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Chapter 4, Solution 55.
Known quantities:
The values of the impedance and the current source shown in
Figure P4.55.
Find:
The voltage.
Analysis:
Assume clockwise currents:
rad
1
ω=2
, I S = 10∠0 o A , Z L = jωL = j 6 Ω , ZC =
= − j 1.5Ω
s
jωC
1
1
1
=
=
= 0.9231− j1.3846 Ω
Z eq =
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
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 68.
Known quantities:
Circuit shown in Figure P4.68, the values of the resistance, R = 9 Ω, capacitance, C = 1 18 F , inductance,

π
L1 = 3 H , L2 = 3 H , L3 = 3 H , and the voltage source v s (t ) = 36 cos 3t −  V .

3
Find:
The voltage across the capacitance
Analysis:
v using phasor tehcniques.
rad
, Vs = 36∠ − 60° V
s
ω=3
Z L2 = jωL2 = j3⋅ 3 = j9 Ω
ZC =
1
1
=
= − j6 Ω
jωC j 3 ⋅ (1 18)
Z L3 = jωL3 = j 3⋅ 3 = j9 Ω
Z eq =
(
1
Z L3 Z L2 + ZC
)
=
j9
1
1
=
=
= 2.25∠90 o Ω
1
1
1
1
4
+
+
Z L3
j 9 j3
Z L2 + ZC
(
)
ZT = Z R + Z L1 + Z eq = 9 + j3 ⋅ 3 + j2.25 = 9 + j11.25 = 14.407∠51.34 o Ω
36∠ − 60 o V
V
= 2.499∠ −111.34 o A
I= S =
ZT 14.407∠51.34 0 Ω
Veq = IZ eq = (2.499∠ −111.34 o )(2.25∠90 o ) = 5.623∠ − 21.34 o V
ZC
− j6
Veq =
5.623∠ − 21.34 o = 11.25∠158.66 o V
V=
j3
Z L + ZC
(
2
(
)
)
v = 11.25cos 3t −158.66 o V
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 75.
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 L3 - jXC 3 + R2 R1 + jX 4
jωL3 + R2 R1 + jX 4
ωC 3
Inverting both sides and equating imaginary parts:

1 
R1R2
R1R2 = −ωL3 +
X4 ⇒ X4 =
 1

ωC 3 

− 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.
a)
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
ωC 3
s
L3C 3
ω
0.098 ⋅ 4.7 ⋅10−6
f = 234.5 Hz
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 61.
Known quantities:
Circuit shown in Figure P7.61, the voltage sources,
˜ = 110∠120° V, V
˜ = 110∠240° V ,
˜ = 110∠0° V, V
V
R
W
B
and the three loads, Z R = 50Ω , ZW = − j20Ω , Z B = j 45Ω .
Find:
a) The current in the neutral wire.
b) The real power.
Analysis:
˜
˜I = VR = 110∠0° = 2.2∠0° A
R
50
ZR
˜
V
110∠240°
˜I = B =
= 2.44∠150° A
B
j45
ZB
˜
˜I = VW = 110∠120° = 5.5∠210° A
W
− j 20
ZW
˜I = I˜ + ˜I + I˜ = 2.2 + 5.5∠210° + 2.44∠150° = 4.92∠ −161.9° A
N
R
W
B
~
b) P = R ⋅ I R2 = 50 ⋅ 2.2 2 = 242W
a)
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 78.
Known quantities:
Circuit shown in Figure P4.78, the values of the impedance,
R = 8 Ω , 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
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 73.
Known quantities:
Circuit shown in Figure P4.73, 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 i 2 (t ).
Analysis:
In the phasor domain:
-j
2000
ZC =
=−j
= − j 666.7 Ω , Z L = j (1500)(0.5) = j750 Ω
-6
3
1500(1 ×10 )
By applying KVL in the first loop, we have
VS = R1I 1 + ZC (I 1 − I 2 )
By applying KVL in the second loop, we have
0 = (ZC )(I 2 − I 1 ) + (Z L + R2 )I 2
That is:

2000 
2000
o 
I2
I 1 + j
15∠0 =  75 − j


3 
3

0 = j 2000 I + 100 + j 250 I
 2
1 


3
3 
By solving above equations, we have
I 1 = 3.8 ⋅10−3 ∠46.6 o A
I 2 = 19.6 ⋅10−3∠ − 83.2 o A
i1 (t ) = 3.8 cos (1,500t + 46.6o ) mA
i2 (t ) = 19.6 cos (1,500t − 83.2o ) mA
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 27.
Known quantities:
Circuit shown in Figure P7.27, the values of the resistances, R1 = 8 Ω , R2 = 6 Ω , the reactances,
~
˜ = 36∠ − π 3 V , V
XC = −12 , X L = 6 , and the voltage sources, V
S1
S 2 = 24∠0.644V .
Find:
a).The active and reactive current for each source b). The total real power.
Analysis:
a)
From Figure P7.13:
VS1 = R1I1 + jX L (I1 − I 2 ) = (8 + j 6)I1 − j6I 2
−VS 2 = − jX L (I1 − I 2 ) + R2 I 2 + XC I 2 = − j 6I1 + (6 − j6)I 2
Substituting the values for the voltages sources gives:
18 − j31.2 = (8 + j 6)I1 − j6 I 2

