-
Problem 2.1 An AWG-14 copper wire has a resistance of 17.1 at
20C. Howlong is it?
Solution: AWG-14 has a diameter of 1.6 mm (Table 2-2), and at
20C, coppersconductivity is = 5.81107 (S/m) [Table 2-1].
R =
A = RA
= Rpi(d/2)2
= 17.15.81107pi(
1.61032
)2 2 km.
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Problem 2.2 A 3-km long AWG-6 metallic wire has a resistance of
approximately6 at 20C. What material is it made of?
Solution:
R =
A, AWG-6 has a diameter of 4.1 mm.
=
RA
=
Rpi(d/2)2 =3103
6pi(4.1103/2)2 = 3.79107 (S/m),
which, according to Table 2-1, is approximately the value of the
conductivity ofaluminum.
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Problem 2.3 A thin-film resistor made of germanium is 2 mm in
length and itsrectangular cross section is 0.2 mm 1 mm, as shown in
Fig. P2.3. Determine theresistance that an ohmmeter would measure
if connected across its:
(a) Top and bottom surfaces(b) Front and back surfaces(c) Right
and left surfaces
0.2 mm 1 mm
2 mm
y
z
x
Figure P2.3: Film resistor of Problem 2.3.
Solution:(a)
R =
A = 0.22 mm, A = 1 mm2 mm = 2106 m2
=2104
2.132106 47 .
(b)
R =
A = 1 mm, A = 2 mm0.2 mm = 4107 m2
=103
2.134107 1,174 .
(c)
R =
A = 2 mm, A = 1 mm0.2 mm = 2107 m2
=2103
2.134107 4,695 .
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Problem 2.4 A resistor of length consists of a hollow cylinder
of radius asurrounded by a layer of carbon that extends from r = a
to r = b, as shown inFig. P2.4.
(a) Develop an expression for the resistance R.(b) Calculate R
at 20C for a = 2 cm, b = 3 cm, and = 10 cm.
Carbon
Hollow 2b2a
l
Figure P2.4: Carbon resistor of Problem 2.4.
Solution:(a) R = A .
The area through which current can flow is the cross section
consisting of carbon.Hence,
A = pib2pia2.
Thus,R =
pi(b2a2) .
(b)R =
0.17.14104pi(0.0320.022) = 0.89 (m).
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Problem 2.5 A standard model used to describe the variation of
resistance withtemperature T is given by
R = R0(1+T ),
where R is the resistance at temperature T (measured in C), R0
is the resistance atT = 0C, and is a temperature coefficient. For
copper, = 4 103 C1. Atwhat temperature is the resistance greater
than R0 by 1%?
Solution: R = 1.01R0. Hence,
RR0
= 1.01 = 1+T
T = 0.01
T =0.01
4103 = 2.5C.
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Problem 2.6 A light bulb has a filament whose resistance is
characterized by atemperature coefficient = 6 103 C1 (see
resistance model given in Problem2.5). The bulb is connected to a
100-V household voltage source via switch. Afterturning on the
switch, the temperature of the filament increases rapidly from
theinitial room temperature of 20C to an operating temperature of
1800C. When itreaches its operating temperature, it consumes 80 W
of power.
(a) Determine the filament resistance at 1800C.(b) Determine the
filament resistance at room temperature.(c) Determine the current
that the filament draws at room temperature and also at
1800C.(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 whenit arrives at its operating temperature?
Solution:(a) R = resistance at 1800C.
p =V 2
R; R =
V 2
p=
(100)2
80 = 125 .
(b) R = R0(1+T ).
R0 =R
1+T=
1251+61031800 = 10.6 @ 0
C.
At T = 20C,
R = R0(1+T ) = 10.6(1+610320) = 11.9 @ 20C.
(c) At T = 20C,I =
VR(20C) =
10011.9 = 8.43 A.
At T = 1800C,I =
pV
=80100 = 0.8 A.
(d) The damage is greater at room temperature because the
current is much closer to10 A.
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Problem 2.7 A 110-V heating element in a stove can boil a
standard-size pot ofwater in 1.2 minutes, consuming a total of 136
kJ of energy. Determine the resistanceof the heating element and
the current flowing through it.
Solution:
W = p t = i t
i =W
t =136103
1101.260 = 17.2 A.
R =
i=
11017.2
= 6.41 .
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Problem 2.8 A certain copper wire has a resistance R
characterized by the modelgiven in Problem 2.5 with = 4 103 C1. If
R = 60 at 20C and the wire isused in a circuit that cannot tolerate
an increase in the magnitude of R by more than10 percent over its
value at 20C, what would be the highest temperature at whichthe
circuit can be operated within its tolerance limits?
Solution: The model is given by
R = R0(1+T ).
We are given that R = 60 @ T = 20C, and the maximum R that can
be toleratedis 66 .
At 20C, 60 = R0(1+20).At unknown T , 66 = R0(1+T ).The ratio
gives
6660 =
1+T1+20 .
Solving for T , we have
T =6+132
60 =6+132+4103
604103 = 27.2C.
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Problem 2.9 The circuit shown in Fig. P2.9 includes two
identical potentiometerswith per-length resistance of 20 /cm.
Determine Ia and Ib.
Ia
80 mA4 cm
6 cm
10 cm 2.5 cm
7.5 cm10 cm
Ib
Figure P2.9: Circuit of Problem 2.9.
Solution: Effectively, the circuit looks as shown in Fig.
