = 800Ω , R2 = 500Ω , L = 200mH , C = 70nF and the voltage applied to
o
the circuit shown in Figure P4.47, v s (t ) = 10 cos(4000t + 60 )V .
The values of the impedance,
Find:
The equivalent impedance of the circuit.
Analysis:
X L = ωL = 800Ω ⇒ Z L = + j ⋅ X L = + j ⋅ 800Ω
1
= 3571Ω ⇒ Z C = − j ⋅ X C = − j ⋅ 3571Ω
ωC
= Z R 2 + Z L = R2 + jX L = 500 + j ⋅ 800Ω = 943.4∠60Ω
XC =
Z eq1
Z eq 2 =
Z eq =
Z eq1 ⋅ R1
Z eq1 + R1
Z eq 2 ⋅ Z C
Z eq 2 + Z C
= 440 + j 219.8
= 540.9∠198.8
Problem 4.48
Determine the equivalent impedance seen by the
source Vs in Figure P4.47 when:
Vs (t) = 5 cos(1000t + 30◦) V
R1 = 300ohm, R2 = 300ohm, L = 100mH, C = 50nF.
Solution:
Known quantities:
R1 = 300Ω , R2 = 300Ω , L = 100mH , C = 50nF and the voltage applied to
o
the circuit shown in Figure P4.47, v s (t ) = 5 cos(1000t + 30 )V .
The values of the impedance,
Find:
The equivalent impedance of the circuit.
Analysis:
X L = ωL = 100Ω ⇒ Z L = + j ⋅ X L = + j ⋅ 100Ω
1
= 20000Ω ⇒ Z C = − j ⋅ X C = − j ⋅ 20000Ω
ωC
= Z R 2 + Z L = R2 + jX L = 300 + j ⋅ 100Ω = 316.22∠19.07Ω
XC =
Z eq1
Z eq 2 =
Z eq =
Z eq1 ⋅ R1
Z eq1 + R1
Z eq 2 ⋅ Z C
Z eq 2 + Z C
= 155.95∠9.37 = 154 + j 24.55
= 156.13∠8.91
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.49
The generalized version of Ohm’s law for
impedance elements is:
V= IZ
Assume the current through a 0.5μF capacitor is given by:
is (t) = Io cos(ωt + π/6)
Io = 13mA ω = 1,000 rad/s
a. Express the source current in phasor notation.
b. Determine the impedance of the capacitor.
c. Determine the voltage across the capacitor, in
phasor notation.
Solution:
Known quantities:
(
)
The current in the circuit, i s (t ) = I 0 cos ωt + 30 o , I 0 = 13 mA, ω = 1000
present in the circuit shown in Figure P4.49 C = 0.5 µF .
rad
, and the value of the capacitance
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
1
1
=−j
= 0 − j2 kΩ = 2∠ − 90 o kΩ
b) ZC = − jXC − j

rad 
ωC
1000
(0.5 ΩF)

s 
c)
(
)(
(t ) = 26 cos(1000t − 60 )V
)
VC = I s ⋅ ZC = 13∠30 o mA ⋅ 2∠ − 90 o kΩ = 26∠ − 60 o V
vC
o
Note that conversion from phasor notation to time notation or vice versa can be done at any time.
Problem 4.50
Determine i2(t) in the circuit in figure P4.50.
Solution:
Known quantities:
i1(t)=100cos(끫븨t+4)mA
i3(t)=80sin(ωt-1.2)mA
i4(t)=150sin(ωt+2)mA
ω=377rad/s
Find:
i2(t)
Analysis:
Simple KCL can be used here to find the missing current. If KCL is used at the top-right node you derive the
following equation:
끫뢬1 − 끫뢬2 − 끫뢬3 − 끫뢬4 = 0
Plug in the known values and solve for i2:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
끫뤔끫뾠 = 끫뾞끫뾜끫뾜 끫렸끫렸끫렸(끫뾪끫뾪 + 끫뾤) − 끫뾞끫뾜 끫렸끫롘끫롘(끫뾪끫뾪 − 끫뾞. 끫뾠) − 끫뾞끫뾞끫뾜 끫렸끫롘끫롘(끫뾪끫뾪 + 끫뾠) 끫뤰끫뤰끫뤰끫뤦끫뤰 끫뾪 = 끫뾢끫뾪끫뾪끫뤦끫뾢끫뤊/끫뤦
Problem 4.51
Determine the voltage across R2 in the circuit of
Figure P4.51.
i (t) = 20 cos(533.33t) A, R1 = 8OHM, R2 = 16ohm,
L = 15mH, C = 117μF.
Solution:
Known quantities:
I (t ) = 20 cos(533.33t )A , R1 = 8Ω, R2 = 16Ω, L = 15mH , C = 117 µF
Find:
VR2 (t )
Analysis:
X L = 8Ω, X C = 16Ω
R2 || C ⇒ Z eq1 =
Z R2 ⋅ Z C
Z R2 + Z C
= 8 − j8
Z eq 2 = Z eq1 + Z L = 8
Using divider current law:
IL = I ⋅
Z R1
8
= 20
= 10
Z R1 + Z eq 2
8+8
VR2 = I L Z eq1 = 80 − j80 ⇒ VR2 (t ) = 113.13 ⋅ cos(533.33t − 0.79 )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.52
Determine the frequency so that the current Ii and
the voltage Vo in Figure P4.52 are in phase.
Zs = 13,000 + jω3ohm
R = 120ohm
L = 19mH C = 220 pF
Solution:
Known quantities:
The values of the impedance in the circuit shown in Figure
P4.52, 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!
V
Z eq = 0 = Z eq ∠0 o = Req + jX eq , X eq (ω ) = 0
Ii
Z + Z L ) ⋅ ZC (R + jX L ) ⋅ (− jXC ) X L XC − jRXC R − j (X L − XC )
(
Z eq = R
=
=
=
Z R + Z L + ZC
R + jX L − jXC
R + j (X L − XC ) R − j (X L − XC )
=
[X L XC R − RXC (X L − XC )]− j [R 2 XC + X L XC (X L − XC )]
R 2 + (X L − XC )
At the resonant frequency the reactive component of this impedance must equal zero:
R 2 XC + X L XC (X L − XC )
X eq (ω ) =
= 0 ⇒ R 2 + X L (X L − XC ) = 0
2
R 2 + (X L − XC )

