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Problem 3.2 Apply nodal analysis to determine Vx in the circuit of Fig. P3.2.
1 Ω
2 Ω
4 Ω
2 Ω
3 A Vx
V
+
_
Figure P3.2: Circuit for Problem 3.2.
Solution: At node V , application of KCL gives
V2+1
−3+V
2+4= 0,
which leads toV = 6 V.
By voltage division,
Vx =V ×42+4
=6×4
6= 4 V.
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Problem 3.3 Use nodal analysis to determine the currentIx and amount of powersupplied by the voltage source in the circuit of Fig. P3.3.
2 Ω
4 Ω
8 Ω
40 V9 A
Ix
V
+_
I
Figure P3.3: Circuit for Problem 3.3.
Solution: At nodeV , application of KCL gives
−9+V2
+V4
+V −40
8= 0
V
(
12
+14
+18
)
= 9+408
7V8
= 9+5
V = 16 V.
The currentIx is then given by
Ix =V4
=164
= 4 A.
To find the power supplied by the 40-V source, we need to first find the current Iflowing into its positive terminal,
I =V −40
8=
16−408
= −3 A.
Hence,P = V I = 40× (−3) = −120 W
(The minus sign confirms that the voltage source is a supplier of power.)
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Problem 3.4 For the circuit in Fig. P3.4:
(a) Apply nodal analysis to find node voltagesV1 andV2.
(b) Determine the voltageVR and currentI.
1 Ω 1 Ω1 Ω
1 Ω 1 Ω
VR+_
16 V
V1 V2 I
+_+_
Figure P3.4: Circuit for Problem 3.4.
Solution: (a) At nodesV1 andV2,
Node 1:V1−16
1+
V1
1+
V1−V2
1= 0 (1)
Node 2:V2−V1
1+
V2
1+
V2
1= 0 (2)
Simplifying Eqs. (1) and (2) gives:
3V1−V2 = 16 (3)
−V1 +3V2 = 0. (4)
Simultaneous solution of Eqs. (3) and (4) leads to:
V1 = 6 V, V2 = 2 V.
(b)
VR = V1−V2 = 6−2 = 4 V
I =V2
1=
21
= 2 A.
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Problem 3.10 The circuit in Fig. P3.10 contains a dependent current source.Determine the voltage Vx.
2Vx3 Ω
2 Ω
6 Ω6 V Vx
Vx
+
_
+_+_
Figure P3.10: Circuit for Problem 3.10.
Solution: In terms of the node voltage Vx, KCL gives
Vx −62
+Vx
3−2Vx +
Vx
6= 0,
whose solution leads toVx = −3 V.
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Problem 3.14 Apply nodal analysis to find the currentIx in the circuit of Fig. P3.14.
0.1 Ω
0.1 Ω
0.5 Ω0.5 Ω
0.2 Ω
0.1 Ω
3 V2 V
4 V
Ix
+_
+_
+_
+_
+ _
V1 V2 V3
Figure P3.14: Circuit for Problem 3.14.
Solution: Application of KCL to the designated node voltagesV1, V2, andV3 gives
V1−20.1
+V1−V2
0.5+
V1−V3−40.2
= 0 (1)
V2−V1
0.5+
V2
0.1+
V2−V3
0.5= 0 (2)
V3−V1 +40.2
+V3−V2
0.5+
V3−30.1
= 0 (3)
Simplification, followed with simultaneous solution, leads to
V1 = 2.865 V, V2 = 0.625 V, V3 = 1.51 V,
and
Ix =V2
0.1=
0.6250.1
= 6.25 A.
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Problem 3.26 Apply mesh analysis to find the mesh currents in the circuit ofFig. P3.26. Use the information to determine the voltageV .
2 Ω
4 Ω
3 Ω
2 Ω
I1 I216 V
12 V
V
+_ +
_
Figure P3.26: Circuit for Problem 3.26.
Solution: Application of KVL to the two loops gives:
Mesh 1: −16+2I1 +3(I1− I2) = 0,
Mesh 2: 3(I2− I1)+(2+4)I2 +12= 0,
which can be simplified to
5I1−3I2 = 16 (1)
−3I1 +9I2 = −12. (2)
Simultaneous solution of (1) and (2) leads to
I1 = 3 A, I2 = −13
A.
Hence,
V = 3(I1− I2) = 3
(
3+13
)
= 10 V.