P2.9(a).Ia
Ib
R1
R2
80 mA
Fig. P2.9 (a)
Resistors R1 and R2 are:
R1 = (4+7.5)20 = 230 ,R2 = (6+2.5)20 = 170 .
Hence, by current division
Ia = (80 mA(
R2R1 +R2
)
= 8102 170400 = 34 mA,
Ib = 8102230400 = 46 mA.
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Problem 2.10 Determine VL in the circuit of Fig. P2.10.
+
_
+
_
12 V VL4 6 6 10
5 5
5 5
Figure P2.10: Circuit of Problem 2.10.
The parallel combination of the 4 and 6 resistors is
R =464+6 = 2.4 .
By voltage division
VL = 12(
2.45+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.
500
I0
+_
500
12 V
6 mA
+_
VL
R
3I0
Figure P2.11: Circuit of Problem 2.11.
Solution: The voltage across the 500- resistor in the right-hand
segment is
VL = (3I0 +6103)500.
Setting VL = 9 V leads to
I0 =13
(VL500 610
3)
=13
(9
500 6103
)= 4 mA.
The left-hand loop has to satisfy KVL:
12+ I0R+500I0 = 0,
which leads to
R =12500I0
I0=
12I0500
=12
4103 500 = 2500 .
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Problem 2.12 A high-voltage direct-current generating station
delivers 10 MWof power at 250 kV to a city, as depicted in Fig.
P2.12. The city is representedby resistance RL and each of the two
wires of the transmission line between thegenerating station and
the city is represented by resistance RTL. The distance betweenthe
two locations is 2000 km and the transmission lines are made of
10-cmdiametercopper wire. Determine (a) how much power is consumed
by the transmission lineand (b) what fraction of the power
generated by the generating station is used by thecity.
+_
RLV0
RTL
RTL
2000 km
Station
(city)
Figure P2.12: Diagram for Problem 2.12.
Solution: For copper, = 1.72108 -m.
For each wire of the transmission line, Eq. (2.2) leads to
RTL =
A=
1.721082106
pi(5102)2
= 4.4 .Delivering 10 MW at 250 kV to the city means that
PL =V 2LRL
= 107 W, with VL = 2.5105 V.
Hence,
RL =V 2LPL
=(2.5105)2
107 = 6.25103 .
Also, the current flowing through RL is
I =VLRL
=2.51056.25103 = 40 A.
(a) The power consumed by transmission lines isPTL = 2I2RTL
= 2 (40)24.4 = 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 = PLPs
=107
1.001408107
= 0.9986 99.86%.
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Problem 2.13 Determine the current I in the circuit of Fig.
P2.13 given that I0 = 0.
3 1
2
1
1
1
24 V
I0 = 0
I
+_
+ _
+_
+_
+_
+_
I2
V2
V1
V4
V5V3
I1
I1I2
a
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,
I11+ I11 = I21+ I21,
or I1 = I2, which is obvious considering that all 4 resistors in
the bridge section arethe same. For the left loop that includes the
24-V source,
24+V1 +V2 +V3 = 0V1 = 3IV2 = I1V3 = I1
and I = I1 + I2 = 2I1. Hence,
24+3I +2I1 = 024+3I + I = 0
or
I =244
= 6 A.
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Problem 2.14 Determine currents I1 to I3 in the circuit of Fig.
P2.14.
7 12
4
2
8 18 V
1 A
3 A
I2
a
bI3
I1
+
_
Figure P2.14: Circuit for Problem 2.14.
Solution: For the loop containing the 18-V source,
18+32+8I1 = 0.
Hence, I1 = 1.5 A.KCL at node a gives
31 I1 I2 = 0I2 = 2 I1 = 21.5 = 0.5 A.
KCL at node b gives
1+ I2 I3 = 0I3 = 1+ I2 = 1+0.5 = 1.5 A.
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Problem 2.15 Determine Ix in the circuit of Fig. P2.15.
Ix
2
5
12 V 1 A+_
I
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 = 177 = 2.43 A.
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Problem 2.16 Determine currents I1 to I4 in the circuit of Fig.
P2.16.
I2I1
I4
I3
6 4
1 8
12 V
4 A
5 V
1 V
+_
+_
+_
+
_
+
_
+
_
Figure P2.16: Circuit for Problem 2.16.
Solution: Application of KVL to the outer-parameter loop
gives
12+41+8I3 +51 = 0,
which givesI3 = 0.5 A.
KVL for the left-most loop is
12+41+4I1 = 0,
which leads toI1 = 2 A.
KCL at the top center node gives
4 I1 I2 I3 = 0
or
I2 = 4 I1 I3 = 420.5 = 1.5 A.
KCL at the bottom left node gives
I4 + I14 = 0,
or
I4 = 4 I1 = 42 = 2 A.
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Problem 2.17 Determine currents I1 to I4 in the circuit of Fig.
P2.17.
I1 I2 I3 I4
2 6 A
4 2 4
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 thatI1 + 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-k
resistor in thecircuit of Fig. P2.18.
103V02 k 3 k10 mA V0+
_
Figure P2.42: Circuit for Problem 2.18.
Solution: In the left loop,
V0 = 101032103 = 20 V.
The dependent current source is I0 = 103V0 = 20 mA.The power
dissipated in the 3-k resistor is
p = I20 R = (20103)23103 = 1.2 W.
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Problem 2.19 Determine Ix and Iy in the circuit of Fig.