1 
L
2 2
2
R 2 + ωL ωL −
= 0 ⇒ ω L = −R

ωC 
C
ω=
1
R2
−
=
LC L2
2
(120 Ω )2 = 239.24G rad − 39.89 M rad = 489.1k rad
1
−
(19mH )(220pF ) (19mH )2
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
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.53
A common model for a practical inductor is a coil
resistance in series with an inductance L. The coil
resistance accounts for the internal losses of an
inductor. Figure P4.53 shows an ideal capacitor in
parallel with a practical inductor. Determine the
current supplied by the source vs . Assume:
vs (t) = Vo cos(ωt + 0)
Vo = 10V ω = 6Mrad/s Rs = 50_
RC = 40ohm L = 20 μH C = 1.25 nF
Solution:
Known quantities:
The values of the impedance, Rs = 50 Ω , Rc = 40 Ω ,
L = 20 µH , C = 1.25 nF , and the voltage applied to the
circuit shown in Figure P4.53,
rad
.
v s (t ) = V0 cos ωt + 0 o , 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 ) = 1203Ω ⇒ Z L = 0 + j120 Ω = 120∠90 Ω

s 
1
1
=
XC =
= 133.3 Ω ⇒ ZC = 0 − j133.3 Ω = 133.3∠ − 90 o Ω
rad 
ωC 
6 M
(1.25 nF)

