Chapter 3 Homework Solution
P3.2-2, 4, 6, 10, 13, 17, 21
P3.3-2, 4, 6, 11
P3.4-1, 3, 6, 9, 12
P3.5-2
P3.6-1, 4, 9, 14, 21, 31, 40
---------------------------------------------------P 3.2-2 Determine the values of i2, i4, v2, v3, and v6 in Figure P 3.2-2.
Solution:
Apply KCL at node a to get
2 = i2 + 6 = 0  i2 = 4 A
Apply KCL at node b to get
3 = i4 + 6  i4 = -3 A
Apply KVL to the loop consisting of elements A and B to get
-v2 – 6 = 0  v2 = -6 V
Apply KVL to the loop consisting of elements C, D, and A to get
-v3 – (-2) – 6 = 0  v4 = -4 V
Apply KVL to the loop consisting of elements E, F and D to get
4 – v6 + (-2) = 0  v6 = 2 V
Check: The sum of the power supplied by all branches is
(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0
P 3.2-4
3.2-4.
Determine the power absorbed by each of the resistors in the circuit shown in Figure P
solution
12
i 
2A
1 6
20
i 
 5A
2
4
i  3i   2 A
3
2
i  i i  3A
4 2 3
Power absorbed by the 4  resistor = 4  i 2 = 100 W
2
Power absorbed by the 6  resistor = 6  i 2 = 24 W
1
Power absorbed by the 8  resistor = 8  i 2 = 72 W
4
P 3.2-6
Answer:
(checked using LNAP 8/16/02)
Determine the power supplied by each voltage source in the circuit of Figure P 3.2-6.
The 2-V voltage source supplies 2 mW and the 3-V voltage source supplies –6 mW.
Figure P 3.2-6
Solution:
P2 mA   3   2 103    6 103  6 mW
P1 mA    7  1103    7 103  7 mW
(checked using LNAP 8/16/02)
P 3.2-10 The circuit shown in Figure P 3.2-10 consists of five voltage sources and four current
sources. Express the power supplied by each source in terms of the voltage source voltages and
the current source currents.
Figure P 3.2-10
Solution:
The subscripts suggest a numbering of the sources. Apply KVL to get
v1  v 2  v 5  v 9  v 6
i 1 and v 1 do not adhere to the passive convention, so
p 1  i1 v 1  i1  v 2  v 5  v 9  v 6 
is the power supplied by source 1. Next, apply KCL to get
i 2    i1  i 4 
i 2 and v 2 do not adhere to the passive convention, so
p 2  i 2 v 2    i1  i 4  v 2
is the power supplied by source 2. Next, apply KVL to get
v3  v6  v5  v9 
i 3 and v 3 adhere to the passive convention, so

p 3  i 3 v 3  i 3 v 6   v 5  v 9 

is the power supplied by source 3. Next, apply KVL to get
v 4  v 2  v5  v8
i 4 and v 4 do not adhere to the passive convention, so
p 4  i4 v 4  i4 v 2  v5  v8 
is the power supplied by source 4. Next, apply KCL to get


i 5  i 3  i 2  i 3    i1  i 4   i1  i 3  i 4
i 5 and v 5 adhere to the passive convention, so
p 5  i 5 v 5    i1  i 3  i 4  v 5
is the power supplied by source 5. Next, apply KCL to get
i 6  i 7   i1  i 3 
i 6 and v 6 adhere to the passive convention, so


p 6  i 6 v 6   i 7   i1  i 3  v 6
is the power supplied by source 6. Next, apply KVL to get
v 7  v 6
i 7 and v 7 adhere to the passive convention, so
p 7  i 7 v 7  i 7  v 6   i 7 v 6
is the power supplied by source 7. Next, apply KCL to get
i 8  i 4
i 8 and v 8 do not adhere to the passive convention, so
p 8  i 8 v 8   i 4  v 8  i 4 v 8
is the power supplied by source 8. Finally, apply KCL to get
i 9  i1  i 3
i 9 and v 9 adhere to the passive convention, so
p 9  i 9 v 9    i1  i 3  v 9
is the power supplied by source 9.
9
(Check:
p
n 1
P 3.2-13
n
 0 .)
Determine the value of the current that is measured by the meter in Figure P 3.2-13.
Figure P 3.2-13
Solution:
We can label the circuit as shown.
The subscripts suggest a numbering of the
circuit elements. Apply KVL to node the left
mesh to get
15 i1  25 i1  20  0  i1 
20
 0.5 A
40
Apply KVL to node the left mesh to get
v 2  25 i1  0  v 2  25 i1  25  0.5   12.5 V
Apply KCL to get i m  i 2 . Finally, apply Ohm’s law to the 50  resistor to get
im  i2 
v2
50