P2.19.
4IxI
2 6
4 10 V+
_
+
_
Ix Iy
Figure P2.19: Circuit for Problem 2.19.
Solution: Application of KVL to the two loops gives
10+2Ix +4I = 04I +6Iy4Ix = 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.
Ia
b
2 2
2
6 V 12 V
+
_
+
_Vab
+
_
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.
3 k
4 k2 k16 V
+
_
12 V
8 V
+
_
+
_
I1 I3
I2 +
_
+
_
+
_
Figure P2.21: Circuit for Problem 2.21.
Solution: Application of KVL to the outer-perimeter loop
gives
16+3103I18+12 = 0,
which leads toI1 = 4 mA.
For the left loop,
16+3103I1 +4103I2 = 0,
I2 =163103I1
4103 =1631034103
4103 = 1 mA.
Consequently,I3 = I1 I2 = 41 = 3 mA.
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Problem 2.22 Find I in the circuit of Fig. P2.22.
2I
3 10 V+
_
I+
_
+
_
+
_
Figure P2.22: Circuit for Problem 2.22.
Solution:10+2I +3I = 0.
Hence,I =
105 = 2 A.
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Problem 2.23 Determine the amount of power supplied by the
independent currentsource in the circuit of Fig. P2.23.
V1
I
2
2
0.2 A4
V1
+
_
Figure P2.23: Circuit for Problem 2.23.
Solution: KCL at top node gives
0.2+ V14 I = 0
Also I = V1/2 (for the 2- resistor).Hence,
0.2+ V14
V12
= 0
V1 = 0.8 V.
The voltage across the 0.2-A source is 2V1 = 1.6 V.Power
dissipated is
P = V I = 1.60.2 = 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.
R1 = 18
1 6 6
6 18 40 V
+
_
V1I1 V2
V3
I2
I3
Figure P2.24: Circuit of Problem 2.24.
Solution:R1 = 18
1 6 6
6 18 40 V
+
_
V1 V2V3
I4
Fig. P2.24 (a)
V1 = 40 (1 ) I1 = 404 = 36 V.
KCL at node V1 leads to
I4 = I1 I2 = 41 = 3 A.
Hence,V2 = V1 +6I4 = 3663 = 18 V.
The voltage across R1 is 18I2.Hence,
V3 = V118I2 = 36181 = 18 V.
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Problem 2.25 After assigning node V4 in the circuit of Fig.
P2.25 as the groundnode, determine node voltages V1, V2, and
V3.
6 6
3 3
12 V3 A
+
_
1 A
6
1 A
V1 V3V2
V4
Figure P2.25: Circuit of Problem 2.25.
Solution:
6 6
3 3
12 V3 A
+
_
I1
1 A
6
1 A
V1 V3V2
V4
Fig. P2.25 (a)
From KCL at node V1, the sum of currents leaving the node is
3+ I11 = 0,
or
I1 =3+1 =2 A.
Node voltages (relative to V4):
V1 =61 =6 V,V2 = V13I1 =63(2) = 0,V3 = 61 = 6 V.
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Problem 2.26 After assigning node V1 in the circuit of Fig.
P2.25 as the groundnode, determine node voltages V2, V3, and
V4.
Solution:
6 6
3 3
12 V3 A
+
_
I1
1 A
6
1 A
V1V3
V2
V4
Fig. P2.26 (a)
From KCL at node V1, the sum of currents leaving the node is
3+ I11 = 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, andI3 = 24/81 A. Determine node voltages V2, V3, and V4
after assigning node V1 asthe ground node.
9
V3 V4
V2
V1
6 9
6
9
6
6 V
+
_
6 V
+
_
+
_
+
_
I3
I1I2
Figure P2.27: Circuit of Problem 2.27.
Solution: At node V3, KCL gives
I2 + I4 I3 = 0,
or
I4 = I3 I2 =2481
4281 =
1881 A.
I4
9
V3
V1 6
V4
V2
V1
6 9
6
9
6
6 V
+
_
6 V
+
_
+
_
+
_
I3
I1I2
Fig. P2.27 (a)
V3 = V169I2
= 06+9 4281 = 1.33 =43 V.
V2 = V36I4
=43 6
(
1881
)= 0,
V4 = V1 +69I1
= 0+69(
4281
)=
43 V.
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Problem 2.28 The independent source in Fig. P2.28 supplies 48 W
of power.Determine I2.
I2 0.25I112 V
I3I1 R
R
RR
+
_
Figure P2.28: Circuit of Problem 2.28.
Solution: From
P = V0I1,
I1 =PV0
=4812
= 4 A.
Current of dependent current source is the same as I3.
Hence,
I3 = 0.25I1 = 0.254 = 1 A.
By KCL,I2 = I1 I3 = 41 = 3 A.
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Problem 2.29 Given that I1 = 1 A in the circuit of Fig. P2.29,
determine I0.
I0
I1 = 1 AI2
1 2 4 8 16
I3I4I5
Figure P2.29: Circuit for Problem 2.29.
Solution: Since the 16- and 8- resistors are connected in
parallel, they have thesame voltage across them, namely
V = 16 I1 = 161 = 16 V.
By KCL, I0 equals the sum of the currents flowing in all five
resistors:
I0 =161
+162
+164
+168 +
1616
= 31 A.
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Problem 2.30 What should R be in the circuit of Fig. P2.30 so
that Req = 4 ?
