Problem 2.1 An AWG-14 copper wire has a resistance of 17.1 Ω at 20◦ C. How
long is it?
Solution: AWG-14 has a diameter of 1.6 mm (Table 2-2), and at 20◦ C, copper’s
conductivity is σ = 5.81 × 107 (S/m) [Table 2-1].
ℓ
σA
ℓ = Rσ A
R=
= Rσ π (d/2)2
1.6 × 10−3
= 17.1 × 5.81 × 10 π ×
2
7
2
≃ 2 km.
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Problem 2.2 A 3-km long AWG-6 metallic wire has a resistance of approximately
6 Ω at 20◦ C. What material is it made of?
Solution:
ℓ
,
AWG-6 has a diameter of 4.1 mm.
σA
ℓ
σ=
RA
3 × 103
ℓ
=
= 3.79 × 107
=
Rπ (d/2)2 6π (4.1 × 10−3 /2)2
R=
(S/m),
which, according to Table 2-1, is approximately the value of the conductivity of
aluminum.
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Problem 2.3 A thin-film resistor made of germanium is 2 mm in length and its
rectangular cross section is 0.2 mm × 1 mm, as shown in Fig. P2.3. Determine the
resistance that an ohmmeter would measure if connected across its:
(a) Top and bottom surfaces
(b) Front and back surfaces
(c) Right and left surfaces
z
2 mm
y
0.2 mm
1 mm
x
Figure P2.3: Film resistor of Problem 2.3.
Solution:
(a)
R=
=
ℓ
σA
ℓ = 0.22 mm,
A = 1 mm × 2 mm = 2 × 10−6 m2
2 × 10−4
≃ 47 Ω.
2.13 × 2 × 10−6
(b)
R=
=
ℓ
σA
ℓ = 1 mm,
A = 2 mm × 0.2 mm = 4 × 10−7 m2
10−3
≃ 1, 174 Ω.
2.13 × 4 × 10−7
(c)
R=
=
ℓ
σA
ℓ = 2 mm,
A = 1 mm × 0.2 mm = 2 × 10−7 m2
2 × 10−3
≃ 4, 695 Ω.
2.13 × 4 × 10−7
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Problem 2.4 A resistor of length ℓ consists of a hollow cylinder of radius a
surrounded by a layer of carbon that extends from r = a to r = b, as shown in
Fig. P2.4.
(a) Develop an expression for the resistance R.
(b) Calculate R at 20◦ C for a = 2 cm, b = 3 cm, and ℓ = 10 cm.
l
Carbon
Hollow
2a 2b
Figure P2.4: Carbon resistor of Problem 2.4.
Solution:
(a) R = σℓA .
The area through which current can flow is the cross section consisting of carbon.
Hence,
A = π b2 − π a2 .
Thus,
R=
ℓ
σ π (b2 − a2 )
.
(b)
R=
0.1
7.14 × 104 π (0.032 − 0.022 )
= 0.89
(mΩ).
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Problem 2.5 A standard model used to describe the variation of resistance with
temperature T is given by
R = R0 (1 + α T ),
where R is the resistance at temperature T (measured in ◦ C), R0 is the resistance at
T = 0◦ C, and α is a temperature coefficient. For copper, α = 4 × 10−3 ◦ C−1 . At
what temperature is the resistance greater than R0 by 1%?
Solution: R = 1.01R0 . Hence,
R
= 1.01 = 1 + α T
R0
α T = 0.01
T=
0.01
= 2.5◦ C.
4 × 10−3
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Problem 2.6 A light bulb has a filament whose resistance is characterized by a
temperature coefficient α = 6 × 10−3 ◦ C−1 (see resistance model given in Problem
2.5). The bulb is connected to a 100-V household voltage source via switch. After
turning on the switch, the temperature of the filament increases rapidly from the
initial room temperature of 20◦ C to an operating temperature of 1800◦ C. When it
reaches its operating temperature, it consumes 80 W of power.
(a) Determine the filament resistance at 1800◦ C.
(b) Determine the filament resistance at room temperature.
(c) Determine the current that the filament draws at room temperature and also at
1800◦ C.
(d) If the filament deteriorates when the current through it approaches 10 A, is the
damage done to the filament greater when it is first turned on or later on when
it arrives at its operating temperature?
Solution:
(a) R = resistance at 1800◦ C.
p=
V2
;
R
R=
V 2 (100)2
=
= 125 Ω.
p
80
(b) R = R0 (1 + α T ).
R0 =
R
125
=
= 10.6 Ω @ 0◦ C.
1 + αT
1 + 6 × 10−3 × 1800
At T = 20◦ C,
R = R0 (1 + α T ) = 10.6(1 + 6 × 10−3 × 20) = 11.9 Ω @ 20◦ C.
(c) At T = 20◦ C,
I=
100
V
=
= 8.43 A.
R(20◦ C) 11.9
At T = 1800◦ C,
p
80
=
= 0.8 A.
V
100
(d) The damage is greater at room temperature because the current is much closer to
10 A.
I=
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Problem 2.7 A 110-V heating element in a stove can boil a standard-size pot of
water in 1.2 minutes, consuming a total of 136 kJ of energy. Determine the resistance
of the heating element and the current flowing through it.
Solution:
W = p ∆t = υ i ∆t
W
136 × 103
=
= 17.2 A.
υ ∆t 110 × 1.2 × 60
υ
110
R= =
= 6.41 Ω.
i
17.2
i=
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Problem 2.8 A certain copper wire has a resistance R characterized by the model
given in Problem 2.5 with α = 4 × 10−3 ◦ C−1 . If R = 60 Ω at 20◦ C and the wire is
used in a circuit that cannot tolerate an increase in the magnitude of R by more than
10 percent over its value at 20◦ C, what would be the highest temperature at which
the circuit can be operated within its tolerance limits?
Solution: The model is given by
R = R0 (1 + α T ).
We are given that R = 60 Ω @ T = 20◦ C, and the maximum R that can be tolerated
is 66 Ω.
At 20◦ C, 60 = R0 (1 + 20α ).
At unknown T , 66 = R0 (1 + α T ).
The ratio gives
66
1 + αT
.
=
60 1 + 20α
Solving for T , we have
T=
6 + 132α
6 + 132 + 4 × 10−3
= 27.2◦ C.
=
60α
60 × 4 × 10−3
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Problem 2.9 The circuit shown in Fig. P2.9 includes two identical potentiometers
with per-length resistance of 20 Ω/cm. Determine Ia and Ib .
Ia
4 cm
7.5 cm
80 mA
10 cm
10 cm
2.5 cm
6 cm
Ib
Figure P2.9: Circuit of Problem 2.9.
Solution: Effectively, the circuit looks as shown in Fig. P2.9(a).
R1
Ia
80 mA
R2
Ib
Fig. P2.9 (a)
Resistors R1 and R2 are:
R1 = (4 + 7.5) × 20 = 230 Ω,
R2 = (6 + 2.5) × 20 = 170 Ω.
Hence, by current division
R2
Ia = (80 mA
R1 + R2
170
= 34 mA,
= 8 × 10−2 ×
400
230
= 46 mA.
Ib = 8 × 10−2 ×
400
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Problem 2.10 Determine VL in the circuit of Fig. P2.10.
5Ω
12 V
+
_
5Ω
+
6Ω
10 Ω
4Ω
5Ω
6Ω
V
_L
5Ω
Figure P2.10: Circuit of Problem 2.10.