s 
Z R c = 40 − j Ω = 40∠0 o Ω , Z R s = 50 − j Ω = 50∠0 o Ω
Equivalent impedances:
Z eq1 = Z R c + Z L = 40 + j120 Ω = 126.5∠71.56 o Ω
Z eq = Z R s +
ZC ⋅ Z eq1
ZC + Z eq1
= 50 + j0 Ω +
OL:
Is =
(133.3∠ − 90 Ω)⋅ (126.5∠71.56 Ω)=
o
= 50 + j 0 Ω +
o
133.3∠ − 90 o Ω + 126.5∠71.56 o Ω
16.87∠ − 18.44 o kΩ 2
42.161∠ − 18.44 o Ω
= 50∠0 o Ω + 400∠0 o Ω = 450∠0 o Ω
(
)
10∠0 o V
Vs
=
= 22.22∠0 o mA ⇒ 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.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.54
Use phasor techniques to solve for the current i (t)
shown in Figure P4.54.
Solution:
Known quantities:
The values of the impedance and the voltage applied to the
circuit shown in Figure P4.54.
Find:
The current in the circuit.
Analysis:
Assume clockwise currents:
rad
, VS = 12∠0 o V
ω=3
s
1
ZC =
= − j Ω , Z L = jωL = j 9 Ω ⇒ Z total = 3 + j9 − 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
(
)
A
Problem 4.55
Use phasor techniques to solve for the voltage v(t)
shown in Figure P4.55.
Solution:
Known quantities:
The values of the impedance and the current source shown in
Figure P4.55.
Find:
The voltage.
Analysis:
Assume clockwise currents:
1
rad
= − j 1.5Ω
, I S = 10∠0 o A , Z L = jωL = j 6 Ω , ZC =
ω=2
jωC
s
1
1
1
Z eq =
=
=
= 0.9231− j1.3846 Ω
1
1
1
1
1
2 0.33+ j0.5
+
+
−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 4.56
Solve for I1 in the circuit shown in Figure P4.56.
I = 20-π/4A, R = 3ohm, Z1 = −j3ohm, Z2 = −j7ohm
Solution:
Known quantities:
The values of the impedance and the current source for
the circuit shown in Figure P4.56.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Find:
The current I1.
Analysis:
Specifying the positive directions of the currents as in figure P4.56:
Z eq1 = C1 + C 2 = − j10Ω
Use Current Division:
끫롸1 =
Solve to get:
끫뢚끫뢤끫뢤1
∗ 끫롸
끫뢊 + 끫뢚끫뢤끫뢤1
끫룠끫뾞 = 끫뾞끫뾞. 끫뾮∠ − 끫뾪끫뾠°
Problem 4.57
Solve for VR shown in Figure P4.57. Assume:
ω = 3 rad/sec V = 13∠0V
R = 15ohm L1 = 7H L2 = 2H
Solution:
Known quantities:
The values of the impedance and the voltage source for circuit
shown in Figure P4.57.
Find:
The voltage V2.
Analysis:
Specifying the positive directions as in figure P4.57:
Z L = jω ( L1 + L2 ) = j 27 Ω
13Ω
V
R=
VR =
15∠0 o V = 6.31∠ − 63o V
(15 + j 27 )Ω
R + ZL
Problem 4.58
Using Figure P4.55, find the value of w for which the current
through the resistor is maximum.
Solution:
Known quantities:
The values of the components in Figure P4.55 and the voltage source.
Find:
Find the value of w for which the current through the resistor is maximum.
Analysis:
Use Current Division to find an equation for w:
3
270 + 끫뢮90 �3끫븨 − �
끫븨
끫롸끫뢊 =
9
9끫븨 2 + 2 − 9
끫븨
Using guess and check the following values can be found:
끫븨 = 3 끫롸끫뢊 = 3.69 + 끫뢮9.86
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
끫븨 = 2 끫롸끫뢊 = 9.23 + 끫뢮13.85
끫븨 = 1 끫롸끫뢊 = 3.3
끫븨 = 1.1 끫롸끫뢊 = 28.94 + 끫뢮5.53
As the value of 1 is approached the current through the resistor increased, however when ω is equal to 1, the current
is very low. There is a discontinuity in the equation at 1 so the value for ω should be slightly higher than 1.
Problem 4.59
Find the current iR(t) through the resistor shown in
Figure P4.59.
Solution:
Known quantities:
The values of the impedance and the current source for circuit
shown in Figure P4.59.
Find:
The current through the resistor.
Analysis:
Specifying the positive directions as in figure P4.59:
By current division:
IR = −
1
R
1 1
+
R ZC
⋅ Is = −
1
1
1− jωRC
⋅ Is = −
⋅ Is = −
⋅ Is =
2
R
1+jωRC
1+(ωRC )
1+
jωC
= −(0.0247 − j0.1552) A ⋅1∠0 o A = 157 ⋅10−3∠99.04 o A
(
)
i R (t ) = 157 cos 200πt + 99.04 o mA
Problem 4.60
Find vout(t) shown in Figure P4.60.
Solution:
Known quantities:
The values of the reactance X L = 1 kΩ , XC = 10 kΩ , and
the current source I = 10∠45o mA for circuit shown in Figure
P4.60.
Find:
The voltage vout .
Analysis:
Specifying the positive directions of the currents as in Figure P4.60:
Vout = Z eq I = (Z L + Z C )I = (0 + jX L + 0 − jX C )I = ( j1kΩ − j10 kΩ ) ⋅ 10∠45 o mA =
= (− j 9 kΩ ) ⋅ 10∠45 o mA = 9∠ − 90 o kΩ ⋅ 10∠45 o mA = 90∠135 o V
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.61
Find the impedance Z shown in Figure P4.61,
assuming ω = 2 rad/s, R1 = R2 = 2ohm C = 0.25 F,
and L = 1H.
Solution:
Known quantities:
The circuit shown in Figure P4.61, the values of the resistances, capacitance, inductance and the frequency.
Find:
The impedance Z .
Analysis:
1
= − j 2Ω
j ωC
1
2
−2j
= ZL +
=2j+
=
Ω
1
1
1− j 1− j
+
Z C R1
Z L = j ωL = j 2 Ω , Z C =
Z eq1 = Z L + Z C Z R2
Z eq =
1
1
Z eq 2
+
1
R2
=
4
2
+ j Ω
5
5
Problem 4.62
Find the sinusoidal steady-state output vout(t) for
each circuit shown in Figure P4.62.
Solution:
Known quantities:
The circuit shown in Figure P4.62, the values of current source and voltage source and components in the circuit.
Find:
The sinusoidal steady-state outputs.
Analysis:
For circuit a):
 1