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Problem 3.31 Apply mesh analysis to determine the amount of power supplied bythe voltage source in Fig. P3.31.
3 Ω 6 Ω
2 Ω
2 Ω 4 Ω
4 A
48 V
+_
I1 I2
I3
Figure P3.31: Circuit for Problem 3.31.
Solution:
Mesh 1: 2I1 +3(I1− I3)+2(I1− I2)+48= 0
Mesh 2: −48+2(I2− I1)+6(I2− I3)+4I2 = 0
Mesh 3: I3 = −4 A.
Solution is:I1 = −8.4 A, I2 = 0.6 A, I3 = −4 A.
Current entering “+” terminal of voltage source is:
I = I1− I2 = −8.4−0.6 = −9 A.
Hence,P = V I = 48× (−9) = −432 W.
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Problem 3.34 Apply mesh analysis to the circuit in Fig. P3.34 to determine Vx.
2Vx3 Ω
2 Ω
6 Ω6 V
+_+_ I1 I2 I3
+
_Vx
+_
Figure P3.34: Circuit for Problem 3.34.
Solution:
Mesh 1: −6+2I1 +3(I1 − I2) = 0Supermesh: 3(I2 − I1)+6I3 = 0Auxiliary 1: I3 − I2 = 2Vx
Auxiliary 2: Vx = 6I3
Solution is:
I1 = 4.5 A, I2 = 5.5 A, I3 = −0.5 A.
Vx = 6I3 = 6× (−0.5) = −3 V.
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Problem 3.52 Apply the by-inspection method to develop a node-voltage matrixequation for the circuit in Fig. P3.52 and then use MATLAB R© or MathScript softwareto solve for V1 and V2.
6 Ω
12 Ω
6 Ω 3 A4 A2 A
V1 V2
Figure P3.52: Circuit for Problem 3.52.
Solution:
Node 1: G11 =
(
16
+1
12
)
= 0.25
G12 = G21 = −112
= −0.083
G22 =
(
16
+1
12
)
= 0.25
It1 = 2+4 = 6 A
It2 = 3−4 = −1 A
Application of Eq. (3.26) gives:[
0.25 −0.083−0.083 0.25
][
V1
V2
]
=
[
6−1
]
Matrix inversion gives
V1 = 25.5 V, V2 = 4.5 V.
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Problem 3.53 Use the by-inspection method to establish a node-voltage matrixequation for the circuit in Fig. P3.53. Solve the matrix equation by MATLABR©
or MathScript software to findV1 to V4.
8 Ω9 Ω
5 Ω
3 Ω2 Ω
6 Ω1 Ω 4 Ω
7 Ω3 A
2 A
V1
V4V2V3
Figure P3.53: Circuit for Problem 3.53.
Solution:
G11 =1
2+1+
13+4
= 0.476
G12 = G21 = −1
2+1= −0.333
G13 = G31 = 0
G14 = G41 = −1
3+4= −0.143
G22 =1
1+2+
17
+16
= 0.643
G23 = G32 = −16
= −0.167
G24 = G42 = 0
G33 =15
+16
+19
= 0.478
G34 = G43 = −15
= −0.2
G44 =1
3+4+
15
= 0.343
Application of Eq. (3.26) gives:
0.476 −0.333 0 −0.143−0.333 0.643 −0.167 0
0 −0.167 0.478 −0.2−0.143 0 −0.2 0.343
V1
V2
V3
V4
=
20−2−3
Matrix inversion gives:
V1 =−8.1689 V, V2 =−8.4235 V, V3 =−16.155 V, V4 =−21.5748 V.
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Problem 3.60 Determine the current Ix in the circuit of Fig. P3.60 by applying thesource-superposition method. Call I′x the component of Ix due to the voltage sourcealone, and I′′x the component due to the current source alone. Show that Ix = I′x + I′′xis the same as the answer to Problem 3.9.
3 Ω 6 Ω
2 Ω
2 Ω 4 Ω
4 A
48 V
+_
Ix
Figure P3.60: Circuit for Problem 3.60.
Solution:
A. Voltage source alone:
3 Ω 6 Ω
2 Ω
2 Ω 4 Ω
48 V
Ix
+_
‘
I1 I2‘ ‘
7I′1 −2I′2 = −48
−2I′1 +12I′2 = 48
I′1 = −6 A, I′2 = 3 A.
HenceI′x = I′2 = 3 A.