The parallel combination of the 4 Ω and 6 Ω resistors is
R=
4×6
= 2.4 Ω.
4+6
By voltage division
2.4
VL = 12
5 + 2.4 + 5
= 2.32 V.
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Problem 2.11 Select the value of R in the circuit of Fig. P2.11 so that VL = 9 V.
12 V
_
I0
R
+
3I0
500 Ω
6 mA
_ VL +
500 Ω
Figure P2.11: Circuit of Problem 2.11.
Solution: The voltage across the 500-Ω resistor in the right-hand segment is
VL = (3I0 + 6 × 10−3 ) × 500.
Setting VL = 9 V leads to
1
I0 =
3
1
=
3
VL
− 6 × 10−3
500
9
− 6 × 10−3
500
= 4 mA.
The left-hand loop has to satisfy KVL:
−12 + I0 R + 500I0 = 0,
which leads to
R=
12 − 500I0 12
=
− 500
I0
I0
12
=
− 500 = 2500 Ω.
4 × 10−3
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Problem 2.12 A high-voltage direct-current generating station delivers 10 MW
of power at 250 kV to a city, as depicted in Fig. P2.12. The city is represented
by resistance RL and each of the two wires of the transmission line between the
generating station and the city is represented by resistance RTL . The distance between
the two locations is 2000 km and the transmission lines are made of 10-cm–diameter
copper wire. Determine (a) how much power is consumed by the transmission line
and (b) what fraction of the power generated by the generating station is used by the
city.
RTL
V0
+
_
RL
(city)
RTL
Station
2000 km
Figure P2.12: Diagram for Problem 2.12.
Solution: For copper,
ρ = 1.72 × 10−8 Ω-m.
For each wire of the transmission line, Eq. (2.2) leads to
RTL = ρ
ℓ
1.72 × 10−8 × 2 × 106
=
A
π (5 × 10−2 )2
= 4.4 Ω.
Delivering 10 MW at 250 kV to the city means that
PL =
VL2
= 107 W,
RL
with VL = 2.5 × 105 V.
Hence,
VL2 (2.5 × 105 )2
=
= 6.25 × 103 Ω.
PL
107
Also, the current flowing through RL is
RL =
VL
2.5 × 105
=
= 40 A.
RL 6.25 × 103
(a) The power consumed by transmission lines is
I=
PTL = 2I 2 RTL
= 2 × (40)2 × 4.4 = 14080 W.
Total power generated by the source is
Ps = PTL + PL = 107 + 14080 = 1.001408 MW,
and the fraction used by the city is
Fraction =
107
PL
=
Ps
1.001408 × 107
= 0.9986
≈ 99.86%.
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Determine the current I in the circuit of Fig. P2.13 given that I0 = 0.
+ V1 _
I
1Ω
V5
_
V3
1Ω
+
I1
+
+
+
_
V4
V2
2 Ω I0 = 0
_
_
24 V
+
1Ω
I2
_
3Ω
a
I1
+
Problem 2.13
I2
1Ω
b
Figure P2.13: Circuit for Problem 2.13.
Solution: Since I0 = 0, the middle branch in the bridge section is of no consequence.
The voltage between nodes a and b is the same following either path between them.
Hence,
I1 × 1 + I1 × 1 = I2 × 1 + I2 × 1,
or I1 = I2 , which is obvious considering that all 4 resistors in the bridge section are
the same. For the left loop that includes the 24-V source,
−24 +V1 +V2 +V3 = 0
V1 = 3I
V2 = I1
V3 = I1
and I = I1 + I2 = 2I1 . Hence,
−24 + 3I + 2I1 = 0
−24 + 3I + I = 0
or
I=
24
= 6 A.
4
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Problem 2.14 Determine currents I1 to I3 in the circuit of Fig. P2.14.
1A
2Ω
3A
a
I2
I1
+
18 V _
12 Ω
8Ω
4Ω
I3
7Ω
b
Figure P2.14: Circuit for Problem 2.14.
Solution: For the loop containing the 18-V source,
−18 + 3 × 2 + 8I1 = 0.
Hence, I1 = 1.5 A.
KCL at node a gives
3 − 1 − I1 − I2 = 0
I2 = 2 − I1 = 2 − 1.5 = 0.5 A.
KCL at node b gives
1 + I2 − I3 = 0
I3 = 1 + I2 = 1 + 0.5 = 1.5 A.
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Problem 2.15 Determine Ix in the circuit of Fig. P2.15.
I
12 V
+
_
Ix
5Ω
1A
2Ω
Figure P2.15: Circuit for Problem 2.15.
Solution:
KVL gives: −12 + 5I + 2Ix = 0.
KCL gives: I + 1 − Ix = 0.
Solution of the two equations yields Ix =
17
7
= 2.43 A.
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Problem 2.16 Determine currents I1 to I4 in the circuit of Fig. P2.16.
4A
1Ω
12 V
8Ω
I1
+
+
_
I3
I2
4Ω
6Ω
1V
+_
+_
+
+
_
5V
I4
Figure P2.16: Circuit for Problem 2.16.
Solution: Application of KVL to the outer-parameter loop gives
−12 + 4 × 1 + 8I3 + 5 − 1 = 0,
which gives
I3 = 0.5 A.
KVL for the left-most loop is
−12 + 4 × 1 + 4I1 = 0,
which leads to
I1 = 2 A.
KCL at the top center node gives
4 − I1 − I2 − I3 = 0
or
I2 = 4 − I1 − I3 = 4 − 2 − 0.5 = 1.5 A.
KCL at the bottom left node gives
I4 + I1 − 4 = 0,
or
I4 = 4 − I1 = 4 − 2 = 2 A.
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Problem 2.17 Determine currents I1 to I4 in the circuit of Fig. P2.17.
I1
I2
2Ω
4Ω
6A
I3
I4
2Ω
4Ω
Figure P2.17: Circuit for Problem 2.17.
Solution: The same voltage exists across all four resistors. Hence,
2I1 = 4I2 = 2I3 = 4I4 .
Also, KCL mandates that
I1 + I2 + I3 + I4 = 6
It follows that I1 = 2 A, I2 = 1 A, I3 = 2 A, and I4 = 1 A.
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Problem 2.18 Determine the amount of power dissipated in the 3-kΩ resistor in the
circuit of Fig. P2.18.
+
V0
_
10 mA
2 kΩ
3 kΩ
10−3V0
Figure P2.42: Circuit for Problem 2.18.
Solution: In the left loop,
V0 = 10 × 10−3 × 2 × 103 = 20 V.
The dependent current source is I0 = 10−3V0 = 20 mA.
The power dissipated in the 3-kΩ resistor is
p = I02 R = (20 × 10−3 )2 × 3 × 103 = 1.2 W.
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Problem 2.19 Determine Ix and Iy in the circuit of Fig. P2.19.
Ix
2Ω
10 V
+
_
6Ω
Iy
I
4Ω
_
+
4Ix
Figure P2.19: Circuit for Problem 2.19.
Solution: Application of KVL to the two loops gives
−10 + 2Ix + 4I = 0
−4I + 6Iy − 4Ix = 0.
Additionally,
I = Ix − Iy .
Solution of the three equations yields
Ix = 3.57 A,
Iy = 2.86 A.
