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Solution manual for Fundamentals of Electric Circuit 5th edition
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Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
http://www.solutionsmanualtestbank.com/products/2013/02/13/fundamentals-of-electric-
circuits-alexander-5th-edition-solutions-manual
Chapter 2, Solution 1. Design a problem, complete with a solution, to help students to
better understand Ohm’s Law. Use at least two resistors and one voltage source. Hint,
you could use both resistors at once or one at a time, it is up to you. Be creative.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
The voltage across a 5-kresistor is 16 V. Find the current through the resistor.
Solution
v = iR i = v/R = (16/5) mA = 3.2 mA
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 2
p = v2/R R = v
2/p = 14400/60 = 240 ohms
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
For silicon, 6.4x102 -m. A r
2 . Hence,
2 2
R L
L
r 2
L
6.4x10 x4x10
0.033953
A r 2 R x240
r = 184.3 mm
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 4
(a) i = 40/100 = 400 mA
(b) i = 40/250 = 160 mA
n = 9; l = 7; b = n + l – 1 = 15
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 6
n = 12; l = 8; b = n + l –1 = 19
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 7
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
6 branches and 4 nodes
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 8. Design a problem, complete with a solution, to help other students
to better understand Kirchhoff’s Current Law. Design the problem by specifying values
of ia, ib, and ic, shown in Fig. 2.72, and asking them to solve for values of i1, i2, and i3.
Be careful specify realistic currents.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72.
Solution
12 A
a
i1
8 A b
i2
i3
12 A
c 9 A d
At node a, 8 = 12 + i1 i 1 = - 4A
At node c, 9 = 8 + i2 i 2 = 1A At node d, 9 = 12 + i3 i 3 = -3A
Chapter 2, Solution 9
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
At A, 1+6–i1 = 0 or i1 = 1+6 = 7 A
At B, –6+i2+7 = 0 or i2 = 6–7 = –1 A
At C, 2+i3–7 = 0 or i3 = 7–2 = 5 A
Chapter 2, Solution 10
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
2 –8A 4A
i2
1 i1 3
–6A
At node 1, –8–i1–6 = 0 or i1 = –8–6 = –14 A
At node 2, –(–8)+i1+i2–4 = 0 or i2 = –8–i1+4 = –8+14+4 = 10 A
Chapter 2, Solution 11
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
V1 15 0
5 2 V2 0
V1 6 V
V2 3 V
Chapter 2, Solution 12
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
+ 30v –
– 50v +
+ 20v –
loop 2
+ v2 –
+
40v
-
+
loop 1 v1
–
+
loop 3 v3
–
For loop 1, –40 –50 +20 + v1 = 0 or v1 = 40+50–20 = 70 V
For loop 2, –20 +30 –v2 = 0 or v2 = 30–20 = 10 V
For loop 3, –v1 +v2 +v3 = 0 or v3 = 70–10 = 60 V
Chapter 2, Solution 13
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
2A
I2 7A I4
1 2 3 4 4A
I1
3A I3
At node 2,
3 7 I2 0
At node 1,
I1 I2 2
At node 4,
2 I4 4
At node 3,
7 I4 I3
Hence,
I2 10 A
I1 2 I2 12 A
I4 2 4 2 A
I3 7 2 5 A
I1 12 A, I2 10 A, I3 5 A, I4 2 A
Chapter 2, Solution 14
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
+ + -
3V V1 I4 V2
- I3 - + 2V - +
- + V3 - + +
4V
I2 - V4 I1 5V + -
For mesh 1,
V4 2 5 0
For mesh 2,
4 V3 V4 0
For mesh 3,
3 V1 V3 0
For mesh 4,
V1 V2 2 0
Thus,
V4 7V
V3 4 7 11V
V1 V3 3 8V
V2 V1 2 6V
V1 8V , V2 6V , V3 11V , V4 7V
Chapter 2, Solution 15
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Calculate v and ix in the circuit of Fig. 2.79.
12 + 16 V –
+
10 V _
+ v – ix
+
4 V
_
+ 3 ix
_
Figure 2.79
For Prob. 2.15.