Req
a
b
R2
5
1
6
Figure P2.30: Circuit for Problem 2.30.
Solution: The parallel combination of R and 2- resistor is
R1 =2R
2+R.
R1 is in series with 5- resistor. Hence
R2 = R1 +5 =2R
2+R+5.
R2 is in parallel with 6- resistor:
R3 =6
(2R
2+R+5
)
6+ 2R2+R
+5,
and
Req = 1+R3 = 1+6
(2R
2+R+5
)
11+2R
2+R
= 4.
Solving for R leads toR = 2 .
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Problem 2.31 Find I0 in the circuit of Fig. P2.31.
3 6
4
12 18 A
I0
Figure P2.31: Circuit for Problem 2.31.
Solution: Combining the 3- and 6- resistors in parallel
gives
R =363+6 =
189 = 2 .
The new circuit becomes
6 12
18 A
I0
Current division leads to
I0 =(
Req12
)18 = 6126+12 18 = 6 A.
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Problem 2.32 For the circuit in Fig. P2.32, find Ix for t < 0
and t > 0.
t = 0
2 3 2
1
4
2 4
4 15 V+
_
Ix
+
_
Figure P2.32: Circuit with SPDT switch for Problem 2.32.
Solution:For t < 0:
2 3 4
2 4
4 15 V+
_
Ix
2 3 4
2 2
15 V
+
_
Ix
2 2 6 || 3 =
2
15 V
+
_
Ix
1
2
15 V
+
_
Ix
= 2 6 3
6 + 3
Ix =15
2+1= 5 A
For t > 0:
2
15 V
+
_
Ix
Ix =152
= 7.5 A.
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Problem 2.33 Determine Req at terminals (a,b) in the circuit of
Fig. P2.33.
8 16
32 8
4 Req
a
b
Figure P2.33: Circuit for Problem 2.33.
Solution:
a
b
8 16
32 8
4 Req
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 finalcircuit below.
R 1 k
2 k5 k5 mA VL
+
_
R 1 k
2 k
5 k
25 VVL
+
_
+
_
1 k
2 k VLR1Is =
R1 = R + 5k
Is
+
_
1 k
VL
R2
Vs
+
_
123
R3+
_
25
R1
Figure P2.34: Circuit
for Problem 2.34.
R2 = R1 2 k =2R1103
R1 +2103=
2103(R+5103)R+7103
Vs = IsR2 =25R2R1
=50103
R+7103
Since no current flows through R3,
VL = Vs =50103
R+7103 .
Setting VL = 5 V leads toR = 3 k.
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Problem 2.35 If R = 12 in the circuit of Fig. P2.35, find I.
Solution:
R
4 20 V
R+
_
I 4 20 V
+
_
I
R/2 R/2
R/2 R/2
20 V
+
_
I4
RR
RR
RR
RR
4 20 V
+
_
I
Figure P2.35: Circuit
for Problem 2.35.
R2
=122
= 6
I =20
4+6 = 2 A.
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Problem 2.36 Use resistance reduction and source transformation
to find Vx in thecircuit of Fig. P2.36. All resistance values are
in ohms.
Solution:
16 16
4
12 16 1664
10 A
Vx+ _
8
4
12 864
10 A
Vx+ _
6 864
10 A
Vx+ _
834
10 A
Vx+ _
83 4
Vx+ _
30 V+
_
Figure P2.36: Circuit
for Problem 2.36.
Vx =304
3+4+8 = 8 V.
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Problem 2.37 Determine A if Vout/Vs = 9 in the circuit of Fig.
P2.37.
Solution:
AI1
I1+
_
12 12 3
3
6 VoutVs
+
_
AI1
+
_
6
3
2 Vs
I
Vout
+
_
Figure P2.37: Circuit
for Problem 2.37.
I =Vs9
I1 =I2
=Vs18
Vout = AI12 =AVs18 2 =
AVs9
VoutVs
=A9 = 9.
HenceA = 81.
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Problem 2.38 For the circuit in Fig. P2.38, find Req at
terminals (a,b).
Solution:
a
b
8 16
32 8
4 Req
Req = 4+5 = 9 .
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Problem 2.39 Find Req at terminals (c,d) in the circuit of Fig.
P2.38.
Solution:
a
b
c
d
5 3 5
5 3 6 6
c
d
5
5 12 6 Req
c
d
5
5 4 Req
Req = 4+5+5 = 14 .
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Problem 2.40 Simplify the circuit to the right of terminals
(a,b) in Fig. P2.40 tofind Req, and then determine the amount of
power supplied by the voltage source. Allresistances are in
ohms.
Solution:
25 V
3 5 8
668
412 12
a
b
+
_
Req
25 V
3 5 8
68
412 4
a
b
+
_
Req
25 V
3 5
68
46 12 || 12 = 6
8 || 8 || 4 = 2
5 + 6 || 6 = 5 + 3 = 8
a
b
+
_
Req
25 V
3
84
8
a
b
+
_
Req
25 V
3
2
a
b
+
_
Req
Figure P2.40: Circuit
for Problem 2.40.
Req = 3+2 = 5
P =V 2
Req=
(25)25 = 125 W.
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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 f
ba
c
d
2
2
2
2
2
2
2
2
2
2
Figure P2.41: Circuit for Problem 2.41.