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Problem 2.20 Find Vab in the circuit in Fig. P2.20.
2Ω
a
2Ω
+
_
6V
I
2Ω
+
Vab
_
12 V
+
_
b
Figure P2.20: Circuit for Problem 2.20.
Solution: For the lower loop, KVL gives
−6 + 4I + 12 = 0,
or
I = −1.5 A.
Moving from a to b via the 12-V supply,
Vab = (−1.5) × 2 + 12 = 9 V.
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Problem 2.21 Find I1 to I3 in the circuit of Fig. P2.21.
+
_ 16 V
I3
8V
_
+
3 kΩ I1
I2
4 kΩ
2 kΩ
+
_ 12 V
Figure P2.21: Circuit for Problem 2.21.
Solution: Application of KVL to the outer-perimeter loop gives
−16 + 3 × 103 I1 − 8 + 12 = 0,
which leads to
I1 = 4 mA.
For the left loop,
− 16 + 3 × 103 I1 + 4 × 103 I2 = 0,
I2 =
16 − 3 × 103 I1 16 − 3 × 103 × 4 × 10−3
=
= 1 mA.
4 × 103
4 × 103
Consequently,
I3 = I1 − I2 = 4 − 1 = 3 mA.
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Problem 2.22 Find I in the circuit of Fig. P2.22.
I
10 V
2I
+_
+
_
3Ω
Figure P2.22: Circuit for Problem 2.22.
Solution:
−10 + 2I + 3I = 0.
Hence,
I=
10
= 2 A.
5
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Problem 2.23 Determine the amount of power supplied by the independent current
source in the circuit of Fig. P2.23.
0.2 A
+
V1
_
2Ω
I
2Ω
V1
4
Figure P2.23: Circuit for Problem 2.23.
Solution: KCL at top node gives
0.2 +
V1
−I = 0
4
Also I = V1 /2 (for the 2-Ω resistor).
Hence,
0.2 +
V1 V1
−
=0
4
2
V1 = 0.8 V.
The voltage across the 0.2-A source is 2V1 = 1.6 V.
Power dissipated is
P = V I = 1.6 × 0.2 = 0.32 W.
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Problem 2.24 Given that in the circuit of Fig. P2.24, I1 = 4 A, I2 = 1 A, and I3 =
1 A, determine node voltages V1 , V2 , and V3 .
I2
I1
1 Ω V1
R1 = 18 Ω
6Ω
+
40 V _
V2
6Ω
V3
6Ω
18 Ω
I3
Figure P2.24: Circuit of Problem 2.24.
Solution:
R1 = 18 Ω
1 Ω V1
I4
6 Ω V2
+
40 V _
6Ω
6Ω
V3
18 Ω
Fig. P2.24 (a)
V1 = 40 − (1 Ω) × I1 = 40 − 4 = 36 V.
KCL at node V1 leads to
I4 = I1 − I2 = 4 − 1 = 3 A.
Hence,
V2 = V1 + 6I4 = 36 − 6 × 3 = 18 V.
The voltage across R1 is 18I2 .
Hence,
V3 = V1 − 18I2 = 36 − 18 × 1 = 18 V.
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Problem 2.25 After assigning node V4 in the circuit of Fig. P2.25 as the ground
node, determine node voltages V1 , V2 , and V3 .
12 V
_
3A
+
3Ω
V1
V2
6Ω
3Ω
6Ω
V4
1A
V3
6Ω
1A
Figure P2.25: Circuit of Problem 2.25.
Solution:
12 V
_
3A
+
3Ω
V1
6Ω
I1
V2
3Ω
6Ω
1A
V4
V3
6Ω
1A
Fig. P2.25 (a)
From KCL at node V1 , the sum of currents leaving the node is
3 + I1 − 1 = 0,
or
I1 = −3 + 1 = −2 A.
Node voltages (relative to V4 ):
V1 = −6 × 1 = −6 V,
V2 = V1 − 3I1 = −6 − 3(−2) = 0,
V3 = 6 × 1 = 6 V.
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Problem 2.26 After assigning node V1 in the circuit of Fig. P2.25 as the ground
node, determine node voltages V2 , V3 , and V4 .
Solution:
12 V
_
3A
+
V1
I1
3Ω
3Ω
V2
6Ω
6Ω
V4
1A
V3
6Ω
1A
Fig. P2.26 (a)
From KCL at node V1 , the sum of currents leaving the node is
3 + I1 − 1 = 0,
or
I1 = −3 + 1 = −2 A.
Hence, relative to node V1 :
V2 = −3I1 = −3(−2) = 6 V.
V3 = 12 V
(because (−) terminal of voltage source is at V1 )
V4 = 6(1) = 6 V.
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Problem 2.27 In the circuit of Fig. P2.27, I1 = 42/81 A, I2 = 42/81 A, and
I3 = 24/81 A. Determine node voltages V2 , V3 , and V4 after assigning node V1 as
the ground node.
I3
9Ω
I2
6Ω
6Ω
V1
6V
_
6V
_
V4
I1
9Ω
+
9Ω
V2
6Ω
V3
+
Figure P2.27: Circuit of Problem 2.27.
Solution: At node V3 , KCL gives
I2 + I4 − I3 = 0,
or
24 42
18
−
= − A.
81 81
81
I4 = I3 − I2 =
I3
9Ω
9Ω
I2
V2
6Ω
6Ω
6Ω
V1
6V
_
6V
_
V4
I1
9Ω
+
I4
V3
+
V1 − 6
Fig. P2.27 (a)
V3 = V1 − 6 − 9I2
= 0−6+9×
−4
42
= 1.33 =
V.
81
3
V2 = V3 − 6I4
18
4
= 0,
= − −6 −
3
81
V4 = V1 + 6 − 9I1
4
42
= V.
= 0+6−9
81
3
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Problem 2.28 The independent source in Fig. P2.28 supplies 48 W of power.
Determine I2 .
I1
+
12 V _
I3
R
R
0.25I1
I2
R
R
Figure P2.28: Circuit of Problem 2.28.
Solution: From
P = V0 I1 ,
I1 =
P
48
=
= 4 A.
V0 12
Current of dependent current source is the same as I3 . Hence,
I3 = 0.25I1 = 0.25 × 4 = 1 A.
By KCL,
I2 = I1 − I3 = 4 − 1 = 3 A.
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Problem 2.29 Given that I1 = 1 A in the circuit of Fig. P2.29, determine I0 .
I5
I0
1Ω
I4
I3
2Ω
4Ω
I2
I1 = 1 A
8Ω
16 Ω
Figure P2.29: Circuit for Problem 2.29.
Solution: Since the 16-Ω and 8-Ω resistors are connected in parallel, they have the
same voltage across them, namely
V = 16 × I1 = 16 × 1 = 16 V.
By KCL, I0 equals the sum of the currents flowing in all five resistors:
16 16 16 16 16
+ + + +
1
2
4
8
16
= 31 A.
I0 =
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Problem 2.30 What should R be in the circuit of Fig. P2.30 so that Req = 4 Ω?
1Ω
a
Req
6Ω
2Ω
R
5Ω
b
Figure P2.30: Circuit for Problem 2.30.
Solution: The parallel combination of R and 2-Ω resistor is
R1 =
2R
.
2+R
R1 is in series with 5-Ω resistor. Hence
R2 = R1 + 5 =
2R
+ 5.