12.5
 0.25 A
50
(Checked: LNAPDC 9/1/04)
P 3.2-17 Determine the current i in Figure P 3.3-17.
Answer: i = 4 A
Figure P 3.3-17
Solution:
Apply KCL at node a to determine the current
in the horizontal resistor as shown.
Apply KVL to the loop consisting of the
voltages source and the two resistors to get
-4(2-i) + 4(i) - 24 = 0  i = 4 A
P3.2-21 Determine the value of the voltage v5 for the circuit shown in Figure P3.2-21.
Figure P3.2-21
Solution:
Apply KVL to the left mesh:
v 2  18  12  0  v 2  6 V
Use the element equation of the dependent
source:
i 6  0.10 v 2  0.10  6   0.6 A
Apply KCL at the right node
v5
25
 i 6  0.25  v 5  25  0.25  i 6   25  0.25  0.6   8.75 V
P 3.3-2 Consider the circuits shown in Figure P 3.3-2.
(a)
Determine the value of the resistance R in Figure P 3.3-2b that makes the circuit in Figure
P 3.3-2b equivalent to the circuit in Figure P 3.3-2a.
(b)
Determine the current i in Figure P 3.3-2b. Because the circuits are equivalent, the
current i in Figure P 3.3-2a is equal to the current i in Figure P 3.3-2b.
(c)
Determine the power supplied by the voltage source.
Solution:
(a ) R  6  3  2  4  15 
28 28

 1.867 A
R 15
p  28  i =28(1.867)=52.27 W
(b) i 
c
(28 V and i do not adhere
to the passive convention.)
P 3.3-4
Determine the voltage v in the circuit shown in Figure P 3.3-4.
Figure P 3.3-4
Solution:
Voltage division
16
12  8 V
16  8
4
v3 
12  4 V
48
v1 
KVL: v3  v  v1  0
v  4 V
(checked using LNAP 8/16/02)
P 3.3-6
The input to the circuit shown in Figure P 3.3-6 is the voltage of the voltage source, va.
The output of this circuit is the voltage measured by the voltmeter, vb. This circuit produces an
output that is proportional to the input, that is
vb = k va
where k is the constant of proportionality.
(a)
Determine the value of the output, vb, when R = 240 Ω and va = 18 V.
(b)
Determine the value of the power supplied by the voltage source when R = 240 Ω and va
= 18 V.
(c)
Determine the value of the resistance, R, required to cause the output to be vb = 2 V when
the input is va = 18 V.
(d)
Determine the value of the resistance, R, required to cause vb = 0.2va (that is, the value of
the constant of proportionality is k  102 ).
Figure P 3.3-6
Solution:
 180 
a.) 
 18  10.8 V
 120  180 
18


b.) 18 
  1.08 W
 120  180 
 R 
c.) 
 18  2  18 R  2 R  2 120   R  15 
 R  120 
R
d.) 0.2 
  0.2 120   0.8 R  R  30 
R  120
P 3.3-11 For the circuit of Figure P 3.3-11, find the voltage v3 and the current i and show that
the power delivered to the three resistors is equal to that supplied by the source.
Figure P 3.3-11
Solution:

3 

 39 
From voltage division v3  12 
then i =
 3V
v3
=1A
3
The power absorbed by the resistors is: 12   6   12   3  12   3  12 W
The power supplied by the source is (12)(1) = 12 W.
P 3.4-1 Use current division to determine the currents i1, i2, i3, and i4 in the circuit shown in
Figure P 3.4-1.
Figure P 3.4-1.
Solution:
i 
1
i 
2
i 
3
i 
4
P 3.4-3
1
1
1
6
4
4 A
1  1  1 1
1 2  3  6
3
6 3 2 1
1
2
3
4  A;
1  1  1 1
3
6 3 2 1
1
2
4 1 A
1  1  1 1
6 3 2 1
1
42A
1  1  1 1
6 3 2
The ideal voltmeter in the circuit shown in Figure P 3.4-3 measures the voltage v.
Figure P 3.4-3
(a)
(b)
(c)
Suppose R2 = 12 Ω. Determine the value of R1 and of the current i.
Suppose, instead, R1 = 12 Ω. Determine the value of R2 and of the current i.
Instead, choose R1 and R2 to minimize the power absorbed by any one resistor.
Solution:
i
8
8
or R1 
R1
i
8  R 2 (2  i )  i  2 
8
8
or R 2 
R2
2i
8 2
8
 A ; R1   12 
2
6 3
3
8 4
8
 b  i   A ; R 2  4  12 
6 3
2
3
1
 c  R1  R 2 will cause i= 2  1 A. The current in both R1 and R 2 will be 1 A.
2
R1 R 2
1
2 
 8 ; R1  R 2  2  R 1  8  R 1  8  R 1  R 2  8 
R1  R 2
2
a
i  2
P 3.4-6 Figure P 3.4-6 shows a transistor amplifier. The values of R1 and R2 are to be selected.
Resistances R1 and R2 are used to bias the transistor, that is, to create useful operating conditions.
In this problem, we want to select R1 and R2 so that vb = 5 V. We expect the value of ib to be
approximately 10 μA. When i1 ≥ 10ib, it is customary to treat ib as negligible, that is, to assume ib
= 0. In that case R1 and R2 comprise a voltage divider.
(a)
(b)
Figure P 3.4-6
Select values for R1 and R2 so that vb = 5 V and the total power absorbed by R1 and R2 is
no more than 5 mW.
An inferior transistor could cause ib to be larger than expected. Using the values of R1 and
R2 from part (a), determine the value of vb that would result from ib = 15 μA.
Solution:
(a) To insure that ib is negligible we require
R1  R 2  150 k
so
To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require
152
 5 103
 R1  R 2  45 k
R1  R 2
Next to cause vb = 5 V we require
5  vb 
R2
R1  R 2
15

R1  2 R 2
For example, R1  40 k, R 2  80 k, satisfy all three requirements.
(b)
80 10  i
3
KVL gives
KCL gives
Therefore
i1 
1
 v b  15  0
vb
 15 106
40 10
 v

80 103   40 b103  15 106   v b  15


3
Finally
3v b  1.2  15

vb 
13.8
 4.6 V
3
P 3.4-9 Determine the power supplied by the
dependent source in Figure P 3.4-9.
Figure P 3.4-9
Solution:
Use current division to get
ia  
so
75
30 103   22.5 mA

25  75
v b  50  22.5 103   1.125 V
The power supplied by the dependent source is
given by
p    30 103   1.125  33.75 mW
P 3.4-12
Determine the value of the current measured by the meter in Figure P 3.4-12.
Figure P 3.4-12
Solution:
Replace the (ideal) ammeter with the equivalent
short circuit. Label the current measured by the
meter.
Apply KCL at the left node of the VCCS to get
1.2 
va
10
 0.2 v a  0.3 v a
 va 
1.2
4V
0.3
Use current division to get
im 
30
30
0.2 v a 
0.2  4   0.6 A
30  10
30  10
P 3.5-2
Determine the power supplied by each source in the circuit shown in Figure P 3.5-2.
Figure P 3.5-2
Solution:
20  5
 4 .
20  5
The 7- resistor is connected in parallel with a short circuit, a 0- resistor. The equivalent
0 7
 0  , a short circuit.
resistance is
07
The 20- and 5- resistors are connected in parallel. The equivalent resistance is
The voltage sources are connected in series and can be
replaced by a single equivalent voltage source.
After doing so, and labeling the resistor currents, we have
the circuit shown.
The parallel current sources can be replaced by an
equivalent current source.
Apply KVL to get
5  v1  4  3.5   0  v1  19 V
The power supplied by each sources is:
Source
8-V voltage source
Power delivered
3-V voltage source
3  3.5   10.5 W
3-A current source
0.5-A current source
3  19  57 W
0.5 19  9.5 W
2  3.5   7 W
(Checked using LNAP, 9/15/04)
P 3.6-1 The circuit shown in Figure P
3.6-1a has been divided into two parts. In
Figure P 3.6-1b, the right-hand part has
been replaced with an equivalent circuit.
The left-hand part of the circuit has not
been changed.
(a)
Determine the value of the
resistance R in Figure P 3.6-1b that
makes the circuit in Figure P 3.61b equivalent to the circuit in
Figure P 3.6-1a.
(b)
Find the current i and the voltage v
shown in Figure P 3.6-1b. Because
of the equivalence, the current i
and the voltage v shown in Figure
P 3.6-1a are equal to the current i
and the voltage v shown in Figure
P 3.6-1b.
(c)
Find the current i2 shown in Figure
Figure P 3.6-1
P 3.6-1a using current division.
Solution:
a
b
c
48  24
 32 
48  24
32  32
v  32  32 24  16 V ;
32  32
8
32  32
16 1
i
 A
32 2
48
1
1
i2 
  A
48  24 2
3
R  16 
P 3.6-4
(a)
Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b
equivalent to the circuit in Figure P 3.6-4a.
(b)
Analyze the circuit in Figure P 3.6-4b to determine the values of the currents ia and ib
(c)
Because the circuits are equivalent, the currents ia and ib shown in Figure P 3.6-4b are
equal to the currents ia and ib shown in Figure P 3.6-4a. Use this fact to determine values
of the voltage v1 and current i2 shown in Figure P 3.6-4a.
Figure P 3.6-4
(a)
1
1
1 1