Solution
For loop 1, –10 + v +4 = 0, v = 6 V
For loop 2, –4 + 16 + 3ix =0, ix = –4 A
Chapter 2, Solution 16
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Determine Vo in the circuit in Fig. 2.80.
16 14
+
10 V +
_
+
Vo _
_
25 V
Figure 2.80
For Prob. 2.16.
Solution
Apply KVL,
–10 + (16+14)I + 25 = 0 or 30I = 10–25 = – or I = –15/30 = –500 mA
Also, –10 + 16I + Vo = 0 or Vo = 10 – 16(–0.5) = 10+8 = 18 V
Chapter 2, Solution 17
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Applying KVL around the entire outside loop we get,
–24 + v1 + 10 + 12 = 0 or v1 = 2V
Applying KVL around the loop containing v2, the 10-volt source, and the
12-volt source we get,
v2 + 10 + 12 = 0 or v2 = –22V
Applying KVL around the loop containing v3 and the 10-volt source we
get,
–v3 + 10 = 0 or v3 = 10V
Chapter 2, Solution 18
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32 I = 4A
-Vab + 5I + 8 = 0 Vab = 28V
Chapter 2, Solution 19
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
p
Applying KVL around the loop, we obtain
–(–8) – 12 + 10 + 3i = 0 i = –2A
Power dissipated by the resistor:
= i2R = 4(3) = 12W
3
Power supplied by the sources:
p12V = 12 ((–2)) = –24W
p10V = 10 (–(–2)) = 20W
p8V = (–8)(–2) = 16W
Chapter 2, Solution 20
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Determine io in the circuit of Fig. 2.84.
io 22
54V + + 5io
–
Figure 2.84
For Prob. 2.20
Solution
Applying KVL around the loop,
–54 + 22io + 5io = 0 io = 4A
Chapter 2, Solution 21
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Applying KVL,
-15 + (1+5+2)I + 2 Vx = 0
But Vx = 5I, -15 +8I + 10I =0, I = 5/6
Vx = 5I = 25/6 = 4.167 V
Chapter 2, Solution 22
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Find Vo in the circuit in Fig. 2.86 and the power absorbed by the dependent
source.
10 V1
10
+ Vo
25A
2Vo
Figure 2.86
For Prob. 2.22
Solution
At the node, KCL requires that [–Vo/10]+[–25]+[–2Vo] = 0 or 2.1Vo = –25
or Vo = –11.905 V
The current through the controlled source is i = 2V0 = –23.81 A
and the voltage across it is V1 = (10+10) i0 (where i0 = –V0/10) = 20(11.905/10) = 23.81 V.
Hence,
pdependent source = V1(–i) = 23.81x(–(–23.81)) = 566.9 W
Checking, (25–23.81)2(10+10) + (23.81)(–25) + 566.9 = 28.322 – 595.2 + 566.9
= 0.022 which is equal zero since we are using four places of accuracy!
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 23
8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3
The circuit is reduced to that shown below.
ix 1
+ vx –
20A 2 3
Applying current division,
ix = [2/(2+1+3)]20 = 6.667 and vx = 1x6.667 = 6.667 V
i 2
(6 A) 2 A, x
2 1 3
vx 1ix 2V
The current through the 1.2- resistor is 0.5ix = 3.333 A. The voltage across the 12- resistor is 3.333 x 4.8 = 16V. Hence the power absorbed by the 12-ohm
resistor is equal to
(16)2/12 = 21.33 W
Chapter 2, Solution 24
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
3
1
4
3
(a) I0 = Vs
R1 R2
V0 I0 R
R = Vs
R 3R 4
V0 R3 R4
R1 R 2 R 3 R 4
Vs R R2 R R4
(b) If R1 = R2 = R3 = R4 = R,
V0
R
10
= 40
VS 2R 2 4
Chapter 2, Solution 25
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
V0 = 5 x 10-3
x 10 x 103
= 50V
Using current division,
I20 5
5 20
(0.01x50) 0.1 A
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Problem 26
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
For the circuit in Fig. 2.90, io = 3 A. Calculate ix and the total power absorbed by
the entire circuit.
ix 10 io
8 4 2 16
Figure 2.90
For Prob. 2.26.