2+R
R2 is in parallel with 6-Ω resistor:
2R
+5
6×
2+R
R3 =
,
2R
6+
+5
2+R
and
2R
+5
6×
2+R
= 4.
Req = 1 + R3 = 1 +
2R
11 +
2+R
Solving for R leads to
R = 2 Ω.
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Problem 2.31 Find I0 in the circuit of Fig. P2.31.
I0
4Ω
18 A
6Ω
3Ω
12 Ω
Figure P2.31: Circuit for Problem 2.31.
Solution: Combining the 3-Ω and 6-Ω resistors in parallel gives
R=
3 × 6 18
=
= 2 Ω.
3+6
9
The new circuit becomes
I0
18 A
6Ω
12 Ω
Current division leads to
I0 =
Req
12
18 =
6 × 12
× 18 = 6 A.
6 + 12
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Problem 2.32 For the circuit in Fig. P2.32, find Ix for t < 0 and t > 0.
Ix
+
_
t=0 1
2Ω
15 V 2 Ω
2
4Ω
3Ω
4Ω
4Ω
Figure P2.32: Circuit with SPDT switch for Problem 2.32.
Solution:
For t < 0:
Ix
+
_
15 V 2 Ω
Ix
+
_
+
_
3Ω
4Ω
4Ω
2Ω
3Ω
4Ω
2Ω
15 V 2 Ω
Ix
4Ω
2Ω
15 V 2 Ω
Ix
+
_
2Ω
2Ω
6 3
6 || 3 = 6 + 3 = 2 Ω
2Ω
15 V 1 Ω
Ix =
15
=5A
2+1
For t > 0:
Ix
15 V
2Ω
+
_
Ix =
15
= 7.5 A.
2
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Problem 2.33 Determine Req at terminals (a, b) in the circuit of Fig. P2.33.
a
Req
4Ω
32 Ω
16 Ω
8Ω
8Ω
b
Figure P2.33: Circuit for Problem 2.33.
Solution:
a
32 Ω
Req
4Ω
8Ω
16 Ω
8Ω
b
Terminals (a, b) are connected together through a short circuit. Hence,
Req = 0.
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Problem 2.34 Select R in the circuit of Fig. P2.34 so that VL = 5 V.
Solution: Multiple application of the source-transformation method leads to the final
circuit below.
1 kΩ
R
+
5 kΩ
5 mA
2 kΩ
_
Figure P2.34: Circuit
for Problem 2.34.
1 kΩ
R
+
5 kΩ
25 V
VL
+
_
2 kΩ
VL
_
1 kΩ
R1 = R + 5k
25
Is =
R1
+
Is
R1
2 kΩ
VL
_
R2
Vs
1 kΩ
123
R3
+
_
_
R2 = R1 k 2 kΩ =
Vs = Is R2 =
+
VL
2R1 × 103
2 × 103 (R + 5 × 103 )
=
R1 + 2 × 103
R + 7 × 103
25R2
50 × 103
=
R1
R + 7 × 103
Since no current flows through R3 ,
VL = Vs =
50 × 103
.
R + 7 × 103
Setting VL = 5 V leads to
R = 3 kΩ.
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Problem 2.35 If R = 12 Ω in the circuit of Fig. P2.35, find I.
Solution:
R
R
R
20 V
_
I
4Ω
+
4Ω
R/2
R
R
R
R/2
20 V
_
I
+
R
R/2
R/2
R
Figure P2.35: Circuit
for Problem 2.35.
20 V
_
I
4Ω
20 V
_
I
+
+
4Ω
R
R
R 12
=
=6Ω
2
2
20
I=
= 2 A.
4+6
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Problem 2.36 Use resistance reduction and source transformation to find Vx in the
circuit of Fig. P2.36. All resistance values are in ohms.
Solution:
+ Vx _
4
Figure P2.36: Circuit
for Problem 2.36.
16
16
12
10 A
4
16
16
+ Vx _
4
8
6
10 A
12
6
4
8
+ Vx _
10 A
6
6
4
8
+ Vx _
10 A
3
4
8
+ Vx _
30 V
Vx =
+
_
3
4
8
30 × 4
= 8 V.
3+4+8
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Problem 2.37 Determine A if Vout /Vs = 9 in the circuit of Fig. P2.37.
Solution:
3Ω
+
_ Vs
I1
12 Ω
3Ω
+
_ Vs
12 Ω
3Ω
AI1
+
6 Ω Vout
_
AI1
2 Ω Vout
Figure P2.37: Circuit
for Problem 2.37.
I
+
6Ω
_
Vs
9
I
Vs
I1 = =
2 18
I=
Vout = AI1 × 2 =
AVs
AVs
×2 =
18
9
Vout A
= = 9.
Vs
9
Hence
A = 81.
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Problem 2.38 For the circuit in Fig. P2.38, find Req at terminals (a, b).
Solution:
a
32 Ω
Req
4Ω
8Ω
16 Ω
8Ω
b
Req = 4 + 5 = 9 Ω.
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Problem 2.39 Find Req at terminals (c, d) in the circuit of Fig. P2.38.
Solution:
a
5Ω
3Ω
6Ω
b
5Ω
6Ω
3Ω
5Ω
5Ω
12 Ω
6Ω
5Ω
5Ω
4Ω
5Ω
c
d
c
Req
d
c
Req
d
Req = 4 + 5 + 5 = 14 Ω.
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Problem 2.40 Simplify the circuit to the right of terminals (a, b) in Fig. P2.40 to
find Req , and then determine the amount of power supplied by the voltage source. All
resistances are in ohms.
Solution:
25 V
+
_
a
Req
3
4
5
8
6
8
12
6
Figure P2.40: Circuit
for Problem 2.40.
12
b
25 V
+
_
a
Req
3
4
5
8
6
8
12
4
b
25 V
+
_
a
Req
3
4
5
8
6
6
12 || 12 = 6
8
5 + 6 || 6 = 5 + 3 = 8
b
25 V
+
_
a
Req
3
4
8
b
25 V
+
_
a
Req
3
2
8 || 8 || 4 = 2
b
Req = 3 + 2 = 5 Ω
P=
V2
(25)2
=
= 125 W.
Req
5
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Problem 2.41 For the circuit in Fig. P2.41, determine Req at
(a) Terminals (a, b)
(b) Terminals (a, c)
(c) Terminals (a, d)
(d) Terminals (a, f )
e
2Ω
2Ω
d
2Ω
2Ω
c
f
2Ω
2Ω
2Ω
2Ω
2Ω
2Ω
a
b
Figure P2.41: Circuit for Problem 2.41.
Solution: All resistances are in ohms.
(a)
2
2
6 2
6 + 2 = 1.5
2
2
2
a
2
2
b
a
2
b
Req
Req
Req = 1.5 + 2 + 2 = 5.5 Ω.
(b)
2
2
c
2
2
2
2
a
Req
2
c
8
2
a
Req
Req = 2 + 2 = 4 Ω.
(c)
d
2
2
2
d
6 2
6 + 2 = 1.5
2
2
Req
2
Req
2
a
2
a
Req = 2 + 2 + 1.5 = 5.5 Ω.
(d)
f
2
Req
2
2
f
2
2
2
4 4
4+4 =2
Req
2
a
2
a
Req = 2 + 2 + 2 = 6 Ω.