Solution: All resistances are in ohms.(a)
ba
2
2
2
2
2
2
Req
ba
22
Req
= 1.56 2
6 + 2
Req = 1.5+2+2 = 5.5 .(b)
Req a
c
2
2
2
222
Req a
c
2
82
Req = 2+2 = 4 .(c)
Req
a
2
2
2
22
d2
Req
a
2
d2
= 1.56 2
6 + 2
Req = 2+2+1.5 = 5.5 .(d)
Req
a
2
2
2
2
2
2
f
2
= 24 4
4 + 4Req
a
2
f
Req = 2+2+2 = 6 .
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Problem 2.42 Find Req for the circuit in Fig. P2.42. All
resistances are in ohms.
Solution:
1010
5
5
10
10
10
Req
Req 10 20 205
5
Req 55
5
Figure P2.42: Circuit
for Problem 2.42.
Req = 15 .
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Problem 2.43 Apply voltage and current division to determine V0
in the circuit ofFig. P2.43 given that Vout = 0.2 V.
Solution:
Vout = 0.2 V
I5
2
1
4
2
4
8
+
_
+
_
V1
+
_V2
+
_
V3
+
_
V4
+
_
V5
+
_
V0
Figure P2.43: Circuit
for Problem 2.43.
I3
I4
I2
I1
I1 =0.21
= 0.2 A
I2 =V22
=I12
(2+1) = 0.3 A
I3 = I1 + I2 = 0.5 A
I4 =V44
=V3 +V2
4=
4I3 +2I24 = 0.65 A
I5 = I3 + I4 = 1.15 AV0 = V4 +V5 = 4I4 +8I5 = 11.8 V.
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Problem 2.44 Apply source transformations and resistance
reductions to simplifythe circuit to the left of nodes (a,b) in
Fig. P2.44 into a single voltage source and aresistor. Then,
determine I.
5 A4
3 A
12 2
10 a
b
I
Figure P2.44: Circuit of Problem 2.44.
Solution:
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5 A
3 A
12 2
10 a
b
30 V
12
10 2 a
b
+_
+
_
10 V
12
12 a
b
+
_
40 V
12 12
a
b
4012
4012
6
a
b
4
6 a
b
+
_
20 V
I
Fig. P2.44 (a)
I =20
6+4 = 2 A.
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Problem 2.45 Determine the open-circuit voltage Voc across
terminals (a,b) inFig. P2.45.
Solution:
Voc
6
2 A
+
_
3 30 V
5 a
b
+
_
Voc
6
2 A
+
_
3 6 A
a
b
5
Voc
6
+
_
8 A
a
b
R = =3 + 5
3 5 15
8
Voc
6
+
_
15 V
a
b
15
8
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:
4
4 2 A
3
6 3 A
30 V
2
+
_
I
4
8 V
3
6
12 V 10 A
2
4
I+
_
+_
8
20 V
= 2
10 A2
I+
_
6 3
9
8 2
20 V
20 V
2
I+
_
+ _
Fig. P2.46 (a)
I =2020
8+2+2 = 0.
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Problem 2.47 Determine currents I1 to I4 in the circuit of Fig.
P2.47.
+
_
12 V
I1
I26
12 I3
I46
3
Figure P2.47: Circuit of Problems 2.47 and 2.48.
Solution:
I1 =1212
= 1 A,
I2 =126 = 2 A,
I3 =123 = 4 A,
I4 =126 = 2 A.
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Problem 2.48 Replace the 12-V source in the circuit of Fig.
P2.47 with a 4-Acurrent source pointing upwards. Then, determine
currents I1 to I4.
Solution:
4 A
I1
I26
12 I3
I46
3
Req =(
112
+16 +
13 +
16
)1=
129 .
I1 =Req124 =
49 A,
I2 =Req6 4 =
89 A,
I3 =Req3 4 =
169 A,
I4 =Req6 4 =
89 A.
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Problem 2.49 Determine current I in the circuit of Fig.
P2.49.
Solution: Resistance combining leads to
Fig. P2.49 (a)
30 60
5
50 V
10 25 40
+
_
10 10
I
= 24
5
50 V
25 40
+
_
5
I
64
5
50 V+
_
30
I
5
50 V+
_
= 20.43
I
40 60
40 + 60
30 64
94
I =50
5+20.43 = 1.97 A.
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Problem 2.50 Determine the equivalent resistance Req at
terminals (a,b) in thecircuit of Fig. P2.50.
Solution:
5
4
6
4
4
5
Req
a
b
10
4
4
5
5
a
b
R = 4 + 4 + (5 || 5 || 10) = 10
a
b
Fig. P2.50 (a)
R = 4+4+(5 5 10) = 10 .
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*2.51 Determine current I in the circuit of Fig. P2.51.
Solution:
1 k
6 mA
2 mA
2 k
2 k
16 V
8 V
2 k 5 mA
+_
+
_
I
1 k
8 mA
2 k
2 k
16 V
8 V
2 k 5 mA
+_
+
_
I
8 mA
= 0.5 k(2k || 2k || 1k)
16 V
8 V
2 k 5 mA
+_
+
_
I
0.5 k
16 V
8 V
2 k 5 mA
+_
+
_
4 V+
_
I
0.5 k2 k 5 mA24 mA
I1
I1
0.5 k2 k 19 mA
I1
Fig. P2.51 (a)
I1 = (19 mA)(
0.5k2.5k
)= 3.8 mA.
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Problem 2.52 Determine voltage Va in the circuit of Fig.
P2.52.