Solution
If i16= io = 3A, then v = 16x3 = 48 V and i8 = 48/8 = 6A; i4 = 48/4 = 12A; and
i2 = 48/2 = 24A.
Thus,
ix = i8 + i4 + i2 + i16 = 6 + 12 + 24 + 3 = 45 A
p = (45)210 + (6)
28 + (12)
24 + (24)
22 + (3)
216 = 20,250 + 288 + 576 +1152 + 144
= 20250 + 2106 = 22.356 kW.
Chapter 2, Problem 27
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Calculate Io in the circuit of Fig. 2.91.
8
+ Io
10V
3 6
Figure 2.91
For Prob. 2.27.
Solution
The 3-ohm resistor is in parallel with the c-ohm resistor and can be replaced by a
[(3x6)/(3+6)] = 2-ohm resistor. Therefore,
Io = 10/(8+2) = 1 A.
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 28 Design a problem, using Fig. 2.92, to help other students better
understand series and parallel circuits.
Although there is no correct way to work this problem, this is an example based on the
same kind of problem asked in the third edition.
Problem
Find v1, v2, and v3 in the circuit in Fig. 2.92.
Solution
We first combine the two resistors in parallel
15 10 6
We now apply voltage division,
v1 = 14
14 6
(40) 28 V
v2 = v3 = 6
14 6
(40) 12 V
Hence, v1 = 28 V, v2 = 12 V, vs = 12 V
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Solution 29
All resistors in Fig. 2.93 are 5 each. Find Req.
Req
Figure 2.93
For Prob. 2.29.
Solution
Req = 5 + 5||[5+5||(5+5)] = 5 + 5||[5+(5x10/(5+10))] = 5+5||(5+3.333) = 5 +
41.66/13.333
= 8.125 Ω
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Chapter 2, Problem 30.
Find Req for the circuit in Fig. 2.94.
25 180
60
Req
60
Figure 2.94
For Prob. 2.30.
Solution
We start by combining the 180-ohm resistor with the 60-ohm resistor which in
turn is in parallel with the 60-ohm resistor or = [60(180+60)/(60+180+60)] = 48.
Thus,
Req = 25+48 = 73 Ω.
Chapter 2, Solution 31
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Req 3 2 // 4 //1 3 1
1/ 2 1/ 4 1
3.5714
i1 = 200/3.5714 = 56 A
v1 = 0.5714xi1 = 32 V and i2 = 32/4 = 8 A
i4 = 32/1 = 32 A; i5 = 32/2 = 16 A; and i3 = 32+16 = 48 A
Chapter 2, Solution 32
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Find i1 through i4 in the circuit in Fig. 2.96.
60 i4 i2 200
40 50
i3 i1
16 A
Figure 2.96
For Prob. 2.32.
Solution
We first combine resistors in parallel.
40 60 40x60
100
24 and
50 200 50x200
250
40
Using current division principle,
i1 i2 24
24 40
(16) 6A,i3
i4 40
(16) 10A 64
i 200
(6) –4.8 A and i 1
250 2
50 250
(6) –1.2 A
i3 60
(10) 100
–6 A and i4
40 (10) –4 A
100
Chapter 2, Solution 33
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Combining the conductance leads to the equivalent circuit below
i 4S i
+
9A v
- 1S 4S
+
9A v 1S 2S -
6S 3S 6x3
2S and 2S + 2S = 4S 9
Using current division,
i 1
1 1
2
(9) 6 A, v = 3(1) = 3 V
Chapter 2, Solution 34
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
160//(60 + 80 + 20)= 80 ,
160//(28+80 + 52) = 80
Req = 20+80 = 100 Ω
I = 200/100 = 2 A or p = VI = 200x2 = 400 W.