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Problem 2.42 Find Req for the circuit in Fig. P2.42. All resistances are in ohms.
Solution:
5
Req
10
10
10
10
Figure P2.42: Circuit
for Problem 2.42.
10
5
5
Req
5
10
20
20
5
Req
5
5
Req = 15 Ω.
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Problem 2.43 Apply voltage and current division to determine V0 in the circuit of
Fig. P2.43 given that Vout = 0.2 V.
Solution:
I5
+
V5 8 I
_
3
V0
+
_
I4
+
V3 4 I
_
1
+
V4 4
_
I2
+
V2 2
_
+
V1
_
Figure P2.43: Circuit
for Problem 2.43.
2
1
+
Vout = 0.2 V
_
0.2
= 0.2 A
1
V2 I1
I2 =
= (2 + 1) = 0.3 A
2
2
I3 = I1 + I2 = 0.5 A
I1 =
V4 V3 +V2 4I3 + 2I2
=
=
A
4
4
4 = 0.65
I5 = I3 + I4 = 1.15 A
I4 =
V0 = V4 +V5 = 4I4 + 8I5 = 11.8 V.
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Problem 2.44 Apply source transformations and resistance reductions to simplify
the circuit to the left of nodes (a, b) in Fig. P2.44 into a single voltage source and a
resistor. Then, determine I.
3A
10 Ω
a
I
5A
2Ω
12 Ω
4Ω
b
Figure P2.44: Circuit of Problem 2.44.
Solution:
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3A
10 Ω
5A
a
12 Ω
2Ω
b
2Ω
30 V
10 Ω
_
a
+
10 V
+
_
12 Ω
b
12 Ω
40 V
a
+
_
12 Ω
b
a
40
12
12 Ω
12 Ω
b
a
40
12
6Ω
b
6Ω
20 V
a
I
+
_
4Ω
b
Fig. P2.44 (a)
I=
20
= 2 A.
6+4
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Problem 2.45 Determine the open-circuit voltage Voc across terminals (a, b) in
Fig. P2.45.
Solution:
5Ω
6Ω
a
+
30 V
+
_
3Ω
2A
Voc
_
b
6Ω
a
+
6A
5Ω
3Ω
2A
Voc
_
b
6Ω
a
+
R=
8A
3 5 15
= Ω
3+5 8
Voc
_
b
15
Ω
8
6Ω
a
+
15 V
+
_
Voc
_
b
Fig. P2.45 (a)
Hence,
Voc = 15 V.
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Problem 2.46 Use circuit transformations to determine I in the circuit of Fig. P2.46.
Solution:
3A
6Ω
4Ω
3Ω
+_
I
2A
30 V
4Ω
2Ω
6Ω
12 V
_
10 A
4Ω
+
4Ω
8V
3Ω
I
+
_
2Ω
6
3
9
=2Ω
8Ω
20 V
I
+
_
10 A 2 Ω
20 V
+ _
8Ω
20 V
2Ω
I
+
_
2Ω
Fig. P2.46 (a)
I=
20 − 20
= 0.
8+2+2
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Problem 2.47 Determine currents I1 to I4 in the circuit of Fig. P2.47.
12 Ω
I1
6Ω
I2
+
_
I3
3Ω
I4
6Ω
12 V
Figure P2.47: Circuit of Problems 2.47 and 2.48.
Solution:
12
12
12
I2 =
6
12
I3 =
3
12
I4 =
6
I1 =
= 1 A,
= 2 A,
= 4 A,
= 2 A.
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Problem 2.48 Replace the 12-V source in the circuit of Fig. P2.47 with a 4-A
current source pointing upwards. Then, determine currents I1 to I4 .
Solution:
12 Ω
I1
I3
3Ω
6Ω
I2
I4
6Ω
4A
Req =
I1 =
I2 =
I3 =
I4 =
1
1 1 1
+ + +
12 6 3 6
Req
4
× 4 = A,
12
9
Req
8
× 4 = A,
6
9
Req
16
×4 =
A,
3
9
Req
8
× 4 = A.
6
9
−1
=
12
Ω.
9
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Problem 2.49 Determine current I in the circuit of Fig. P2.49.
Solution: Resistance combining leads to
10 Ω
40 Ω
25 Ω
I
5Ω
30 Ω
60 Ω
+
_
50 V
40 Ω
10 Ω
10 Ω
25 Ω
I
5Ω
40 60
= 24 Ω
40 + 60
5Ω
+
_
50 V
I
5Ω
64 Ω
+
_
30 Ω
50 V
I
5Ω
+
_
30 64
= 20.43 Ω
94
50 V
Fig. P2.49 (a)
I=
50
= 1.97 A.
5 + 20.43
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Problem 2.50 Determine the equivalent resistance Req at terminals (a, b) in the
circuit of Fig. P2.50.
Solution:
5Ω
5Ω
4Ω
a
Req
4Ω
4Ω
b
6Ω
5Ω
5Ω
4Ω
a
10 Ω
4Ω
b
a
R = 4 + 4 + (5 || 5 || 10) = 10 Ω
b
Fig. P2.50 (a)
R = 4 + 4 + (5 k 5 k 10) = 10 Ω.
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*2.51 Determine current I in the circuit of Fig. P2.51.
Solution:
+
_
16 V
2 kΩ
6 mA
2 kΩ
2 kΩ
+
_
5 mA
2 mA
0.5 kΩ
2 kΩ
1 kΩ
I
_
+
8V
+
_
16 V
16 V
+
_
4V
_
+
8V
5 mA
I
2 kΩ
8 mA
2 kΩ
5 mA
2 kΩ
2 kΩ
0.5 kΩ
I1
24 mA
5 mA
I1
1 kΩ
I
_
+
8V
+
_
16 V
2 kΩ
0.5 kΩ
I1
2 kΩ
5 mA
8 mA
(2k || 2k || 1k)
= 0.5 kΩ
I
_
+
8V
Fig. P2.51 (a)
I1 = (19 mA)
0.5k
2.5k
= 3.8 mA.
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19 mA
Problem 2.52 Determine voltage Va in the circuit of Fig. P2.52.
Solution:
4Ω
2Ω
2A
_
V
a
+
2A
5A
2.5 A
4Ω
2Ω
2Ω
4Ω
+
2A
V
a
_
_
+
+
_
4Ω
10 V
_
+
4V
2Ω
10 V
4Ω
V
a
+
_
2A
8Ω
_
+
16 V
4Ω
+
a
2A
V
_
2A
8Ω
I
4Ω
V
_
a
+
4A
8Ω
Fig. P2.52 (a)
By current division,
I=
4×8
= 2.67 A
4+8
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and
V = 4I = 10.67 V.
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Problem 2.53 Convert the circuit in Fig. P2.53(a) from a ∆ to a Y configuration.
Solution:
a
c
6Ω
3Ω
a
R2
c
R3
1Ω
b
R1
d
b
d
Figure P2.53(a)
6×3
= 1.8 Ω
6+3+1
6×1
= 0.6 Ω
R2 =
10
3×1
= 0.3 Ω.
R3 =
10
R1 =
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Problem 2.54 Convert the circuit in Fig. P2.53(b) from a T to a Π configuration.
