Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 9(8).
Find Vx in the network shown in Fig. 13.78.
Figure 13.78
Chapter 13, Solution 9(8).
Consider the circuit below.
2Ω
8∠30o
+
–
I1
2Ω
j4
j4
I2
-j1
-j2V
+
–
For loop 1,
8∠30° = (2 + j4)I1 – jI2
For loop 2,
(1)
((j4 + 2 – j)I2 – jI1 + (–j2) = 0
or
Substituting (2) into (1),
I1 = (3 – j2)i2 – 2
8∠30° + (2 + j4)2 = (14 + j7)I2
I2 = (10.928 + j12)/(14 + j7) = 1.037∠21.12°
Vx = 2I2 = 2.074∠21.12°
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 21(14).
Find I1 and I2 in the circuit of Fig. 13.90. Calculate the power absorbed by the
4-Ω resistor.
Figure 13.90
Chapter 13, Solution 21(14).
For mesh 1, 36∠30° = (7 + j6)I1 – (2 + j)I2
For mesh 2,
(1)
0 = (6 + j3 – j4)I2 – 2I1 jI1 = –(2 + j)I1 + (6 – j)I2
Placing (1) and (2) into matrix form,
36∠30°  7 + j6 − 2 − j  I1 
 0  = − 2 − j 6 − j  I 

 
 2 
∆ = 48 + j35 = 59.41∠36.1°,
∆1 = (6 – j)36∠30° = 219∠20.54°
∆2 = (2 + j)36∠30° = 80.5∠56.56°,
I1 = ∆1/∆ = 3.69∠–15.56°,
I2 = ∆2/∆ = 1.355∠20.46°
Power absorbed fy the 4-ohm resistor, = 0.5(I2)24 = 2(1.355)2 = 3.672 watts
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
1
Chapter 13, Problem 22(15).
Find current Io in the circuit of Fig. 13.91.
Figure 13.91
Chapter 13, Solution 22(15).
With more complex mutually coupled circuits, it may be easier to show the effects of the
coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure
13.91 then becomes,
-j50
Io
I3
j20Ic
+ −
j40
j10Ib
j60
+ −
Ia
j30Ic
− +
−
+
Ix
j30Ib
− +
j20Ia
50∠0° V
+
−
j80
I1
I2
Ib
−
+
j10Ia
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100 Ω
2
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Note the following,
Ia = I 1 – I3
Ib = I2 – I1
Ic = I3 – I2 and
Io = I3
Now all we need to do is to write the mesh equations and to solve for Io.
Loop # 1,
-50 + j20(I3 – I2) j 40(I1 – I3) + j10(I2 – I1) – j30(I3 – I2) + j80(I1 – I2) – j10(I1 – I3) = 0
j100I1 – j60I2 – j40I3 = 50
Multiplying everything by (1/j10) yields 10I1 – 6I2 – 4I3 = - j5
(1)
Loop # 2,
j10(I1 – I3) + j80(I2–I1) + j30(I3–I2) – j30(I2 – I1) + j60(I2 – I3) – j20(I1 – I3) + 100I2 = 0
Loop # 3,
-j60I1 + (100 + j80)I2 – j20I3 = 0
(2)
-j50I3 +j20(I1 –I3) +j60(I3 –I2) +j30(I2 –I1) –j10(I2 –I1) +j40(I3 –I1) –j20(I3 –I2) = 0
-j40I1 – j20I2 + j10I3 = 0
Multiplying by (1/j10) yields,
-4I1 – 2I2 + I3 = 0
(3)
Multiplying (2) by (1/j20) yields -3I1 + (4 – j5)I2 – I3 = 0
Multiplying (3) by (1/4) yields
-I1 – 0.5I2 – 0.25I3 = 0
(4)
(5)
Multiplying (4) by (-1/3) yields I1 – ((4/3) – j(5/3))I2 + (1/3)I3 = -j0.5 (7)
Multiplying [(6)+(5)] by 12 yields
(-22 + j20)I2 + 7I3 = 0
(8)
Multiplying [(5)+(7)] by 20 yields
-22I2 – 3I3 = -j10
(9)
(8) leads to I2 = -7I3/(-22 + j20) = 0.2355∠42.3o = (0.17418+j0.15849)I3
(9) leads to I3 = (j10 – 22I2)/3, substituting (1) into this equation produces,
I3 = j3.333 + (-1.2273 – j1.1623)I3
or
I3 = Io = 1.3040∠63o amp.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 26(19).
Find Io in the circuit of Fig. 13.95. Switch the dot on the winding on the right and
calculate Io again.
Fig. 13.95.
Chapter 13, Solution (26)19.
M = k L1L 2
ωM = k ωL1ωL 2 = 0.6 20x 40 = 17
The frequency-domain equivalent circuit is shown below.
j17
50 Ω
200∠60
+
−
–j30
I1
Io
j2
j4
I2
10 Ω
For mesh 1,
200∠60° = (50 – j30 + j20)I1 + j17I2 = (50 – j10)I1 + j17I2
For mesh 2,
0 = (10 + j40)I2 + j17I1
j17   I1 
200∠60° 50 − j10
=
In matrix form,