Chapter 2, Solution 35
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
-
i
70 +
V1
30
200V +
i1 - Io
a b +
20
i2
Vo 5 -
Combining the resistors that are in parallel,
70 30 70x30 21 ,
100
20 5 20x5 4
25
i = 200
21 4
8 A
v1 = 21i = 168 V, vo = 4i = 32 V v1 vo
i1 = 2.4 A, i2 = 1.6 A
70 20
At node a, KCL must be satisfied
i1 = i2 + Io 2.4 = 1.6 + Io Io = 0.8 A
Hence,
vo = 32 V and Io = 800 mA
Chapter 2, Solution 36
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
20//(30+50) = 16, 24 + 16 = 40, 60//20 = 15
Req = 80+(15+25)40 = 80+20 = 100 Ω
i = 20/100 = 0.2 A
If i1 is the current through the 24-resistor and io is the current through the 50-
resistor, using current division gives
i1 = [40/(40+40)]0.2 = 0.1 and io = [20/(20+80)]0.1 = 0.02 A or
vo = 30io = 30x0.02 = 600 mV.
Chapter 2, Solution 37
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Applying KVL,
-20 + 10 + 10I – 30 = 0, I = 4
10 RI
R 10
2.5 I
Chapter 2, Solution 38
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4
The circuit is reduced to that shown below.
2.5 4
15 16
60
Req
(4 + 16)//60 = 20x60/80 = 15
Req = 2.5+15||15 = 2.5+7.5 = 10 Ω and
io = 35/10 = 3.5 A.
Chapter 2, Solution 39
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm
resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second
2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the
second 1k-ohm resistor. This now leads to,
Req = [(1x2.667)/3.667]k = 727.3 Ω.
(b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor.
This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in
parallel with the 4k-ohm resistor producing,
Req = [(4x12)/16]k = 3 kΩ.
Chapter 2, Solution 40
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Req = 8 4 (2 6 3) 8 2 10
I = 15 R eq
15
10
1.5 A
Chapter 2, Solution 41
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Let R0 = combination of three 12resistors in parallel
1
1
R o 12
1
1 12 12
Ro = 4
R eq 30 60 (10 R 0 R) 30 60 (14 R)
50 30 60(14 R)
74 R 74 + R = 42 + 3R
or R = 16
Chapter 2, Solution 42
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) Rab = 5 (8 20 30) 5 (8 12)
5x20
25
4
(b) Rab = 2 4 (5 3) 8 5 10 4 2 4 4 5 2.857 2 2 1.8181 5.818
Chapter 2, Solution 43
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
1
1
(a) Rab = 5 20 10 40
5x20
25
400
50
4 8 12
(b) 60 20 30 1 1
60 10
60 20 30 6
Rab = 80 (10 10) 80 20
100 16
Chapter 2, Solution 44
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b.
5 20
a
2 3
b
Figure 2.108
For Prob. 2.44
Solution
First we note that the 5 Ω and 20 Ω resistors are in parallel and can be replaced by
a 4 Ω [(5x20)/(5+20)] resistor which in now in series with the 2 Ω resistor and
can be replaced by a 6 Ω resistor in parallel with the 3 Ω resistor thus,
Rab = [(6x3)/(6+3)] = 2 Ω.
Chapter 2, Solution 45
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab 5 50 4.8 59.8
(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm
and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give
30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus,
Rab 5 12.8 15 32.5
Chapter 2, Solution 46
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Req = 12 + 5||20 + [1/((1/15)+(1/15)+(1/15))] + 5 + 24||8 = 12 + 4 + 5 + 5 + 6 = 32 Ω
I = 80/32 = 2.5 A
Chapter 2, Solution 47
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
5 20
6 3
5x20 4
25
6x3 2
9
10 8
a b
4 2
Rab = 10 + 4 + 2 + 8 = 24
Chapter 2, Solution 48
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) Ra = R1R 2 R 2 R 3 R 3 R1
100 100 100
30
R 3 10
Ra = Rb = Rc = 30
(b) R 30x 20 30x50 20x50
3100
103.3
a 30
R 3100
155, b 20
30
R 3100
62c 50
(a) R1 =
R a R b R c
36
R1 = R2 = R3 = 4
Chapter 2, Solution 49
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
R1
60 30 10 R
60x10 6
R 30x10
3
Ra = 103.3 , Rb = 155 , Rc = 62
R a R c 12 *12
4
Chapter 2, Solution 50
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(b)
60x30 18
2 100
3 100
R1 = 18, R2 = 6, R3 = 3
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
2.50 Design a problem to help other students better understand wye-delta transformations using
Fig. 2.114.
Although there is no correct way to work this problem, this is an example based on the same
kind of problem asked in the third edition.