B. Current source alone:
3 Ω 6 Ω
2 Ω2 Ω 4 Ω
4 A
Ix‘‘‘‘
‘‘I1 ‘‘I2
I3
I′′3 = −4 A
7I′′1 −2I′′2 −3I′′3 = 0
−2I′′1 +12I′′2 −6I′′3 = 0
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Hence,
I′′1 = −2.4 A, I′′2 = −2.4 A.
I′′x = I′′2 − I′′3 = −2.4+4 = 1.6 A.
Ix = I′x + I′′x = 3+1.6 = 4.6 A.
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Problem 3.61 Apply the source-superposition method to the circuit in Fig. P3.61to determine:
(a) I′x, the component of Ix due to the voltage source alone
(b) I′′x , the component of Ix due to the current source alone
(c) The total current Ix = I′x + I′′x(d) P′, the power dissipated in the 4-Ω resistor due to I′x(e) P′′, the power dissipated in the 4-Ω resistor due to I′′x(f) P, the power dissipated in the 4-Ω resistor due to the total current I. Is P =
P′+P′′? If not, why not?
2 Ω
4 Ω
8 Ω
40 V9 A
Ix +_
Figure P3.61: Circuit for Problem 3.61.
Solution:(a) Voltage source alone:
2 Ω
4 Ω
8 Ω
40 V
+_
V1
Ix‘
‘
V ′1
2+
V ′1
4+
V ′1 −40
8= 0.
Hence,
V ′1 =
407
V.
I′x =40
7×4=
107
A.
(b) Current source alone:
2 Ω
4 Ω
8 Ω
9 A
Ix
V1
‘‘
‘‘
V ′′1
2+
V ′′1
4+
V ′′1
8= 9.
V ′′1 =
727
V.
I′′x =V ′′
1
4=
724×7
=187
A.
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(c)
Ix = I′x + I′′x =107
+187
=287
= 4 A.
(d)
P′= (I′x)
2R =
(
107
)2
×4 =40049
W = 8.16 W.
(e)
P′′= (I′′x )
2R =
(
187
)2
×4 = 26.45 W.
(f)P = I2
x R = 42 ×4 = 64 W.
P 6= P′+P′′ because the superposition theorem does not apply to power.
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Problem 3.64 Find the Thevenin equivalent circuit at terminals(a,b) for the circuitin Fig. P3.64.
1 Ω
2 Ω 3 Ω
4 Ω
2 Ω
3 A
a
b
Voc
V
+
_
Figure P3.64: Circuit for Problem 3.64.
Solution:
V3
+V6
= 3
V = 6 V.
Voltage division gives
VTh = Voc =V6×4 = 4 V.
Suppressing the current source:
1 Ω
2 Ω 3 Ω
4 Ω
2 Ωa
b
RTh
RTh
RTh
5 Ω
3 Ω
4 Ω
a
b
3 Ω
Ω
a
b
20
9
RTh = 3+209
=479
= 5.2 Ω.
Thevenin equivalent circuit:
5.2 Ωa
b
4 V
+_
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Problem 3.73 Find the Norton equivalent circuit at terminals(a,b) for the circuitin Fig. P3.73.
0.2 Ω
0.2 Ω
0.1 Ω 0.25 Ω
a
b
I0
0.2I0
Iex
Vex
+_I1 I2 I3
+_
Figure P3.73: Circuit for Problem 3.73.
Solution: The circuit contains no independent sources. Hence,
VTh = 0.
To determineRTh, we add an external voltage sourceVex and proceed to findIex.
0.1I1 +0.2(I1− I2)−0.2I0 = 0
0.2I0 +0.2(I2− I1)+0.2I2 +0.25(I2− I3) = 0
0.25(I3− I2)+Vex = 0
Additionally, I0 = I1.Solution is:
I1 = −5Vex, I2 = −2.5Vex, I3 = −6.5Vex,
Iex = −I3 = 6.5Vex
RTh =Vex
Iex=
16.5
= 0.15 Ω
Hence,
a
b
0.15 Ω
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Problem 3.81 What value of the load resistorRL will extract the maximum amountof power from the circuit in Fig. P3.81, and how much power will that be?
2 Ω
4 Ω 6 Ω
8 Ω
4 Ω
3 A RL
a
b
Figure P3.81: Circuit for Problem 3.81.