Solution:
V a
4
2.5 A
2
2 A
2 +
_
2 A
5 A
4
4
V a
4
10 V 10 V
16 V
2
2
+_
2 A
2 A
4
8
V a
+_
4 V
+_
2 A 2 A
4
8
V a
+_
4 A
4
8 V a
+_
+_
+_
+_
I
Fig. P2.52 (a)
By current division,I =
484+8 = 2.67 A
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andV = 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
b
c
d
3 1
6 a
b
c
d
R1 R2
R3
Figure P2.53(a)
R1 =63
6+3+1 = 1.8
R2 =61
10 = 0.6
R3 =31
10 = 0.3 .
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Problem 2.54 Convert the circuit in Fig. P2.53(b) from a T to a
configuration.
Solution:
8 2
4
Ra
Rc
Rb
Figure P2.53(b)
Ra =28+24+48
8 =568 = 7
Rb =28+24+48
4=
564
= 14
Rc =28+24+48
2=
562
= 28 .
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Problem 2.55 Find the power supplied by the generator in Fig.
P2.55.
Solution:
R1 = 18
Y
1 6 6
6 18 20 V+
_
1 18
18 18 18
18
20 V
+
_
1 9
18 9 20 V+
_
1
9 20 V+
_
I
I =2010 = 2 A
P = V I = 202 = 40 W.
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Problem 2.56 Repeat Problem 2.55 after replacing R1 with a short
circuit.
Solution:
Y
1 6 6
6 18 20 V+
_
1 18
18 18 18 20 V+
_
1
18 9 20 V+
_
1
20 V
+
_
I
= 6 9 18
9 + 18
I =20
6+1 =207
P = V I = 20 207
= 57.1 W.
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Problem 2.57 Find I in the circuit of Fig. P2.57.
Solution:
9 6 9
6 9 6
3 V
+
_
3 V
+
_
I
I
9
18 18
18 9
9
3 V 3 V
+
_+_
I
9 18 18
6 9
1/3 A
3 V
+
_
I
12 18
9
1/6 A
3 V
+
_
+_
I
36/5 9
6/5 V
3 V
+
_
2 V
I
18
6 9 6
3 V
+
_
+_
T Transformation
9 || 8 = 6
18 || 12 =36
5
I =3+ 659+ 365
0.26 A.
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Problem 2.58 Find the power supplied by the voltage source in
Fig. P2.58.
Solution:
6 R = 6 6
3 3
4 V
+
_
6 Rb = 15 Rc = 15
Ra = 7.5
6
4 V
+
_
Ra =33+36+36
6 = 7.5
Rb =33+36+36
3 = 15
Rc = Rb = 15 .
4.3 4.3 7.5
4.006
4 V
+
_
4 V
+
_
I
R =(4.3+4.3)7.54.3+4.3+7.5 = 4.006
I =4
4.006 = 0.998 A
P = I2R = (0.998)24.006 = 3.99 4 W.
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Problem 2.59 Repeat Problem 2.58 after replacing R with a short
circuit.
Solution:
6 6
3 3
4 V
+
_
6 3 3 6
4 V
+
_
2
4 V
+
_
I
= 2 6 3
6 + 3
I =4
2+2= 1 A
P = V I = 41 = 4 W.
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Problem 2.60 Find I in the circuit of Fig. P2.60. All
resistances are in ohms.
Solution:
14
2 2
2 2
12 V
I+_
1
2 2
12 V
I
+_
R1 = 0.5
R3 = 1R2 = 1
Y
R1 =22
2+2+4= 0.5
R2 =24
2+2+4= 1
R3 = R2 = 1 .
1
3 3
12 V
I
+_
0.5
1
12 V
I
+_
0.5
1.5
I =12
1+0.5+1.5 = 4 A.
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Problem 2.61 Find Req for the circuit in Fig. P2.61.
Solution:
Req
18
6
6 18
1 6
18
9
18 18
18
1
18
18
18
9
Req
9
18
1
9
18
9
Req
18
1 9
9
Req
9
1
Req
Req = 9+1 = 10 .
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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.
e
d
a bf
c
3
3 3
3 3
3Req
Figure P2.62: Circuit for Problem 2.62.
Solution:(a)
d
c
a33 3
3 3
b 3
d
c
a9
9 9
9 9
9b a4.5 4.5b
Req = 9
(b)d
c
ae f
3
3
b 3
3 3
3
a3 3b
Req = 6
(c)
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dc
ae
3
3
b 3
3 3
3
d
c
a3
3
b 3
3
3
a3 3b
Req = 3 + 3 + 2 = 8
= 26 3
6 + 3
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Problem 2.63 For the Wheatstone bridge circuit of Fig. 2-30,
solve the followingproblems:
(a) If R1 = 1 , R2 = 2 , and Rx = 3 , to what value should R3 be
adjusted soas to achieve a balanced condition?
(b) If V0 = 6 V, Ra = 0.1 , and Rx were then to deviate by a
small amount toRx = 3.01 , what would be the reading on the
ammeter?
Solution:(a) From Eq. (2.45),
R3R1
=RxR2
, R3 =R1RxR2
=13
2= 1.5 .
(b)
R1
I1 I2
I4I3 R3
RaIa
R2
Rx
V0 V1 V2
V0
+
KCL equations at nodes V1 and V2 are
I1 = I3 + IaI4 = I2 + Ia
KVL for the left loop and outside perimeter loop are
V0 + I1R1 + I3R3 = 0,V0 + I2R2 + I4Rx = 0.