Problem
What value of R in the circuit of Fig. 2.114 would cause the current source to deliver 800 mW to
the resistors.
Solution
Using R = 3RY = 3R, we obtain the equivalent circuit shown below:
30mA
R 3R
3R 3R
R
30mA 3R
3R/2
3R R 3RxR
3 R
4R 4
3Rx 3
R 3R 3
R 3
R 3R 3
R 2 = R
4 4
2 3R
3 R
2
P = I2R 800 x 10
-3 = (30 x 10
-3)2
R
R = 889
Chapter 2, Solution 51
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) 30 30 15 and 30 20 30x20 /(50) 12
Rab = 15 (12 12) 15x24 /(39) 9.231
a
30
30
30
30
a
20
20
15
12
12
b b
(b) Converting the T-subnetwork into its equivalent network gives
Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70
Rb'c' = 350/(10) = 35, Ra'c' = 350/(20) = 17.5
Also 30 70 30x70 /(100) 21 and 35/(15) = 35x15/(50) = 10.5
Rab = 25 + 17.5 (21 10.5) 25 17.5 31.5
Rab = 36.25
30 30
25 70
25 a
10
5
20
15
a a’ b’
17.5 35 15
b b
c’ c’
Chapter 2, Solution 52
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below.
3 9
9 3 9
3 3
3
3 6
3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the
circuit below. 3
2.25
3
9 2.25
3 2
We now convert the wye-subnetwork to the delta-subnetwork.
Ra = [(2.25x3+2.25x3+2.25x2.25)/3] = 6.188 Ω
Rb = Rc = 18.562/2.25 = 8.25 Ω
This leads to the circuit below.
3
8.25
3
6.188 9
8.25
2
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
R = 9||6.188+8.25||2 = 3.667+1.6098 = 5.277
Req = 3+3+8.25||5.277 = 9.218 Ω.
Chapter 2, Solution 53
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) Converting one to T yields the equivalent circuit below:
20 a
b
30 a’ 4
60
5 b’
20
c’
80
Ra'n = 40x10
40 10 50
4, R b'n
10x50
100
5, R c'n
40x50
100
20
Rab = 20 + 80 + 20 + (30 4) (60 5) 120 34 65
Rab = 142.32
(b) We combine the resistor in series and in parallel.
30 (30 30) 30x60
2090
We convert the balanced s to Ts as shown below:
a
30
b
30
30
30
30
20
30
a
10
10
10 10
10
b
10 20
Rab = 10 + (10 10) (10 20 10) 10 20 20 40
Rab = 33.33
Chapter 2, Solution 54
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) Rab 50 100 / /(150 100 150) 50 100 / /400 130
(b) Rab 60 100 / /(150 100 150) 60 100 / /400 140
Chapter 2, Solution 55
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
We convert the T to .
24 V
+ -
I0
Req
a
20
40
10
20
b
60
50
24 V
+ -
I0
Req
a
35
b
140
70
60
70
Rab =
R1R 2 R 2 R 3 R 3 R1
20x40 40x10 10x20
1400
35
R 3 40 40
Rac = 1400/(10) = 140, Rbc = 1400/(20) = 70
70 70 35 and 140 160 140x60/(200) = 42
Req = 35 (35 42) 24.0625
I0 = 24/(Rab) = 997.4mA
Chapter 2, Solution 56
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
We need to find Req and apply voltage division. We first tranform the Y network to .