Solution: We start by obtaining the Thevenin equivalent circuit at terminals(a,b),as if RL were not there. We first findVoc:
V6−3+
V12
= 0
V = 12 V.
Hence,
2 Ω
4 Ω 6 Ω
8 Ω
4 Ω
3 A
a
bVoc
V
+
_
Voltage division gives:
VTh = Voc =
(
84+8
)
V =812
×12= 8 V.
Next, we suppress the current source to findRTh:
2 Ω
4 Ω 6 Ω
8 Ω
4 Ωa
b
RTh
Simplification leads to:RTh = 10.44 Ω.
Equivalent circuit:
RL
10.44 Ω
8 V
I
+_
For maximum power transfer toRL ,
RL = RTh = 10.44 Ω
I =8
2×10.44= 0.38 A
Pmax = I2RL = (0.38)2×10.44= 1.53 W.
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Problem 3.82 For the circuit in Fig. P3.82, choose the value ofRL so that the powerdissipated in it is a maximum.
2 kΩ
6 kΩ
4 kΩ
8 kΩ
2 mA
RL
a
b
Figure P3.82: Circuit for Problem 3.82.
Solution: We need to find the Thevenin equivalent circuit at terminals(a,b), as ifRL
were not present.
2 kΩ
6 kΩ
4 kΩ
8 kΩ
2 mA
a
b
Voc
I1
I2
+
_
The current source will divide amongI1 andI2 such that
(4+2)I1 = (8+6)I2
Also, I1 + I2 = 2 mA.The solution yields:
I1 = 1.4 mA, I2 = 0.6 mA.
Voc = (−4I1 +8I2)×103
= −4×1.4+8×0.6 = −0.8 V.
To find RTh, we suppress the current source and simplify the circuit:
2 kΩ
6 kΩ
4 kΩ
8 kΩ
a
b
RTh = 8 kΩ || 12 kΩ = 4.8 kΩ
Hence,RL should be 4.8 kΩ for maximum power transfer to it.
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Problem 3.83 Determine the maximum power that can be extracted by the loadresistor from the circuit in Fig. P3.83.
2000Ix
6 kΩ
3 kΩ
4 kΩ
RL
Ix
15 V
+_
+_
Figure P3.83: Circuit for Problem 3.83.
Solution: To find the Thevenin equivalent circuit, we start by determiningVTh = Voc.
2000Ix
6 kΩ
3 kΩ
4 kΩ
Ix
15 V
+_
+_
Voc
a
b
V1
+
_
Voltage division:
V1 =15
(3+6)k×6k = 10 V
Ix =V1
6k=
106
mA.
The dependent voltage source is:
2000Ix = 2×106×103×10−3
=206
V.
With (a,b) an open circuit, no current flows through the 4-kΩ resistor. Hence, thereis no voltage drop across it.
VTh = Voc = V1−2000Ix = 10−206
=406
= 6.67 V.
2000Ix
6 kΩ
3 kΩ
4 kΩ
15 V
+_
+_
Isc
a
b
I1 I2
Next, we findIsc:
−15+3kI1 +6k(I1− I2) = 0
6k(I2− I1)+4kI2 +2000Ix = 0
Also,Ix = I1− I2
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Solution yields:
I1 = 2.5 mA, I2 = 1.25 mA.
Isc = I2 = 1.25 mA.
RTh =Voc
Isc=
6.671.25×10−3 = 5.33 kΩ.
Hence,RL = 5.33 kΩ extracts maximum power.
RL
RTh = 5.33 kΩ
6.67 V
I
+_ 5.33 kΩ
I =6.67
2×5.33= 0.625 mA
Pmax = I2RL = (0.625×10−3)2×5.33×103
= 2.09 (mW).
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Problem 3.87 Determine the maximum power extractable from the circuit inFig. P3.87 by the load resistor RL.
200I0
1 kΩ
2 kΩ
2 kΩ
I0
RL
+_
Figure P3.87: Circuit for Problem 3.87.
Solution: The circuit has no independent sources. Hence,
VTh = 0.
Consequently, RL cannot extract any power from the circuit.
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Problem 3.88 In the circuit of Fig. P3.88, what value ofRs would result inmaximum power transfer to the 10-Ω load resistor?
2 A RLRs 10 Ω
Figure P3.88: Circuit for Problem 3.88.
Solution: Maximum power transfer toRL occurs when all of the 2 A flowsthroughRL , requiringRs to be∞.
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