Solution:
2Ω
8Ω
4Ω
Rb
Ra
Rc
Figure P2.53(b)
2 × 8 + 2 × 4 + 4 × 8 56
=
=7Ω
8
8
2 × 8 + 2 × 4 + 4 × 8 56
Rb =
=
= 14 Ω
4
4
2 × 8 + 2 × 4 + 4 × 8 56
=
= 28 Ω.
Rc =
2
2
Ra =
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Problem 2.55 Find the power supplied by the generator in Fig. P2.55.
Solution:
R1 = 18 Ω
1Ω
6Ω
6Ω
+
20 V _
6Ω
Y
18 Ω
∆
18 Ω
1Ω
18 Ω
+
20 V _
18 Ω
1Ω
+
20 V _
18 Ω
18 Ω
9Ω
18 Ω
9Ω
1Ω I
20 V
+
_
9Ω
20
=2A
10
P = V I = 20 × 2 = 40 W.
I=
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Problem 2.56 Repeat Problem 2.55 after replacing R1 with a short circuit.
Solution:
1Ω
20 V
6Ω
+
_
6Ω
6Ω
Y
1Ω
+
20 V _
18 Ω
∆
18 Ω
18 Ω
18 Ω
18 Ω
9Ω
18 Ω
1Ω
+
20 V _
1Ω I
20 V
+
_
9 18
9 + 18 = 6 Ω
20
20
=
6+1
7
20
P = V I = 20 ×
= 57.1 W.
7
I=
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Problem 2.57 Find I in the circuit of Fig. P2.57.
Solution:
9Ω
3V
_
6Ω
9Ω
6Ω
9Ω
3V
_
I
+
6Ω
+
T
Π Transformation
9Ω
18 Ω
9Ω
3V
9Ω
18 Ω
_
+
18 Ω
I
+
_
6Ω
1/3 A
9Ω
18 Ω
3V
9Ω
18 Ω
I
+
_
3V
9 || 8 = 6
6Ω
2V
6Ω
_
+
9Ω
18 Ω
I
+
_
3V
9Ω
1/6 A
12 Ω
18 Ω
I
+
_
3V
36
18 || 12 = 5
36/5 Ω
6/5 V
9Ω
_
+
I
+
_
3V
I=
3 + 65
9 + 36
5
≃ 0.26 A.
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Problem 2.58 Find the power supplied by the voltage source in Fig. P2.58.
Solution:
4V
_
+
3Ω
3Ω
R=6Ω
_
6Ω
4V
+
6Ω
Ra = 7.5 Ω
Rb = 15 Ω
6Ω
Rc = 15 Ω
6Ω
3×3+3×6+3×6
= 7.5 Ω
6
3×3+3×6+3×6
= 15 Ω
Rb =
3
Rc = Rb = 15 Ω.
Ra =
7.5 Ω
_
4V
4.3 Ω
I
+
4.3 Ω
4V
+
_
4.006 Ω
(4.3 + 4.3) × 7.5
= 4.006 Ω
4.3 + 4.3 + 7.5
4
I=
= 0.998 A
4.006
P = I 2 R = (0.998)2 × 4.006 = 3.99 ≃ 4 W.
R=
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Problem 2.59 Repeat Problem 2.58 after replacing R with a short circuit.
Solution:
4V
_
+
3Ω
3Ω
6Ω
6Ω
4V
_
+
6Ω
3Ω
3Ω
4V
_
6Ω
I
+
6 3
6+3 =2Ω
2Ω
4
=1A
2+2
P = V I = 4 × 1 = 4 W.
I=
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Problem 2.60 Find I in the circuit of Fig. P2.60. All resistances are in ohms.
Solution:
12 V
1
+
_
2
I
2
∆
1
12 V
2
12 V
2
R1 = 0.5
R2 = 1
1
2
Y
I
+
_
4
I
R3 = 1
2
2×2
= 0.5 Ω
2+2+4
2×4
R2 =
=1Ω
2+2+4
R3 = R2 = 1 Ω.
R1 =
0.5
+
_
3
1
3
I
0.5
12 V
+
_
1.5
I=
12
= 4 A.
1 + 0.5 + 1.5
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Problem 2.61 Find Req for the circuit in Fig. P2.61.
Solution:
18 Ω
6Ω
18 Ω
6Ω
6Ω
1Ω
9Ω
18 Ω
Req
18 Ω
18 Ω
18 Ω
1Ω
18 Ω
18 Ω
18 Ω
Req
9Ω
9Ω
9Ω
1Ω
18 Ω
18 Ω
Req
9Ω
9Ω
1Ω
18 Ω
Req
9Ω
1Ω
9Ω
Req
Req = 9 + 1 = 10 Ω.
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Problem 2.62 Find Req at terminals (a, b) in Fig. P2.62 if
(a) Terminal c is connected to terminal d by a short circuit
(b) Terminal e is connected to terminal f by a short circuit
(c) Terminal c is connected to terminal e by a short circuit
All resistance values are in ohms.
d
3
e
3
3
Req
3
f
a b
3
3
c
Figure P2.62: Circuit for Problem 2.62.
Solution:
(a)
d
3
3
a b
3
3
3
3
c
d
9
9
a b
9
9
4.5
9
a b 4.5
Req = 9 Ω
9
c
(b)
e
3
d
3
3
a b
3
3
3
f
3
a b
3
Req = 6 Ω
c
(c)
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e
3
d
3
3
a b
3
3
3
c
3
d
3
3
a b
3
3
6 3
6+3 =2
3
a b
3
Req = 3 + 3 + 2 = 8 Ω
c
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Problem 2.63 For the Wheatstone bridge circuit of Fig. 2-30, solve the following
problems:
(a) If R1 = 1 Ω, R2 = 2 Ω, and Rx = 3 Ω, to what value should R3 be adjusted so
as to achieve a balanced condition?
(b) If V0 = 6 V, Ra = 0.1 Ω, and Rx were then to deviate by a small amount to
Rx = 3.01 Ω, what would be the reading on the ammeter?
Solution:
(a) From Eq. (2.45),
Rx
R3
=
,
R1 R2
R1 Rx 1 × 3
=
= 1.5 Ω.
R2
2
R3 =
(b)
V0
I1
I2
R1
V0
+
−
R2
Ra
V1
I3
Ia
V2
I4
R3
Rx
KCL equations at nodes V1 and V2 are
I1 = I3 + Ia
I4 = I2 + Ia
KVL for the left loop and outside perimeter loop are
−V0 + I1 R1 + I3 R3 = 0,
−V0 + I2 R2 + I4 Rx = 0.
Also, for the upper triangle of the bridge,
I1 R1 + Ia Ra − I2 R2 = 0.
Simultaneous solution of the five equations leads to
Ia =
(R3 R2 − Rx R1 )V0
.
R2 Rx (R1 + R3 ) + (R2 + Rx )[R1 R3 + Ra (R1 + R3 )]
For R1 = 1 Ω, R2 = 2 Ω, R3 = 1.5 Ω, Rx = 3.01 Ω, and V0 = 6 V,
Ia = −2.5 mA.
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Problem 2.64 If V0 = 10 V in the Wheatstone-bridge circuit of Fig. 2-31 and
the minimum voltage Vout that a voltmeter can read is 1 mV, what is the smallest
resistance fraction (∆R/R) that can be measured by the circuit?
Solution:
V0
Vout =
4
∆R
R
∆R 4Vout 4 × 10−3
=
= 4 × 10−4 ,
=
R
V0
10
or 4 parts in 10,000.