−19.2 − j14.4 = − j6 I1 + (6 − j 6)I 2
Solving for I1 and I2 yields:
I1 = 0.398 − j 3.38 A

I 2 = 1.091− j0.911A
Therefore, the active and reactive currents for each source are:
I A 1 = 0.398 A
I R 1 = 3.38 A
and 

I A 2 = 1.091A
I R 2 = 0.91 A
b)
P = R2 I 22 + R1I12 = 6 ⋅1.4212 + 8 ⋅ 3.4032 = 105 W
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 74.
Known quantities:
Circuit shown in Figure P4.74, the values of the resistance, R1 = 40 Ω , R2 = 10 Ω ,
capacitance, C = 500 µF , inductance, L = 0.2 H ,
and the current source i s (t ) = 40 cos(100t ) A.
Find:
The voltages in the circuit v1(t ) and v 2 (t ).
Analysis:
ZC =
1
-j
=
= -j20 Ω ,
jω C 100 ⋅ 500 ⋅ 10- 6
Z L = jω L = j100 ⋅ 0.2 = j20 Ω
Applying KCL at node 1, we have:
1
1
V V − V2
1 
1
j 
j
⇒ IS =  +
V2 ⇒ 40∠0 o =  + V1 − V2
IS = 1 + 1
V1 −
 40 20 
ZC
20
R1
ZC
 R1 ZC 
Applying KCL at node 2, we have
 1
V
1
1 
V 1
1
1 
V1 − V2 V2 V2
=
+
⇒ 1 =
+
+
+ j V2
V2 ⇒ j 1 =  − j
20
20 
ZC
R2 Z L
Z C  R2 Z L Z C 
20  10
Therefore:

j 
j
o  1

1
V2
j 
j
40∠0 =  + V1 −

40∠0 o =  + (− j2V2 ) −
V2
 40 20 
20
⇒ 
⇒

 40 20 
20
 j V1 =  1 V
V = − j 2V
 1
2
 20  10  2

1
j
1
j
j
40∠0 o = − V2 + V2 −
V2 =  − V2

 10 10 
20
10
20
V = − j2V
 1
2
V2 =
o
40∠0 o
= 282.84∠45o V, V1 = − j 2V2 = 565.68∠ − 45 V
1
j 
 − 
 10 10 
v 2 (t ) = 282.84 cos (100t + 45o ) V, v1 (t ) = 568.68 cos (100t - 45o ) V
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 84.
Known quantities:
Circuit shown in Figure P4.84, the values of the resistance, RL = 120 Ω , the capacitance, C = 12.5 µF , and
the inductance, L = 60 mH , and the voltage source,
(
)
v i = 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.
1
1
=−j
= − j80 Ω = 80∠ − 90 o Ω

rad 
ωC
1, 000 k
(12.5 ΩF)