Also, for the upper triangle of the bridge,
I1R1 + IaRa I2R2 = 0.
Simultaneous solution of the five equations leads to
Ia =(R3R2RxR1)V0
R2Rx(R1 +R3)+(R2 +Rx)[R1R3 +Ra(R1 +R3)].
For R1 = 1 , R2 = 2 , R3 = 1.5 , Rx = 3.01 , 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 andthe minimum voltage Vout that a voltmeter can read is
1 mV, what is the smallestresistance fraction (R/R) that can be
measured by the circuit?
Solution:
Vout =V04
(RR
)
RR
=4VoutV0
=4103
10 = 4104
,
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 theWheatstone-bridge sensor of Fig. 2-31 with V0 = 2 V, = 1
109 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 =V04
FL
WH2.
Solving for F , we have
F = 4(
VoutV0
)WH2
L
= 4(22
)0.2102 (0.2103)2
(1109)0.5102
= 64 N.
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Problem 2.66 A touch sensor based on a piezoresistor built into
a micromechanicalcantilever made of silicon is connected in a
Wheatstone-bridge configuration with aV0 = 1 V. If L = 1.44 cm and
W = 1 cm, what should the thickness H be so thatthe touch sensor
registers a voltage magnitude of 10 mV when the touch pressure is10
N?
Solution: From Example 2-16:
Vout =V04
FL
WH2.
Solving for H, we have
H =[(
V04Vout
)
FLW
]1/2
=
[1
4 (10103) (1109)101.44102
102
]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 Vfor both diodes.
I2I153 53
6 V
+_
0.7 V 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 =60.7
53 = 0.1 A.
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Problem 2.68 Determine V1 in the circuit of Fig. P2.68. Assume
VF = 0.7 V for alldiodes.
50
25
100
9 V
+_
+
_
V1
Figure P2.68: Circuit for Problem 2.68.
Solution:
I =93(0.7)
50+100+25 0.04 A,
V1 = 100I = 4 V.
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Problem 2.69 If the voltage source in the circuit of Fig. P2.69
generates a singlesquare wave with an amplitude of 2 V, generate a
plot for out for the same timeperiod.
100
+
_
out
+
_
s(t)
2 V
-2 V
tT
s(t)
Figure P2.69: Circuit and voltage waveform for Problem 2.69.
Solution: Current will flow through the loop only when s(t) is
greater than 0.7 V.Hence,
out =
{20.7 = 1.3 V for 0 t T/20 for T/2 t T.
1.3 V
tTT/2
vout
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Problem 2.70 If the voltage source in the circuit of Fig.
P2.70(a) generates thesingle square waveform shown in Fig.
P2.70(b), generate plots for i1(t) and i2(t).
Solution:
+
_
i1
s(t)
i2
73 146
s(t)
t (s)4
8 V
8 V
2
(b) Square wave
(a)
t (s)
0.1 A
0.05 Ai2
i1
8 0.7= 0.05 A
146
8 + 0.7= 0.1 A
73
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Problem 2.71 If the voltage source in the circuit of Fig.
P2.70(a) generates thesingle triangular waveform shown in Fig.
P2.70(c), generate plots for i1(t) and i2(t).
Solution:
+
_
i1
s(t)
i2
73 146
s(t)
t (s)4
8 V
8 V
2
(c) Triangular wave
(a)
31 2 4t (s)
0.1 A
0.05 Ai2
i1
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Problem 2.72 Use the DC Operating Point Analysis in Multisim to
solve forvoltage Vout in the circuit of Fig. P2.72. Solve for Vout
by hand and compare withthe value generated by Multisim. See the
solution for Exercise 2-14 (on CDROM ) for howto incorporate
circuit variables into algebraic expressions.
Vout10 10
25 15
2.5 V
+
_
+
_
Circuit for Problem 2.72.
Solution:
A. By-Hand Solution
By voltage division:
Vout =(
1010+25
)2.5 = 0.714 V.
Vout
25
10
25
2.5 V
+
_
+
_
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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 DCOperating Point Analysis in Multisim. See the
Multisim Tutorial included on theCD on how to reference currents in
ABM sources (you should not just type in I(V1)).
Vin
Iin
Vout100Iin10 k
1 k
1 k
+
_
+
_
Figure P2.73: Circuit for Problem 2.73.
Solution:
Vout = (100Iin)1000
= 105 Vin104 = 10Vin.
Hence,VoutVin
= 10.
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 sixlabeled resistor currents in the circuit of Fig.
P2.74.
I1
I3
I5
1
1
1
I2
I4
I6
1
1
1
1 A2 V
3 V
+ _
+ _
Figure P2.74: Circuit for Problem 2.74.
Solution:
Circuit in MultiSIM Schematic Capture DC Operating Point
solution
I1 =V5V3
1= 1.52 =0.5 A,
I2 =V5V4
1= 1.50 = 1.5 A,
I3 =V3V1
1= 22 = 0,
I4 =V4V2
1= 0+1 = 1 A,
I5 =V11
= 2 A,
I6 =V21
=1 A.
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Problem 2.75 Find the voltages across R1, R2, and R3 in the
circuit of Fig. P2.75using the DC Operating Point Analysis tool in
Multisim.
R1
R2V1
10
R3
15
30 15 V 1.5I
I+
_
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 onthe 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 thecircuit in Fig. P2.76 using a test voltage source
and current probes in the InteractiveSimulation in Multisim.