0
10 + j40 I 2 

  j17
(1)
(2)
∆ = 900 + j100, ∆1 = 2000∠60°(1 + j4) = 8246.2∠136°, ∆2 = 3400∠–30°
I2 = ∆2/∆ = 3.755∠–36.34°
Io = I2 = 3.755∠–36.34° A
Switching the dot on the winding on the right only reverses the direction of Io.
This can be seen by looking at the resulting value of ∆2 which now becomes
3400∠150°. Thus,
Io = 3.755∠143.66° A
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 43(30).
Obtain V1 and V2 in the ideal transformer circuit of Fig. 13.107.
Chapter 13, Solution 43(30).
Transform the two current sources to voltage sources, as shown below.
10 Ω
+
20 V
+
–
I1
Using mesh analysis,
12 Ω
1:4
+
v1
−
v2
−
I2
12V
+
–
–20 + 10I1 + v1 = 0
20 = v1 + 10I1
12 + 12I2 – v2 = 0 or 12 = v2 – 12I2
At the transformer terminal, v2 = nv1 = 4v1
I1 = nI2 = 4I2
(1)
(2)
(3)
(4)
Substituting (3) and (4) into (1) and (2), we get,
Solving (5) and (6) gives
20 = v1 + 40I2
(5)
12 = 4v1 – 12I2
(6)
v1 = 4.186 V and
v2 = 4v = 16.744 V
Copyright ©2004 The McGraw-Hill Companies Inc.
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 44(31).
In the ideal transformer circuit of Fig. 13.108, find i1(t) and i2(t).
Chapter 13, Solution 44(31).
We can apply the superposition theorem.
Let i1 = i1’ + i1” and i2 = i2’ + i2” where the single prime (i1’, i2’) is due to the
DC source and the double prime (i1”, i2”) is due to the AC source.
Since we are looking for the steady-state values of i1 and i2,
i1’ =V0/R
i2’ = 0. (No induction on the secondary at DC)
For the AC source, consider the circuit below.
R
1:n
+
i1”
V2/V1 = –n,
+
v1
v2
−
−
i2”
+
–
I2”/I1” = +1/n
But V2 = Vm, V1 = –Vm/n
I1” = - V1/R= Vm/(Rn)
I2” = I1”/n = –Vm/(Rn2)
Hence,
i1(t) = V0/R+ (Vm/Rn)cosωt A, (DC and AC together)
i2(t) = (Vm/(n2R))cosωt A
(Only AC)
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Vm∠0°
Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 56(40).
Find the power absorbed by the 10-Ω resistor in the ideal transformer circuit of
Fig. 13.120.
Chapter 13, Solution 56(40).
We apply mesh analysis to the circuit as shown below.
2Ω
1:2
+
+
v1
46V
+
−
I1
v2
−
−
I2
10 Ω
5Ω
For mesh 1,
46 = 7I1 – 5I2 + v1
(1)
For mesh 2,
v2 = 15I2 – 5I1
(2)
At the terminals of the transformer,
v2 = nv1 = 2v1
I1 = nI2 = 2I2
(3)
(4)
Substituting (3) and (4) into (1) and (2),
46 = 9I2 + v1
v1 = 2.5I2
Combining (5) and (6),
46 = 11.5I2 or I2 = 4
P10 = 0.5I22(10) = 80 watts.
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Fundamentals of Electric Circuits, Second Edition - Alexander/Sadiku
Chapter 13, Problem 61(45).
For the circuit in Fig. 13.125 below, find I1, I2, and Vo.
Chapter 13, Solution 61(45).
We reflect the 160-ohm load to the middle circuit.
ZR = ZL/n2 = 160/(4/3)2 = 90 ohms, where n = 4/3
2Ω
1:5
I1
+
24∠0°
+
–
v1
−
Io
14 Ω
Io ’
+
vo
−
14 + 60||90 = 14 + 36 = 50 ohms
We reflect this to the primary side.
ZR’ = ZL’/(n’)2 = 50/52 = 2 ohms when n’ = 5
I1 = 24/(2 + 2) = 6A
24= 2I1 + v1 or v1 = 24 – 2I1 = 12 V
vo = –nv1 = –60 V,
Io = –I1 /n1 = –6/5 = –1.2
Io‘ = [60/(60 + 90)]Io = –0.48A
I2 = –Io’/n = 0.48/(4/3) = 0.36 A
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60 Ω
90 Ω
1