∠-90° 
Vout = 10∠0° × 
−3
 π10

10000
=
∠-90° = 3183∠-90°
π
vout (t ) = 3183 cos(100πt −90° ) V
For circuit b):
Vout = 20∠0° × 1∠90° = 20∠90°
vout (t ) = 20 sin (10t + 90°) = 20 cos10t V
For circuit c):
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
− j1000
× 50∠0° ≈ 50∠0°
j 0.1 − j1000
vout (t ) = 50.0 sin 100 t V
Vout
=
Problem 4.63
Determine the voltage vL (t) across the inductor
shown in Figure P4.63.
Solution:
Known quantities:
The circuit shown in Figure P4.63, the values of voltage source
and components in the circuit.
Find:
The voltage across the inductor.
Analysis:
We have
jϖL = j (1000 )(0.003) = j 3
By voltage division:
j3
(24)
4 + j3
= 8.64 + j11.52 = 14.4∠53.13°
VL =
Problem 4.64
Determine the current iR(t) through the capacitor
shown in Figure P4.64.
Solution:
Known quantities:
The circuit shown in Figure P4.64, the values of current source and components in the circuit.
Find:
The current through the capacitor.
Analysis:
We have C = 100 μF , R = 100 Ω
By current division:
100
× 1∠0°
IR = −
− j106
100 +
100(100π )
= -0.9080 - 0.2890j = 0.953∠ − 162.3°
iR (t ) = 2.859 cos(100π t − 162.3°) mA
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.65
Find the frequency that causes the equivalent
impedance Zeq in Figure P4.65 to be purely resistive.
Solution:
Known quantities:
The circuit shown in Figure P4.65, the values of resistance,
inductor and capacitor.
Find:
The frequency causes the equivalent impedance to appear to be purely resistive
Analysis:

1
1 
= 15 + j  0.001ω − − 6 
jω
10 ω 

This impedance is purely resistive if its imaginary part (the reactance) is zero.
1
. So ω = 31,622.77 rad/s
Therefore, we solve for the frequency, ω, at which 0.001ω =
−6
10 ω
The series impedance of the circuit is Z eq = R + jω +
Problem 4.66
a. Find the equivalent impedance Z0 seen by the
source in Figure P4.66(a). Assume the frequency is
377 rad/s.
b. What capacitance should be placed between the
terminals a and b, as shown in Figure P4.66(b), to
make the equivalent impedance Z0 purely resistive?
[Hint: Find C so that the phase angle of Z0 is zero.]
c. What is the amplitude of Z0 when the capacitor is
included?
Solution:
Known quantities:
The circuit shown in Figure P4.66, the values of resistance and
inductor.
Find:
a)
The equivalent impedance seen by the source, if the
frequency is 377 rad/s
b)
The value of capacitor to make the equivalent impedance
purely resistitive
c)
The actual impedance with the capacitor included
Analysis:
(
)
a) Z L = R + jX L = 1 + j 377 × 13.26 × 10 −3 = 1 + j 5
b) Z L = Z L||Z C , Z C =
1
1
=
jωC j 377C
1
(1+j5)
(1+j5) =
1+j 5
j 377C
such that Z L=
=
1
1+j 377C (1+j 5) 1-1885C+j 377C
+(1+j 5)
j 377C
 5
 377C 
The impedance angle is: ∠Z L' = arctan  − arctan