s 

rad 
o
Z L = jX L = jωL = j 1, 000 k
(60 mH ) = j 60 Ω = 60∠90 Ω

s 
ZC = − jXC = − j
Reference phasor: Vi = 4∠30 o V
V0 − 0 V0 − 0 V0 − Vi
KCL:
+
+
=0
ZL
Z RL
ZC
Vi
ZL
Vi
4∠30 o V
V0 =
=
=
=
o
o
1
ZL
1
1
Z
+
+
+ L +1 60∠90 Ω + 60∠90 Ω +1
Z R L ZC
Z R L ZC Z L
120∠0 o Ω 80∠ − 90 o Ω
=
4∠30 o V
0.5∠90 o + 0.75∠180 o +1
=
4∠30 o V
=
(0 + j0.5) + (−0.75+ j 0) + (1+ j0)
4∠30 o V
4∠30 o V
=
= 7.155∠ − 33.43o V
0.25+ j 0.5 0.559∠63.43o
v 0 (t ) = 7.155cos ωt − 33.43o V
=
(
)
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 65.
Known quantities:
Circuit shown in Figure P7.65, the voltage sources, v s1 (t ) = 170 cos(ωt ) V , v s2 (t ) = 170 cos(ωt +120°) V ,
v s3 (t ) = 170 cos(ωt − 120°) V, and the impedances, Z1 = 0.5∠20°Ω , Z 2 = 0.35∠0° Ω , Z 3 = 1.7∠ − 90° Ω ,
the frequency, f = 60 Hz .
Find:
The current through Z1, using:
a) Loop/mesh analysis.
b) Node analysis.
c) Superposition.
Analysis:
a)
Applying KVL in the upper mesh:
(
)
~
~
~
~ ~
~
~
~
~
Vs 2 − Vs1 + I1 Z 1 + I1 − I 2 Z 2 = 0 ⇒ I1 (Z 1 + Z 2 ) + I 2 (− Z 2 ) = Vs1 − Vs 2
Applying KVL in the lower mesh:
˜ + ˜I − I˜ Z + I˜ Z = 0 ⇒ ˜I (−Z ) + I˜ (Z + Z ) = V
˜ −V
˜
˜ −V
V
s3
s2
2
1 2
2 3
1
2
2 2
3
s2
s3
For each mesh equation:
˜ = 170∠0° −170∠120° = 170 − (−85+ j147) = 294∠ − 30° V
˜ −V
V
s1
s2
˜ = 170∠120° − 170∠ −120° = (−85+ j147) − (−85− j147) = 294∠90° V
˜ −V
V
(
s2
)
s3
Z1 + Z 2 = 0.47 + j0.171+ 0.35 = 0.838∠11.8°Ω
Z 2 + Z 3 = 0.35− j1.7 = 1.74∠ − 78.4° Ω
Therefore, the current through Z1 is:
~
~
Vs1 − Vs 2
− Z2
294∠ − 30°
− 0.35∠0°
~
~
Vs 2 − Vs 3 Z 2 + Z 3
294∠90° 1.74∠ − 78.4°
~
I1 =
=
0.838∠11.8°
− 0.35∠0°
Z1 + Z 2
− Z2
− 0.35∠0° 1.74∠ − 78.4°
− Z2
Z2 + Z3
512∠ − 108.4° + 103∠90° 415.9∠ − 112.9°
=
= 293∠ − 41.8°A
1.46∠ − 66.6° − 0.123∠0° 1.416∠ − 71.2°
Choose the ground at the center of the three voltage source, and let a be the center of the three
b)
loads. The voltage between the node a and the ground is unknown.
Applying KCL at the node a:
˜
˜
˜
˜
˜
˜ −V
V
a
s1 + Va − Vs2 + Va − Vs3 = 0
Z1
Z2
Z3
Rearranging the equation:
˜
˜
˜
V
170∠0° 170∠120° 170∠ − 120°
s1 + Vs2 + Vs3
+
+
Z
Z
Z
0.5∠20°
1
2
3
0.35∠10° 1.7∠ − 90°
˜ =
V
=
a
1
1
1
1
1
1
+
+
+
+
Z1 Z 2 Z 3
0.5∠20° 0.35∠10° 1.7∠ − 90°
=
340∠ − 20° + 486∠120° + 100∠330°
303∠57.3°
=
= 63.9∠58.5° V
2∠ − 20° + 2.86∠0° + 0.59∠90°
4.74∠ −1.2°
Applying KVL, the current through Z1 is:
˜
˜
˜ − ˜I Z = 0 ⇒ ˜I = Vs1 − Va = 170∠0° − 63.9∠58.5° = 293∠ − 41.8° A
˜ −V
V
a
s1
1 1
1
Z1
0.5∠20°
Superposition is not the method of choice in this case.
=
c)
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 7, Solution 19.
Known quantities:
Circuit as shown in Figure P7.19.
Find:
The value of capacitor when circuit is at unity power
factor.
Analysis:
(a) The source current in the parallel circuit is
The reactive power in the inductor is
QL
IS
=
100 / 2
= 10∠-45° A
5 + j5
2
2
= I S X L = 10 × 5 = 500
VAR
Thus, the capacitive reactance required to cancel the reactive power in the inductor is
The required capacitor is C =
1
= 265.3 µF
377XC
(b) In the series circuit, we can cancel the inductive reactance by setting jωL +
resulting in C =
1
2
ω L
=
1
ωX L
=
1
= 530.5 µF
377 × 5
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
1
= 0,
jω C
XC
=
VS2
= 10
QL
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Chapter 7, Solution 21.
Known quantities:
Circuit as shown in Figure P7.21.
Find:
a) The average power dissipated in the load
b) Motor’s power factor
c) What value of capacitor will change the power factor to 0.9
(lagging).
Analysis:
The current is
120
120
I=
=
= 8.47∠ − 32.14° A
-3
14.17∠32.14°
12 + j 377 × 20 × 10
(a) The average power dissipated in the load is Pav
=
I 2 R = 8.47 2 × 10 = 717.4 W
(b) The power factor of the motor is pf = cos 32.14° = 0.847 lagging
(c) θ = cos −1 0.9 = 25.84°
S NEW ∠25.84° = 717.4 W + j(QL - QC ) S NEW
QL
=
=
797.1
QNEW
450.7 VAR QC = 103.3 VAR
QC =
V 2 120 2
=
= 103.3
XC
XC
X C = 139.4 Ω
C=
1
= 19µF
ωX C
Principles and Applications of Electrical Engineering, 5/e, Giorgio Rizzoni
© 2007 The McGraw-Hill Companies.
= 347.4