 
 R2  4 
R2 24 12 8
and
(b)
First, apply KVL to the left mesh to get 27  6 ia  3 ia  0  ia  3 A . Next,
apply KVL to the left mesh to get 4 ib  3ia  0  ib  2.25 A .
(c)
P 3.6-9 Determine the value of the
current i in Figure 3.6-9.
Answer: i = 0.5 mA
Figure 3.6-9
Solution:
P 3.6-14 All of the resistances in the circuit shown in Figure P 3.6-14 are multiples of R.
Determine the value of R.
Figure P 3.6-14
Solution:
R

R  2R
So the circuit is equivalent to
4
6
R  R  2R
5
5
 R   2R 2R   2R
4 R    2 R 3R  
 R   2R
2R 

Then
12  0.1 R   2 R 2 R    0.1 2 R 

R  60 
P 3.6-21 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given
that Req = 9 Ω.
Answer: R = 15 Ω
Figure P 3.6-22
Solution:
Replace parallel resistors by an equivalent
resistor:
8 || 24 = 6 
A short circuit in parallel with a resistor is
equivalent to a short circuit.
Replace series resistors by an equivalent
resistor:
4+6 = 10 
Now
9  R eq  5  12 || R ||10 
so
60
11
4
60
R
11
R
P 3.6-31
 R  15 
The voltmeter in Figure P 3.6-31 measures the voltage across the current source.
Figure P 3.6-31
(a)
(b)
Determine the value of the voltage measured by the meter.
Determine the power supplied by each circuit element.
Solution:
Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by
the meter. Label the element voltages and currents as shown in (b).
Using units of V, A,  and W:
Using units of V, mA, k and mW:
a.) Determine the value of the voltage measured by
the meter.
a.) Determine the value of the voltage
measured by the meter.
Kirchhoff’s laws give
Kirchhoff’s laws give
12  v R  v m and i R  i s  2 mA
12  v R  v m and i R  i s  2 103 A
Ohm’s law gives
Ohm’s law gives


v R  25 i R
v R   25 103 i R
Then
Then




v R  25 i R  25  2   50 V

v R   25 103 i R   25 103 2 103  50 V
v m  12  v R  12  50  62 V
v m  12  v R  12  50  62 V
b.) Determine the power supplied by each element.
voltage source

 
12 i s  12 2 103

 24 10
3
62 2 10
resistor
vR iR
3
3
0
voltage source
 
12 i s  12  2 
 24 mW
W
current source
62  2   124 mW
resistor
v R i R  50  2 
3
 100  103 W
total

W
  124 10
 50  2 10 
current source
b.) Determine the power supplied by each
element.
 100 mW
total
0
P3.6-40 Consider the circuit shown in Figure P3.6-40. Given that the voltage of the dependent
voltage source is v a  8 V , determine the values of R1 and v o .
Figure P3.6-40
Solution:
First,
vo  
20
8  3.2 V
20  30
Next,



8
40
40 
10
40 
10
 ib 
ic 

 
40 R1
20
40  R1
40  R1  12  40 || R1  40  R1 
 12 
40  R1

then
400  20 
8
400

 480  52 R1 
 1000 
20 480  52 R1
8


400
400


 12 40  R  40 R
480  52 R1
1
1




1000  480
 10 
52