1 
 1- 1885C 
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
In order to have a strictly real impedance we set ∠Z L' = 0 , or:
5
 5   377C 
= 510.1 μF
 , which leads to: 5 = (377 + 9425)C or C =
 =
9802
 1   1-1885C 
1
(1+j5)
j 0.1923
c) Z L=
= 26
1
+(1+j5)
j 0.1923
Problem 4.67
A common model for a practical capacitor has a
”leakage” resistance, RC , in parallel with an ideal
capacitor, as shown in Figure P4.67. The effects of
lead wires are also represented by resistances R1 and
R2 and inductances L1 and L2.
a. Assume C = 1μF, RC = 100 Mohm,
R1 = R2 = 1μ_ and L1 = L2 = 0.1μH and find
the equivalent impedance Zab seen across terminals
a-b as a function of frequency ω.
b. Find the range of frequencies for which Zab is
capacitive.
[Hint: Assume that RC is is much greater than 1/ωC such
that RC can be ignored in part b.]
Solution:
Known quantities:
The circuit shown in Figure P4.67, the values of resistance and
capacitor.
Find:
a)
b)
Equivalent impedance
The range of frequency for which Z ab is capacitive
Analysis:
a) Let Z1 denote the impedance of the parallel R-C combination:
1
RC
RC
1
jwC
=
=
Z1 = RC||
1
jwRC + 1
jwC
RC +
jwC
Let R1 = R2 = R/ 2 and L1 = L2 = L/ 2 . The total equivalent
resistance, Zab, is given by:
Z ab = R1 + jwL1 + Z + R2 + jwL2
= R + jwL +
[
RC
jwRC + 1
]
= R + RC − w 2 LRC + j[w(R RC
+
L )]
= Re(Z ab ) + j Im(Z ab )
b)
If we set Im(Zab ) = 10Re(Z ab ) and solve for ω, we will find the minimum frequency at which the impedance of the
physical capacitor may be considered pure imaginary. Using the result of part a),
[
]
10 R + RC − w 2 LRC = [w(R RC
10 LRC w 2
+
+
L )] Thus, we need to solve the quadratic equation
(RRC + L ) w - 10(R + RC ) = 0
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
or
20 w 2 + 20 w − 1 × 108 = 0
which has roots w = -2 ,236.6 and w = 2 ,236.6 . Selecting the positive root for physical reason, we conclude that
physical capacitor will act very nearly like an ideal capacitor for frequencies above 2.2356×103 rad/s (355.6 Hz).
4.47
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition 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) to phasor form.
Represent each circuit element by its impedance.
Solve the resulting phasor circuit, using appropriate circuit analysis tools.
Convert the (phasor-form) answer to its time-domain equivalent, using equation 4.50.
Problem 4.68
Using phasor techniques, solve for VR2 shown in
Figure P4.68.
i (t) = 3 cos(200t) A, R1 = 3ohm, R2 = 5ohm, L = 18 mH, C = 170 μF.
Solution:
Known quantities:
Circuit shown in Figure P4.68, the values of the resistance, capacitance,
inductance and the current source.
Find:
R2 using phasor tehcniques.
The voltage across the
Analysis:
ω = 200 Rad / s
Z L = jωL = j 3.6 Ω
ZC =
1
= − j 29.41Ω
jωC
Z eq1 = R2 || Z C =
Z R2 ⋅ Z C
Z R2 + Z C
=
− j147
Ω = 4.85 − j 0.82
5 − j 29.41
Z eq 2 = Z eq1 + Z L = 4.85 + j 2.77
IL = I ⋅
Z R1
Z R1 + Z eq 2
= 1.02 − j 0.36
V R2 = I L ⋅ Z eq1 = 4.65 − j 2.59V
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.69
Use phasor techniques to solve for iL in the circuit
shown in Figure P4.69.
i1(t) = 5 cos(500t) A, i2(t) = 5cos(500t) A, R = 5ohm,
C = 2mF, L = 2mH.
Solution:
Known quantities:
Circuit shown in Figure P4.69, the values of the resistances,
capacitance, inductance and the current source.
Find:
The current through the inductance i L .
Analysis:
1
rad
, Z L = j ωL = j Ω , Z C =
= − j Ω , Z R L = j ωL + R = 5 + j Ω
s
jωC
5 − 25 j
Z RC
26
=5
= 0.18 − j 0.97
I L' = I 1 ⋅
5 − 25 j
Z RC + Z RL
+ j+5
26
5 − 25 j
Z RC
26
=5
= 0.18 − j 0.97
I L'' = I 2 ⋅
5 − 25 j
Z RC + Z RL
+ j+5
26
I L = I L' + I L'' = 0.36 − j1.94 ⇒ i L (t ) = 1.97 cos(500t − 79.49°) A
ω = 2500
Problem 4.70
Determine the Thévenin equivalent network seen
by the load Ro in Figure P4.70. When:
RS = Ro = 500ohm L = 10mH R = 1kohm
and:
a. vS(t) = 10 cos(1,000t)
b. vS(t) = 10 cos(1,000,000t)
Solution:
Known quantities:
Circuit shown in Figure P4.70 the values of the resistance,
RS = RL = 500 Ω , R = 1 kΩ and the inductance,
L = 10 mH .
Find:
a) The Thèvenin equivalent circuit if the voltage applied
to the circuit is v S (t ) = 10 cos(1,000t ).
b) The Thèvenin equivalent circuit if the voltage applied
to the circuit is v S (t ) = 10 cos(1,000,000t ).
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Analysis:
rad
⋅10 mH = j10 Ω ,
s
The equivalent impedance is:
Z ⋅R
( j10)1000 + 500 = 500 + j10 3 = 500.1 + j 9.999 Ω
ZT = L
+ RS =
j10 + 1000
100 + j
(Z L + R)
Z L = jωL = j1000
a)
The equivalent Thèvenin voltage is: VT = VS = 10∠0 o V
rad
⋅10 mH = j10 4 Ω ,
s
The equivalent impedance is:
Z L = jωL = j10 6
b)
ZT =
( )
j10 4 1000
ZL ⋅ R
j10 4
+ RS =
+ 500 = 500 +
= 1490.1 + j99.01 Ω
4
1 + j10
(Z L + R)
j10 + 1000
The equivalent Thèvenin voltage is: VT = VS = 10∠0 o V
Problem 4.71
Determine the Norton equivalent network seen by
the capacitor in Figure P4.71. Use the result and
current division to find iC (t). Assume
i (t) = 0.5 cos(300t) A, R1 = R2 = 40ohm,
L1 = L2 = 200 mH, and C = 15μF.
Solution:
Known quantities:
Circuit shown in Figure P4.71 the values of the impedances
and the current source.
Find:
Using Thèvenin theorem iC (t )
Analysis:
Z TH = ( R1 + Z L 2 ) || ( Z L 2 + R2 ) = (40 + j 60) || (40 + j 60) =
=
− 50 + j120
2 + j3
I
( R1 − Z L2 ) = 10 − j15
2
E0
65
=
= 0.0827 + j 0.0435 = 0.0934e j 27.74 ⇒
IC =
Z TH + Z C 616 − j 324
E0 = VR1 − VZ L 2 =
⇒ iC (t ) = 0.0934 cos(300t + 270.74) A
Problem 4.72
Use phasor techniques to solve for iL(t) in Figure
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
P4.72. Assume vS(t) = 2 cos2t V, R1 = R2 = 4ohm,
L = 2H, and C = 0.25 F.
Solution:
Known quantities:
Circuit shown in Figure P4.72 the values of the resistance, R1 = 4 Ω , R2 = 4Ω , capacitance,
inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V .
C =1 4 F,
Find:
The current in the circuit i L (t ) using phasor techniques.
Analysis:
VS (t ) = 2∠0 o V
1
1
ZC =
=
= − j2 Ω
jωC j 2 1
4
Z L = jωL = j 2 ⋅ 2 = j4 Ω
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:
V
1.05∠18.4°
IL = L =
= 0.2635∠ − 71.6 o A
ZL
4∠90°
(
i L (t ) = 0.2635cos 2t − 71.6 o
)A
Problem 4.73
Use mesh analysis to determine the currents I1(t)
and I2(t) in Figure P4.73. Assume:
V1 = 10e−j40 V, V2 = 12e j40 V, R1 = 8ohm, R2 = 4ohm
R3 = 6ohm,XL = 10ohm, XC = −14_ìohm
Solution:
Known quantities:
Circuit shown in Figure P4.73, the values of the resistances, capacitance, inductance and the voltage source.
Find:
The currents in the circuit
I 1 (t ) and I 2 (t ) .
Analysis:
Mesh 1 :
끫뾤끫뾞 − 끫룠끫뾞 ∗ 끫룲끫뾞 − 끫룠끫뾞 ∗ 끫뤂끫룔끫뾞 − (끫룠끫뾞 − 끫룠끫뾠 ) ∗ 끫룲끫뾠 + 끫뾤끫뾠 = 끫뾜
끫롸1 =
(16.85 + 끫뢮1.28) + 끫롸2 (4)
(12 − 끫뢮14)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Mesh 2
−끫뾤끫뾠 − 끫룠끫뾠 ∗ 끫룲끫뾢 − 끫룠끫뾞 ∗ 끫뤂끫룦끫뾞 − (끫룠끫뾠 − 끫룠끫뾞 ) ∗ 끫룲끫뾠 = 끫뾜
끫롸1 =
(9.19 + 끫뢮7.71) + 끫롸2 (10 + 끫뢮10)
(4)
Solving the system, we find
I 1 = 0.45 + j 0.68
I 2 = −0.61 + j 0.12
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.74
Use nodal analysis to determine the node voltages
va(t) and vb(t) shown in Figure P4.74. Assume:
i (t) = 2 cos(300t) A, v(t) = 7 cos(300t+π/4) V, R1 = 4ohm,R2 = 3ohm, R3 = 5phm, L = 300 mH, and C = 300 μF.
Solution:
Known quantities:V
Circuit shown in Figure P4.74, the values of the resistances, capacitance, inductance and the current and voltage
source.
Find:
The voltages in the circuit v a (t ) and vb (t ) .
Analysis:
-j
1
=
= -j3.7 Ω ,
jωC 300 ⋅ 900 ⋅ 10 - 6
Z L = jωL = j 300 ⋅ 0.3 = j90 Ω
ZC =
Applying KCL at node 1, we have:
−I +
V2 V2 − V1 V2 − V1
+
+
=0
R1
R2
ZC
Applying KCL at node 2, we have
V1 − V2 V1 V1 − V2 V1 − V
+
+
+
=0
R2
ZL
ZC
R3
Therefore, solving the linear system :
 V1 = 4.25e j 42.91 ⇒ v1 (t ) = 4.25 cos(300t + 42.91°)V