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Problem 2.65 Suppose the cantilever system shown in Fig. 2-38 is used in the
Wheatstone-bridge sensor of Fig. 2-31 with V0 = 2 V, α = −1 × 10−9 m2 /N,
L = 0.5 cm, W = 0.2 cm, and H = 0.2 mm. If the measured voltage is Vout = −2 V,
what is the force applied to the cantilever?
Solution: From Example 2-16,
Vout =
V0
FL
α
.
4
W H2
Solving for F, we have
Vout W H 2
V0
αL
−2 0.2 × 10−2 × (0.2 × 10−3 )2
=4
2
(−1 × 10−9 ) × 0.5 × 10−2
F =4
= 64 N.
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Problem 2.66 A touch sensor based on a piezoresistor built into a micromechanical
cantilever made of silicon is connected in a Wheatstone-bridge configuration with a
V0 = 1 V. If L = 1.44 cm and W = 1 cm, what should the thickness H be so that
the touch sensor registers a voltage magnitude of 10 mV when the touch pressure is
10 N?
Solution: From Example 2-16:
Vout =
V0
FL
.
α
4
W H2
Solving for H, we have
H=
V0
4Vout
FL
α
W
1/2
1
(−1 × 10−9 ) × 10 × 1.44 × 10−2
=
×
4 × (−10 × 10−3 )
10−2
1/2
= 0.6 mm.
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Problem 2.67 Determine I1 and I2 in the circuit of Fig. P2.67. Assume VF = 0.7 V
for both diodes.
I1
53 Ω
_
6V
0.7 V
+
I2
53 Ω
+
_
+
_0.7 V
Figure P2.67: Circuit for Problem 2.67.
Solution: The diode in the left-hand loop is reverse biased, so
I1 = 0.
In the right-hand loop, the diode is forward biased. Hence,
I2 =
6 − 0.7
= 0.1 A.
53
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Problem 2.68
diodes.
Determine V1 in the circuit of Fig. P2.68. Assume VF = 0.7 V for all
50 Ω
9V
+
_
100 Ω
+
_
V1
25 Ω
Figure P2.68: Circuit for Problem 2.68.
Solution:
9 − 3(0.7)
≃ 0.04 A,
50 + 100 + 25
V1 = 100I = 4 V.
I=
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Problem 2.69 If the voltage source in the circuit of Fig. P2.69 generates a single
square wave with an amplitude of 2 V, generate a plot for υout for the same time
period.
υs(t)
+
_
100 Ω
+
υout
_
υs(t)
2V
T
t
-2 V
Figure P2.69: Circuit and voltage waveform for Problem 2.69.
Solution: Current will flow through the loop only when υs (t) is greater than 0.7 V.
Hence,
(
2 − 0.7 = 1.3 V for 0 ≤ t ≤ T /2
υout =
0
for T /2 ≤ t ≤ T.
vout
1.3 V
t
T/2
T
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Problem 2.70 If the voltage source in the circuit of Fig. P2.70(a) generates the
single square waveform shown in Fig. P2.70(b), generate plots for i1 (t) and i2 (t).
Solution:
i1
i2
73 Ω
146 Ω
υs(t)
+
_
(a)
υs(t)
8V
2
t (s)
4
−8 V
(b) Square wave
0.05 A
8 − 0.7
146 = 0.05 A
i2
t (s)
i1
−0.1 A
−8 + 0.7
73 = −0.1 A
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Problem 2.71 If the voltage source in the circuit of Fig. P2.70(a) generates the
single triangular waveform shown in Fig. P2.70(c), generate plots for i1 (t) and i2 (t).
Solution:
i1
i2
73 Ω
146 Ω
υs(t)
+
_
(a)
υs(t)
8V
4
2
t (s)
−8 V
(c) Triangular wave
0.05 A
i2
t (s)
1
2
3
4
i1
−0.1 A
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Problem 2.72 Use the DC Operating Point Analysis in Multisim to solve for
voltage Vout in the circuit of Fig. P2.72. Solve for Vout by hand and compare with
the value generated by Multisim. See the solution for Exercise 2-14 (on ) for how
to incorporate circuit variables into algebraic expressions.
C
M
O
DR
+
+
2.5 V _
10 Ω
10 Ω
Vout
_
15 Ω
25 Ω
Circuit for Problem 2.72.
Solution:
A. By-Hand Solution
By voltage division:
Vout =
10
× 2.5 = 0.714 V.
10 + 25
+
+
2.5 V _
10 Ω
Vout
_
25 Ω
25 Ω
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B. By Multisim
Circuit in MultiSIM Schematic Capture
DC Operating Point Solution. The value in the last row, V(1)-V(3), is the specific
solution to the problem.
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Problem 2.73 Find the ratio Vout /Vin for the circuit in Fig. P2.73 using DC
Operating Point Analysis in Multisim. See the Multisim Tutorial included on the
CD on how to reference currents in ABM sources (you should not just type in I(V1)).
1 kΩ
Iin
Vin
+
_
+
10 kΩ
100Iin
Vout
1 kΩ
_
Figure P2.73: Circuit for Problem 2.73.
Solution:
Vout = (100Iin ) × 1000
= 105
Hence,
Vin
= 10Vin .
104
Vout
= 10.
Vin
Circuit in MultiSIM Schematic Capture
DC Operating Point solution
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Problem 2.74 Use DC Operating Point Analysis in Multisim to solve for all six
labeled resistor currents in the circuit of Fig. P2.74.
I2
I1
1Ω
1Ω
+_
I3
1A
I4
2V
1Ω
1Ω
+_
I5
1Ω
I6
3V
1Ω
Figure P2.74: Circuit for Problem 2.74.
Solution:
Circuit in MultiSIM Schematic Capture
DC Operating Point solution
V5 −V3
= 1.5 − 2 = −0.5 A,
1
V5 −V4
= 1.5 − 0 = 1.5 A,
I2 =
1
V3 −V1
I3 =
= 2 − 2 = 0,
1
V4 −V2
I4 =
= 0 + 1 = 1 A,
1
V1
= 2 A,
I5 =
1
V2
= −1 A.
I6 =
1
I1 =
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Problem 2.75 Find the voltages across R1 , R2 , and R3 in the circuit of Fig. P2.75
using the DC Operating Point Analysis tool in Multisim.
R1
V1
10 Ω
+
_ 15 V
R3
I
R2
15 Ω
30 Ω
1.5I
Figure P2.75: Circuit for Problem 2.75.
Solution: When built, the circuit should look like that shown below:
To set the current source up so that it refers to the proper current, double-click on
the current source. This will bring up the window shown below:
In ABM syntax, this becomes: 1.5*V(2)/30. Enter this as shown in the figure above.
Using DC Operating Analysis, select the following outputs:
V(1)
V(2)
V(3)
V(1) − V(2)
V(2) − V(3)
The numerical results are shown in the grapher window:
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Problem 2.76 Find the equivalent resistance looking into the terminals of the
circuit in Fig. P2.76 using a test voltage source and current probes in the Interactive
Simulation in Multisim. Compare the answer you get to what you obtain from series
and parallel combining of resistors carried out by hand.
Figure P2.76: Circuit for Problem 2.76.
Solution: The two 12-kΩ resistors in parallel are equal to 6 kΩ because for any
parallel combination of identically valued resistors,
R
Req = ,
N
where Req is the total equivalent resistance of the combination, R is the resistance of
the elements, and N is the number of resistors in the parallel network.