Compare the answer you get to what you obtain from seriesand
parallel combining of resistors carried out by hand.
Figure P2.76: Circuit for Problem 2.76.
Solution: The two 12-k resistors in parallel are equal to 6 k
because for anyparallel combination of identically valued
resistors,
Req =RN
,
where Req is the total equivalent resistance of the combination,
R is the resistance ofthe 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-k
resistor, a 10-kresistor, and a 3.9-k resistor. Together these
equal 19.9 k.
This new value is in parallel with a 1-k resistor. Therefore19.9
1 = 952.153 .
This is then in series with a 1-k resistor, so1.952153 k.
This is then in parallel with another 1-k resistor, so1.952 1 =
661.25 .
Finally, this value is in series with a 4.7-k resistor, soReq =
4.7 k+661.25 = 5.361 k.
When you construct the circuit, add a voltage source across the
two pins, and add thecurrent probe right above the test source. Run
the Interactive Simulation to get thefollowing screen shot:
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Note that you can set the test voltage to anything you want.
Unity test voltage (andcurrent) sources usually just make things
easier to solve.
Take the values from this simulation, to get
Req =1.00
187106 = 5.348 k,
which is pretty close (0.24% off) from what we obtained by hand.
The differencescan be explained by rounding.
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m2.1 Kirchhoffs Laws: Determine currents I1 to I3 and the
voltage V1 in thecircuit of Fig. m2.1 with component values Isrc =
1.8 mA, Vsrc = 9.0 V, R1 = 2.2 k,R2 = 3.3 k, and R3 = 1.0 k.
+
_
+
_
Vsrc
Isrc
R2
R1
I2
R3
V1
I1
I3
Figure m2.1: Circuit for Problem m2.1.
Solution:
Summary Comparison:
Results: I1 (mA) I2 (mA) I3 (mA) V1 (V)
Analysis -0.56 2.36 -1.80 9.59
Simulaon -0.56 2.36 -1.80 9.58
Measurement -0.51 2.43 -1.92 9.80
Rela ve Dierences:
Simulaon -- Analysis 0.0% 0.0% 0.0% -0.1%
Measurement -- Analysis -8.9% 3.0% 6.7% 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.
Explanaon: The current source maintains a constant current in
the lower branch and therefore the remaining
circuit sees no dierence. 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 followingterminal 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 k, R2 = 33 k, R3 = 15 k, R4
= 47 k,and R5 = 2 k.
R2
R4d
R1 R5R3
c
ba
Figure m2.2: Circuit for Problem m2.2.
Solution:
Summary Comparison:
Results: R_AB (kohms) R_AD (kohms) R_BC (kohms) R_AD (kohms)
Analysis 16.72 9.77 21.45 9.025
Simulaon 16.72 9.766 21.45 9.025
Measurement 16.64 9.79 21.4 9.04
Rela ve Dierences:
Simulaon -- Analysis 0.0% 0.0% 0.0% 0.0%
Measurement -- Analysis -0.5% 0.2% -0.2% 0.2%
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Analytical Solution:
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Multisim Results:
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myDAQ Results:
See summary comparison table for DMM ohmmeter measurements.
Further Exploration Results:
Measurement Voltage (V) Current (mA) Calculated resistance
(k)
R_AB 4.89 0.29 16.9
R_AD 4.88 0.50 9.76
R_BC with wire 4.91 0.23 21.4
R_AD with wire 4.88 0.54 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 voltagesV1 to V3 in the circuit of Fig.
m2.3. Use these component values: Vsrc = 12 V,R1 = 1.0 k, R2 = 10
k, R3 = 1.5 k, R4 = 2.2 k, R5 = 4.7 k, and R6 = 3.3 k.
+
_
V2 R4
+
_
V1Vsrc
R5
R3
R2
R1
R6
I1
I2
I3
+ _V3
+
_
Figure m2.3: Circuit for Problem m2.3.
Solution:
Summary Comparison:
Results: I1 (mA) I2 (mA) I3 (mA) V1 (V) V2 (V) V3 (V)
Analysis 1.82 1.40 0.45 4.19 2.09 -5.99
Simulaon 1.82 1.40 0.45 4.19 2.09 -5.99
Measurement 1.83 1.41 0.45 4.21 2.10 -5.96
Rela ve Dierences:
Simulaon -- Analysis 0.2% 0.2% 0.0% 0.0% 0.0% 0.0%
Measurement -- Analysis 0.8% 0.9% 1.1% 0.5% 0.3% -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 ofFig. m2.4 and (b) the power delivered by each of
the two voltage sources. Use thesecomponent values: V1 = 15 V, V2 =
15 V, R1 = 3.3 k, R2 = 1.5 k, R3 = 4.7 k,R4 = 5.6 k, R5 = 1.0 k,
and R6 = 2.2 k.
R4
R3
R1
I1
V1 V2
R2 R5
R6
+_ +_
I2
Figure m2.4: Circuit for Problem m2.4.
Solution:
Summary Comparison:
Results: I1 (mA) I2 (mA) P1 (mW) P2 (mW)
Analysis 4.05 4.45 60.7 66.7
Simulaon 4.05 4.45 60.7 66.7
Measurement 4.05 4.45 60.6 66.5
Rela ve Dierences:
Simulaon -- Analysis 0.0% 0.0% 0.0% 0.0%
Measurement -- Analysis 0.0% 0.0% -0.2% -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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