j 57.27
⇒ v 2 (t ) = 2.94 cos(300t + 57.37°)V
V2 = 2.94e
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.75
The circuit shown in Figure P4.75 is a Wheatstone
bridge, which can be used to determine the reactance
X4 of an inductor or capacitor. R1 and R2 are adjusted
until vab is zero.
a. Assume a balanced bridge (vab = 0) and determine
X4 in terms of the other circuit elements.
b. Assume a balanced bridge and let C3 = 4.7μF,
L3 = 0.098 H, R1 = 100ohm, R2 = 1ohm, and
vS(t) = 24 sin(2,000t). What is the reactance of
the unknown circuit element? Is it a capacitor or an
inductor? What is its value?
c. What frequency should be avoided by the source in
this circuit, and why?
Solution:
Known quantities:
The circuit called Wheatstone bridge shown in Figure P4.75.
a) The balanced status for the bridge: v ab = 0 .
b) The values of the resistance, 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 .
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:
Assuming a balanced circuit, we have v ab = 0 , that is, v a = vb
R2
jX 4
R2
jX 4
From the voltage divider:
=
⇒
=
j
jX L 3 - jXC 3 + R2 R1 + jX 4
jωL3 + R2 R1 + jX 4
ωC 3
Inverting both sides and equating imaginary parts:

1 
R1R2
R1R2 = −ωL3 +
X ⇒ X 4 =

 1
ωC 3  4

− ωL 3 


 ωC 3
b)
100 ⋅1
= −1.116 Ω
X4 =


1
2000
⋅
0.098



 2000 ⋅ 4.7 ⋅10−6
Negative reactance implies that the component is a capacitor.
1
1
= 1.116Ω ⇒ C =
= 448 µF
ωC
ω ⋅1.116
c)
If the reactances of L3 and C3 cancel, the bridge cannot measure X4. Thus, the condition to be avoided is:
1
1
1
1
rad
ωL3 −
= 0 ⇒ L3C 3 = 2 ⇒ ω =
=
= 1473
−6
s
ωC 3
C
L
ω
3 3
0.098 ⋅ 4.7 ⋅10
f = 234.5 Hz
a)
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.76
Find the Thévenin equivalent network seen by the
capacitor C in Figure P4.76. Use the result and voltage
division to determine VC (t). Assume:
v(t) = cos(300t) V, i (t) = 2 cos(300t) A, R1 = 8ohm,
R2 = 8ohm, L = 3μH, and C = 5μF.
Solution:
Known quantities:
Circuit shown in Figure P4.72, the values of the resistances,
capacitance, inductance and the voltage and current
sources.
Find:
Using equivalent Thévenin circuit find Vc (t ) .
Analysis:
1
= − j 666Ω
ωC
Z L = jωL = j 0.0009Ω
ZC = . j
64 + j 0.0072
16 + j 0.0009
Z L + R2
8 + j 0.0009
=
Vo' = V
Z L + R1 + R2 16 + j 0.0009
136 + j 0.0009
'
''
 Vo = Vo + Vo =
R2 ⋅ R1
16 + j 0.0009
128
=
Vo'' = I
Z L + R1 + R2 16 + 0.0009
Z TH = ( R2 + Z L ) || R1 =
VC = Z C ⋅
Vo
= 8.4997 − j 0.0515 = 8.49e − j 0.3469 ⇒ vc (t ) = 8.49 cos(300t − 0.34°)V
Z TH + Z C
Problem 4.77
Determine the Thévenin equivalent circuit seen by
the load Zo shown in Figure P4.77. Assume:
Vs = 10∠0◦ V, RS = 40ohm, XL = 40ohm, and
XC = −2000ohm.
Solution:
Known quantities:
Circuit shown in Figure P4.74, the values of the
resistances, capacitance, inductance and the voltage
source.
Find:
The Thévenin circuit seen by the load
Analysis:
We compute the Thèvenin equivalent impedance with the
load removed:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Z T = Z S + Z L || Z C = 40 + j 39.22 = 56.02e j 44.44
Thèvenin equivalent voltage is equal to the source voltage, since once the load impedance is removed, no current
flows in the circuit:
VT = VS = 10V
Problem 4.78
Find the Thévenin equivalent network seen across
terminals a-b in Figure P4.77.
Solution:
Known quantities:
Circuit shown in Figure P4.78, the values of the impedance,
R = 8Ω , ZC = − j8 Ω , Z L = j8 Ω , and the voltage source Vs = 5∠ − 30 o V .
Find:
The Thévenin equivalent circuit seen from the terminals a-b.
Analysis:
The Thévenin equivalent circuit is given by:
 8 + j8 
o
VTH = 
5∠ − 30° = (1+ j )5∠ − 30 = 7.07∠15 V
 8 + j8 − j8 
(8 + j8)(− j8) = 8 − j8 = 8 2∠ − 45o Ω
ZTH =
(
)
8 + j8 − j8
Problem 4.79
Determine the Norton equivalent network seen by
the capacitor in Figure P4.79. Use the result and
current division to find iC (t). Assume:
vs (t) = 4 cos(100t) V, R1 = 7ohm, R2 = 8ohm, L = 30 mH,
and C = 10 mF.
Solution:
Known quantities:
Circuit shown in Figure P4.79, the values of the
resistances, capacitance, inductance and the voltage
source.
Find:
iC (t ) .
Analysis:
1
= − jΩ
j ωC
Z L = jωL = j 3Ω
ZC =
Find Zrl:
Find Zeq:
끫뢚끫뢊끫뢊 = 8 + 끫뢮3
끫뢚끫뢤끫뢤 = 끫뢚끫뢊끫뢊 ||끫뢊 = 3.86 + 끫뢮0.63
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Find Norton Equivalent current using resistance Zt:
4
= 1 − 끫뢮0.16끫롨
끫뢚끫뢤끫뢤
Zn=Zt so use current division to find current through the capacitor:
끫뤂끫뤰끫뤌
끫룠끫룔 = 끫룠끫룪 ∗
= 끫뾜. 끫뾜끫뾪 + 끫뤖끫뾜. 끫뾜끫뾞
끫뤂끫뤰끫뤌 + 끫뤂끫룔
끫롸끫뢂 =
Problem 4.80
Find the Thévenin equivalent network seen by R2
in Figure P4.80. Use the result and voltage
division to
determine the voltage v2(t) across R2. Assume:
v(t) = 70 cos(275t) V, R1 = R2 = 42ohm, L =
1mH, and
C = 12 μF.
Solution:
Known quantities:
Circuit shown in Figure P4.72, the values of
the resistances, capacitance, inductance and
the voltage source.
Find:
TheThevenin equivalent circuit seen by the resistor R2 and the Voltage across it.
Analysis:
Z L = jωL = j 0.275Ω
1
= − j 300Ω
ωC
Z L ( Z C + R1 )
= 0.27e j 93°
=
Z L + ( Z C + R1 )
ZC = − j
Z eq
Use Voltage division:
끫뾤끫룲 = 끫뾤끫룶 ∗
끫룲끫뾠
= 끫뾪끫뾮. 끫뾮끫뾮 + 끫뤖끫뾜. 끫뾤끫뾪
끫뤂끫뤰끫뤌 + 끫룲끫뾠
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Problem 4.81
Use mesh analysis to solve for the phasor mesh
currents shown in Figure P4.81.
Solution:
Known quantities:
Circuit shown in Figure P4.81.
Find:
The equations required to solve for the loop currents in the
circuit in:
a. Integral-differential form;
b. Phasor form.
Analysis:
1 t
∫ (i1 − i 2 )dt + (i1 − i 2 )R1 = 0
C 0
di
1
KVL: (i 2 − i1)R1 − v c (0) + ∫ 0t (i 2 − i1)dt + L 2 + i 2 R2 = 0
dt
C
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 + I1RS + (I1 − I 2 )ZC + (I1 − I 2 )R1 = 0
KVL: (I 2 − I1)R1 + (I 2 − I1)ZC + +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!
KVL:
−v S + i1RS + v c (0) +
Problem 4.82
Write the node equations required to solve for all
voltages and currents in the circuit of Figure P4.81.
Assume all impedances and the two source voltages
are known.
Solution:
Known quantities:
Circuit shown in Figure P4.81.
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 (ZC + R1) + I 3 (Z L + R2 )
Problem 4.83
Find the Voltage across the resistor using Figure P4.83.
Assume:
v(t) = 4cos(1000t+π/6) V, Ro 120ohm, L = 60mH, and
C = 12.5 μF.
4.59
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 4
Solution:
Known quantities:
Circuit shown in Figure P4.83, the values of the resistances, capacitance, inductance and the voltage source.
Find:
The Voltage across Ro
Analysis:
Find Zc and Zl:
끫뢮
= −끫뢮80
1000 ∗ 12.5 ∗ 10−6
−3
끫뢚끫뢊 = 1000 ∗ (60 ∗ 10 ) = 끫뢮60
끫뢚끫롬 = −
Find Equivalent resistance seen by the Voltage source:
끫뢚끫뢤끫뢤 = (끫뢊끫뢸 �|끫뢚끫뢊 + 끫뢚끫롬 )|�끫뢚끫롬 = 0.32 − 끫뢮19.46 끫뢄ℎ끫뢴끫룀
Find total current:
끫뢒끫뢬
끫롸끫뢎 =
= 0.21 − 끫뢮15.65끫롨
끫뢚끫뢤끫뢤
Use Current Division twice to find the current through Ro:
끫뢚끫롬
끫롸끫뢊 = 끫롸끫뢎 ∗
= −6.39 − 끫뢮13.55
끫뢚끫롬 + 끫뢚끫뢊||끫롬+끫뢊
끫뢚끫롬
= −8.21 − 끫뢮1.26끫롨
끫롸끫뢊 = 끫롸끫뢊 ∗
끫뢚끫롬 + 끫뢚끫뢊
Multiply Current by R to find Voltage:
끫뾤끫룲 = 끫룠끫룲 ∗ 끫룲 = 끫뾮끫뾞끫뾞. 끫뾠 − 끫뤖끫뾞끫뾞끫뾞. 끫뾠끫뾤
Problem 4.84
Using figure P4.84 find the value of one of the four unknown passive components.
Solution:
Known quantities:
Circuit shown in Figure P4.84
i1(t) = 3.127 cos(ωt − 0.825) A
i2(t) = 3.914 cos(ωt − 1.78) A
i3(t) = 1.900 cos(ωt + 0.655) A
vS1(t) = 130.0 cos(ωt + 0.176) V
vS2(t) = 130.0 cos(ωt − 0.436) V
Find:
The value of one of the following: L1, L3, C2, R3
Analysis:
Use mesh analysis:
4.60
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