30 30
+
100 V
-
Req
16 35
15
12
10
20
+
100 V
-
Req
16 a
35
37.5 b
30
45
c
20
Rab = 15x10 10x12 12x15
12
450
12
37.5
Rac = 450/(10) = 45, Rbc = 450/(15) = 30
Combining the resistors in parallel,
30||20 = (600/50) = 12 ,
37.5||30 = (37.5x30/67.5) = 16.667
35||45 = (35x45/80) = 19.688
Req = 19.688||(12 + 16.667) = 11.672
By voltage division,
v = 11.672
11.672 16
100
= 42.18 V
Chapter 2, Solution 57
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
4 a
18
b
d
2
27
36
7
1
c e
14
10 28
f
Rab = 6x12 12x8 8x6
12
216
12
18
Rac = 216/(8) = 27, Rbc = 36
Rde = 4x2 2x8 8x4
8
56 7
8
Ref = 56/(4) = 14, Rdf = 56/(2) = 28
Combining resistors in parallel,
10 28 280
7.368, 38
27 3 27x3
2.730
36 7 36x7
5.86843
4
7.568
18
5.868
2.7
14
4
1.829
3.977 0.5964
7.568 14
R 18x 2.7
18x 2.7
1.829
an 18 2.7 5.867 26.567
R bn
R cn
18x5.868
3.977
26.567
5.868x2.7
0.5904
Chapter 2, Solution 58
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
26.567
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
R eq 4 1.829 (3.977 7.368) (0.5964 14)
5.829 11.346 14.5964
i = 20/(Req) = 1.64 A
12.21
Chapter 2, Solution 58
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
The resistance of the bulb is (120)2/60 = 240
40 2 A 1.5 A
+
VS -
+ 90 V - 0.5 A
240
+ 120 V
-
80
Once the 160and 80resistors are in parallel, they have the same voltage
120V. Hence the current through the 40resistor is equal to 2 amps.
40(0.5 + 1.5) = 80 volts.
Thus
vs = 80 + 120 = 200 V.
Chapter 2, Solution 59
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Three light bulbs are connected in series to a 120-V source as shown in Fig.
2.123. Find the current I through each of the bulbs. Each bulb is rated at 120
volts. How much power is each bulb absorbing? Do they generate much light?
Figure 2.123
For Prob. 2.59.
Solution
Using p = v2/R, we can calculate the resistance of each bulb.
R30W = (120)2/30 = 14,400/30 = 480 Ω
R40W = (120)2/40 = 14,400/40 = 360 Ω
R50W = (120)2/50 = 14,400/50 = 288 Ω
The total resistance of the three bulbs in series is 480+360+288 = 1128 Ω.
The current flowing through each bulb is 120/1128 = 0.10638 A.
p30 = (0.10638)2480 = 0.011317x480 = 5.432 W.
p40 = (0.10638)2360 = 0.011317x360 = 4.074 W.
p50 = (0.10638)2288 = 0.011317x288 = 3.259 W.
Clearly these values are well below the rated powers of each light bulb so we
would not expect very much light from any of them. To work properly, they need
to be connected in parallel.
Chapter 2, Solution 60
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source,
calculate the current through each bulb.
Solution
Using p = v2/R, we can calculate the resistance of each bulb.
R30W = (120)2/30 = 14,400/30 = 480 Ω
R40W = (120)2/40 = 14,400/40 = 360 Ω
R50W = (120)2/50 = 14,400/50 = 288 Ω
The current flowing through each bulb is 120/R.
i30 = 120/480 = 250 mA.
i40 = 120/360 = 333.3 mA.
i30 = 120/288 = 416.7 mA.
Unlike the light bulbs in 2.59, the lights will glow brightly!
Chapter 2, Solution 61
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
There are three possibilities, but they must also satisfy the current range of 1.2 +
0.06 = 1.26 and 1.2 – 0.06 = 1.14.