On the right side there is now a series combination of a 6-kΩ resistor, a 10-kΩ
resistor, and a 3.9-kΩ resistor. Together these equal 19.9 kΩ.
This new value is in parallel with a 1-kΩ resistor. Therefore
19.9 k 1 = 952.153 Ω.
This is then in series with a 1-kΩ resistor, so
1.952153 kΩ.
This is then in parallel with another 1-kΩ resistor, so
1.952 k 1 = 661.25 Ω.
Finally, this value is in series with a 4.7-kΩ resistor, so
Req = 4.7 kΩ + 661.25 Ω = 5.361 kΩ.
When you construct the circuit, add a voltage source across the two pins, and add the
current probe right above the test source. Run the Interactive Simulation to get the
following screen shot:
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Note that you can set the test voltage to anything you want. Unity test voltage (and
current) sources usually just make things easier to solve.
Take the values from this simulation, to get
Req =
1.00
= 5.348 kΩ,
187 × 10−6
which is pretty close (0.24% off) from what we obtained by hand. The differences
can be explained by rounding.
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m2.1 Kirchhoff’s Laws: Determine currents I1 to I3 and the voltage V1 in the
circuit of Fig. m2.1 with component values Isrc = 1.8 mA, Vsrc = 9.0 V, R1 = 2.2 kΩ,
R2 = 3.3 kΩ, and R3 = 1.0 kΩ.
Vsrc
R1
I1
R2
I2
R3
I3
+
_
_
Isrc
V1
+
Figure m2.1: Circuit for Problem m2.1.
Solution:
Summary Comparison:
Results:
Analysis
Simulaon
Measurement
Rela ve Differences:
Simulaon -- Analysis
Measurement -- Analysis
I1 (mA) I2 (mA) I3 (mA)
-0.56
2.36
-1.80
-0.56
2.36
-1.80
-0.51
2.43
-1.92
0.0%
-8.9%
0.0%
3.0%
0.0%
6.7%
V1 (V)
9.59
9.58
9.80
-0.1%
2.2%
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Analytical Solution:
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Multisim Results:
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myDAQ Results:
Further Exploration:
No change observed in I1, I2, and I3, but V1 varied from 7.88 V to 9.88 V.
Explana!on: The current source maintains a constant current in the lower branch and therefore the remaining
circuit sees no difference. However, as the voltage changes across R3 the current source voltage adapts its own
voltage V1.
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m2.2 Equivalent Resistance: Find the equivalent resistance between the following
terminal pairs in the circuit of Fig. m2.2 under the stated conditions:
(a) a-b with the other terminals unconnected,
(b) a-d with the other terminals unconnected,
(c) b-c with a wire connecting terminals a and d, and
(d) a-d with a wire connecting terminals b and c.
Use these component values: R1 = 10 kΩ, R2 = 33 kΩ, R3 = 15 kΩ, R4 = 47 kΩ,
and R5 = 2 kΩ.
a
R1
c
R2
R3
R4
b
R5
d
Figure m2.2: Circuit for Problem m2.2.
Solution:
Summary Comparison:
Results:
Analysis
Simulaon
Measurement
Rela ve Differences:
Simulaon -- Analysis
Measurement -- Analysis
R_AB (kohms) R_AD (kohms) R_BC (kohms) R_AD (kohms)
16.72
9.77
21.45
9.025
16.72
9.766
21.45
9.025
16.64
9.79
21.4
9.04
0.0%
-0.5%
0.0%
0.2%
0.0%
-0.2%
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0.0%
0.2%
Analytical Solution:
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Multisim Results:
All rights reserved. Do not reproduce or distribute. c 2013 National Technology and Science Press
All rights reserved. Do not reproduce or distribute. c 2013 National Technology and Science Press
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myDAQ Results:
See summary comparison table for DMM ohmmeter measurements.
Further Exploration Results:
Measurement
R_AB
R_AD
R_BC with wire
R_AD with wire
Voltage (V)
4.89
4.88
4.91
4.88
Current (mA)
0.29
0.50
0.23
0.54
Calculated resistance (kΩ)
16.9
9.76
21.4
9.04
These calculated resistance values agree with the myDAQ ohmmeter measurements within 2%.
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m2.3 Current and Voltage Dividers: Apply the concepts of voltage dividers,
current dividers, and equivalent resistance to find the currents I1 to I3 and the voltages
V1 to V3 in the circuit of Fig. m2.3. Use these component values: Vsrc = 12 V,
R1 = 1.0 kΩ, R2 = 10 kΩ, R3 = 1.5 kΩ, R4 = 2.2 kΩ, R5 = 4.7 kΩ, and R6 = 3.3 kΩ.
I1
R1
R3
Vsrc
I2
+
+
_
V1
_
R2
+
V2 R4
_
R5
I3
R6
+ V _
3
Figure m2.3: Circuit for Problem m2.3.
Solution:
Summary Comparison:
Results:
Analysis
Simulaon
Measurement
Rela ve Differences:
Simulaon -- Analysis
Measurement -- Analysis
I1 (mA)
1.82
1.82
1.83
I2 (mA)
1.40
1.40
1.41
I3 (mA)
0.45
0.45
0.45
0.2%
0.8%
0.2%
0.9%
0.0%
1.1%
V1 (V)
4.19
4.19
4.21
V2 (V)
2.09
2.09
2.10
V3 (V)
-5.99
-5.99
-5.96
0.0%
0.5%
0.0%
0.3%
0.0%
-0.6%
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Analytical Solution:
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Multisim Results:
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myDAQ Results:
See summary comparison table for DMM voltmeter and ammeter measurements.
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m2.4 Wye-Delta Transformation: Find (a) the currents I1 and I2 in the circuit of
Fig. m2.4 and (b) the power delivered by each of the two voltage sources. Use these
component values: V1 = 15 V, V2 = 15 V, R1 = 3.3 kΩ, R2 = 1.5 kΩ, R3 = 4.7 kΩ,
R4 = 5.6 kΩ, R5 = 1.0 kΩ, and R6 = 2.2 kΩ.
R3
R2
R1
R5
R4
V1
_
R6
V2
_
+
+
I1
I2
Figure m2.4: Circuit for Problem m2.4.
Solution:
Summary Comparison:
Results:
Analysis
Simulaon
Measurement
Rela ve Differences:
Simulaon -- Analysis
Measurement -- Analysis
I1 (mA)
4.05
4.05
4.05
I2 (mA)
4.45
4.45
4.45
P1 (mW)
60.7
60.7
60.6
P2 (mW)
66.7
66.7
66.5
0.0%
0.0%
0.0%
0.0%
0.0%
-0.2%
0.0%
-0.3%
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Analytical Solution:
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Multisim Results:
Interac ve simula on results: I1 = 4.05 mA and I2 = 4.4 5mA
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Interac ve simula on results: PL = 60.703 mW delivered, PR = 66.687 mW delivered
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myDAQ Results:
DMM ammeter on 20mA range se!ng measures I1 = 4.06 mA and I2 = 4.45 mA.
DMM voltmeter on 20V range se!ng measures 14.91 V for the le" source and 14.85 V for the right
source. Calculate power as product of voltage and current: PL = 60.5 mW and PR = 66.1 mW.
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