(a) Use R1 and R2:
R = R1 R 2 80 90 42.35
p = i2R = 70W
i2
= 70/42.35 = 1.6529 or i = 1.2857 (which is outside our range)
cost = $0.60 + $0.90 = $1.50
(b) Use R1 and R3:
R = R1 R 3 80 100 44.44
i2
= 70/44.44 = 1.5752 or i = 1.2551 (which is within our range),
cost = $1.35
(c) Use R2 and R3:
R = R 2 R 3 90 100 47.37
i2
= 70/47.37 = 1.4777 or i = 1.2156 (which is within our range),
cost = $1.65
Note that cases (b) and (c) satisfy the current range criteria and (b) is the
cheaper of the two, hence the correct choice is:
R1 and R3
Chapter 2, Solution 62
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
pA = 110x8 = 880 W, pB = 110x2 = 220 W
Energy cost = $0.06 x 365 x10 x (880 + 220)/1000 = $240.90
Chapter 2, Solution 63
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
n
m
Use eq. (2.61),
Rn = Im R 2x10
3 x100
0.04
I Im 5 2x103
In = I - Im = 4.998 A
p = I2 R (4.998)
2 (0.04) 0.9992
1 W
Chapter 2, Solution 64
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
When Rx = 0, i x 10A R = 110 10
11
110
When Rx is maximum, ix = 1A
i.e., Rx = 110 - R = 99
Thus, R = 11 , Rx = 99
R R x 1
110
Chapter 2, Solution 65
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
n
R Vfs
I fs
R m
50
10mA
1 k
4 k
Chapter 2, Solution 66
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
fs
n
n
20 k/V = sensitivity =
1
1
Ifs
i.e., Ifs = k/ V 50 A 20
V The intended resistance Rm = fs
I fs
10(20k/ V) 200k
(a) R Vfs
i fs
R m 50 V
50 A
200 k800 k
(b) p = I2 R (50 A)
2 (800 k) 2 mW
Chapter 2, Solution 67
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
i
v
i
0
0
(a) By current division,
i0 = 5/(5 + 5) (2 mA) = 1 mA
V0 = (4 k) i0 = 4 x 103
x 10-3
= 4 V
(b) 4k 6k 2.4k. By current division,
' 5 0
1 2.4 5
(2mA) 1.19 mA
' (2.4 k)(1.19 mA) 2.857 V
v v'
1.143
(c) % error = 0 0 x 100% = v0
x100 28.57% 4
(d) 4k 36 k3.6 k. By current division,
' 5 0
1 3.6 5
(2mA) 1.042mA
v' (3.6 k)(1.042 mA) 3.75V
% error =
v v' 0 x100%
v0
0.25x100
4
6.25%
Chapter 2, Solution 68
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) 40 24 60
i = 4
16 24
100 mA
(b)
i'
4
16 1 24
97.56 mA
(c) % error = 0.1 0.09756 x100% 2.44%
0.1
Chapter 2, Solution 69
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
V
With the voltmeter in place,
V0 R 2 R m
S
R1 R S R 2 R m
where Rm = 100 kwithout the voltmeter, R 2
V0 VS R1 R 2 R S
(a) When R2 = 1 k, R m
100
R 100
k2 101
V0 = 101
100 30
101
1
(40)
1.278 V (with)
V0 = 1 30
(40) 1.29 V (without)
1000
(b) When R2 = 10 k, R 2
9.091
R m
110 9.091k
V0 = 9.091 30
10
(40) 9.30 V (with)
V0 = 10 30
(40) 10 V (without)
(c) When R2 = 100 k, R 2 R m 50k
V0 50
50 30 100
(40) 25 V (with)
V0 = 100 30
(40) 30.77 V (without)
Chapter 2, Solution 70
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) Using voltage division,
v 12
(25) 15V a
12 8
v 10
(25) 10V b
10 15
vab va vb 15 10 5V
(b) c
+ 8k 15k
25 V
– a b
12k 10k
va = 0; vac = –(8/(8+12))25 = –10V; vcb = (15/(15+10))25 = 15V.
vab = vac + vcb = –10 + 15 = 5V.
vb = –vab = –5V.
Chapter 2, Solution 71
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
V
i 1
R1
iL
+ s − RL
Given that vs = 30 V, R1 = 20 , IL = 1 A, find RL.
v i ( R R )
R
vs R 30
20 10
s L 1 L L 1 L
Chapter 2, Solution 72
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Converting the delta subnetwork into wye gives the circuit below.
1
⅓
1 ⅓
+
10 V _
⅓ Zin
1
Z 1 (1
1) //(1
1)
1
1 ( 4
) 1 in
3 3 3 3 2 3
Vo Zin
1Zin
(10) 1
11
(10) 5 V
Chapter 2, Solution 73
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
By the current division principle, the current through the ammeter will be
one-half its previous value when
R = 20 + Rx
65 = 20 + Rx Rx = 45
Chapter 2, Solution 74
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
With the switch in high position,
6 = (0.01 + R3 + 0.02) x 5 R3 = 1.17
At the medium position,
6 = (0.01 + R2 + R3 + 0.02) x 3 R2 + R3 = 1.97
or R2 = 1.97 - 1.17 = 0.8
At the low position,
6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 + R2 + R3 = 5.97
R1 = 5.97 - 1.97 = 4
Chapter 2, Solution 75
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Converting delta-subnetworks to wye-subnetworks leads to the circuit below. 1
⅓
1 ⅓
⅓
1 1
1 1
⅓
1 ⅓
1 ⅓
1 (1
1) //(1
1)
1
1 ( 4
) 1 3 3 3 3 2 3 With this combination, the circuit is further reduced to that shown below.
1 1
1
1 1 1
Z 11 (1
1) //(1
1) 11 = 2
ab 3 3 3
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
1
Chapter 2, Solution 76
Zab= 1 + 1 = 2
Chapter 2, Solution 77
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) 5 = 10 10 20 20 20 20
i.e., four 20 resistors in parallel.
(b) 311.8 = 300 + 10 + 1.8 = 300 + 20 20 1.8
i.e., one 300resistor in series with 1.8resistor and
a parallel combination of two 20resistors.
(c) 40k= 12k+ 28k= (24 24k ) (56k 56k)
i.e., Two 24kresistors in parallel connected in series with two
56kresistors in parallel.
(a) 42.32k= 42l + 320
= 24k + 28k = 320
= 24k = 56k 56k 300 20
i.e., A series combination of a 20resistor, 300resistor,
24kresistor, and a parallel combination of two 56k
resistors.
Chapter 2, Solution 78
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
The equivalent circuit is shown below:
R
VS + -
+
V0 (1-)R
-
V0 =
(1 )R V
R (1 )R S
1
V 2
S
V0 1
VS 2
Chapter 2, Solution 79
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Since p = v2/R, the resistance of the sharpener is
R = v2/(p) = 6
2/(240 x 10
-3) = 150
I = p/(v) = 240 mW/(6V) = 40 mA
Since R and Rx are in series, I flows through both.
IRx = Vx = 9 - 6 = 3 V
Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75
Chapter 2, Solution 80
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
2 2 1
The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:
V + V +
- R1 - R2
Case 1 Case 2
Hence p V
, p2 R1 p
R1 p 10
(12) 30 W
R p1 R 2 R 2 4
Chapter 2, Solution 81
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
Let R1 and R2 be in k.
R eq R1 R 2 5 (1)
V0
VS
5 R 2
5 R 2 R1
(2)
From (1) and (2), 0.05
5 R1 2 = 5 R 2
5R 2
or R2 = 3.333 k
40
From (1), 40 = R1 + 2 R1 = 38 k
5 R 2
Thus,
R1 = 38 k, R2 = 3.333 k
Chapter 2, Solution 82
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
(a) 10
10
40
80
1 2
R12
50 R12 = 80 + 10 (10 40) 80
6
88.33
(b)
10
3
20
40
10 R13
80
1
R13 = 80 + 10 (10 40) 20 100 10 50 108.33
(c) 4
R14
20
10
40
10
80
1
R14 = 80 0 (10 40 10) 20 80 0 20 100
Chapter 2, Solution 83
Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual
The voltage across the fuse should be negligible when compared with 24
V (this can be checked later when we check to see if the fuse rating is
exceeded in the final circuit). We can calculate the current through the
devices.
I1 = p1 45mW
5mA
I2 =
V1
p2
V2
9V
480mW
24
20mA
+
24 V -
Ifuse
R1
R2
iR1
i1 = 5 mA
i2 = 20 mA
iR2
Let R3 represent the resistance of the first device, we can solve for its value from
knowing the voltage across it and the current through it.
R3 = 9/0.005 = 1,800 Ω
This is an interesting problem in that it essentially has two unknowns, R1 and R2 but only
one condition that need to be met and that the voltage across R3 must equal 9 volts.
Since the circuit is powered by a battery we could choose the value of R2 which draws
the least current, R2 = ∞. Thus we can calculate the value of R1 that give 9 volts across
R3.
9 = (24/(R1 + 1800))1800 or R1 = (24/9)1800 – 1800 = 3 kΩ
This value of R1 means that we only have a total of 25 mA flowing out of the battery
through the fuse which means it will not open and produces a voltage drop across it of
0.05V. This is indeed negligible when compared with the 24-volt source.