CHAPTER 2
Solutions for Exercises
E2.1
(a) R2, R3, and R4 are in parallel. Furthermore R1 is in series with the
combination of the other resistors. Thus we have:
1
Req  R1 
3
1 / R2  1 / R3  1 / R4
(b) R3 and R4 are in parallel. Furthermore, R2 is in series with the
combination of R3, and R4. Finally R1 is in parallel with the combination of
the other resistors. Thus we have:
1
Req 
5
1 / R1  1 /[R2  1 /(1 / R3  1 / R4 )]
(c) R1 and R2 are in parallel. Furthermore, R3, and R4 are in parallel.
Finally, the two parallel combinations are in series.
1
1
Req 

 52.1 
1 / R1  1 / R2 1 / R3  1 / R4
(d) R1 and R2 are in series. Furthermore, R3 is in parallel with the series
combination of R1 and R2.
1
Req 
 1.5 k
1 / R3  1 /(R1  R2 )
E2.2
(a) First we combine R2, R3, and R4 in parallel. Then R1 is in series with
the parallel combination.
Req 
1
 9.231 
1 / R2  1 / R3  1 / R4
v eq  Req i1  9.600 V
i1 
20 V
20

 1.04 A
R1  Req 10  9.231
i2  v eq / R2  0.480 A
i4  v eq / R4  0.240 A
1
i3  v eq / R3  0.320 A
(b) R1, and R2 are in series. Furthermore, R3, and R4 are in series.
Finally, the two series combinations are in parallel.
Req 1  R1  R2  20 
v eq  2  Req  20 V
Req 2  R3  R4  20  Req 
i1  v eq / Req 1  1 A
1 / Req 1
1
 10 
 1 / Req 2
i2  veq / Req 2  1 A
(c) R3, and R4 are in series. The combination of R3 and R4 is in parallel
with R2. Finally the combination of R2, R3, and R4 is in series with R1.
Req 1  R3  R4  40  Req 2 
v 2  i1Req 2  20 V
E2.3
(a) v 1  v s
1
 20 
1 / Req 1  1 / R2
i1 
vs
1A
R1  Req 2
i2  v 2 / R2  0.5 A i3  v 2 / Req 1  0.5 A
R1
R2
 10 V . v 2  v s
 20 V .
R1  R2  R3  R4
R1  R2  R3  R4
Similarly, we find v 3  30 V and v 4  60 V .
2
(b) First combine R2 and R3 in parallel: Req  1 (1 / R2  1 R3 )  2.917 .
Then we have v 1  v s
v2  vs
E2.4
Req
R1  Req  R4
R1
 6.05 V . Similarly, we find
R1  Req  R4
 5.88 V and v 4  8.07 V .
(a) First combine R1 and R2 in series: Req = R1 + R2 = 30 . Then we have
Req
R3
15
30
i1  is

 1 A and i3  is

 2 A.
R3  Req 15  30
R3  Req 15  30
(b) The current division principle applies to two resistances in parallel.
Therefore, to determine i1, first combine R2 and R3 in parallel: Req =
Req
5
1/(1/R2 + 1/R3) = 5 . Then we have i1  is

 1A.
R1  Req 10  5
Similarly, i2 = 1 A and i3 = 1 A.
E2.5
Write KVL for the loop consisting of v1, vy , and v2. The result is -v1 - vy +
v2 = 0 from which we obtain vy = v2 - v1. Similarly we obtain vz = v3 - v1.
E2.6
Node 1:
E2.7
Using determinants we can solve for the unknown voltages as follows:
6  0.2
v1  v3 v1  v2
v  v1 v2 v2  v3

 ia


0
Node 2: 2
R1
R2
R2
R3
R4
v
v  v2 v3  v1

 ib  0
Node 3: 3  3
R5
R4
R1
v1 
1
0.5
0.7  0.2
 0.2 0.5

3  0.2
 10.32 V
0.35  0.04
0.7 6
 0.2 1
0.7  1.2
v2 

 6.129 V
0.7  0.2 0.35  0.04
 0.2 0.5
Many other methods exist for solving linear equations.
3
E2.8
First write KCL equations at nodes 1 and 2:
v 1  10 v 1 v 1  v 2
 
0
2
5
10
v  10 v 2 v 2  v 1
Node 2: 2
 
0
10
5
10
Node 1:
Then simplify the equations to obtain:
8v 1 v 2  50 and  v 1  4v 2  10
Solving, we find v1 = 6.77 V and v2 = 4.19 V.
E2.9
(a) Writing the node equations we obtain:
v1  v3 v1 v1  v2
 
0
20
5
10
v v3
v  v1
 10  2
0
Node 2: 2
10
5
v  v1 v3 v3  v2


0
Node 3: 3
20
10
5
Node 1:
(b) Simplifying the equations we obtain:
0.35v 1  0.10v 2  0.05v 3  0
 0.10v 1  0.30v 2  0.20v 3  10
 0.05v1  0.20v2  0.35v3  0
(c) Solving yields v1 = -27.27 V, v2 = -72.73 V, and v3 = -45.45 V.
(d) Finally, ix  (v 1  v 3 ) 20  0.909 A.
E2.10
The equation for the supernode enclosing the 15-V source is:
v3  v2 v3  v1 v1 v2



R3
R1
R2 R4
This equation can be readily shown to be equivalent to Equation 2.34 in
the book. (Keep in mind that v3 = -15 V.)
4
E2.11
Write KVL from the reference to node 1 then through the 10 V source to
node 2 then back to the reference node:
 v 1  10  v 2  0
Then write KCL equations. First for a supernode enclosing the 10-V
source we have:
v1 v1  v3 v2  v3


1
R1
R2
R3
Node 3:
v3 v3  v1 v3  v2


0
R4
R2
R3
Reference node:
v1 v3

1
R1 R4
An independent set consists of the KVL equation and any two of the KCL
equations.
E2.12
(a) Select the
reference node at the
left-hand end of the
voltage source as shown
at right.
Then write a KCL
equation at node 1.
v 1 v 1  10

1  0
R1
R2
Substituting values for the resistances and solving, we find v1 = 3.33 V.
10  v 1
Then we have ia 
 1.333 A.
R2
(b) Select the
reference node and
assign node voltages as
shown.
Then write KCL
equations at nodes 1
and 2.
5
v 1  25 v 1 v 1  v 2


0
R2
R4
R3
v 2  25 v 2  v 1 v 2


0
R1
R3
R5
Substituting values for the resistances and solving, we find v1 = 13.79 V
v v2
 -0.259 A.
and v 2 = 18.97 V. Then we have ib  1
R3
E2.13
(a) Select the
reference node and
node voltage as
shown. Then write a
KCL equation at node
1, resulting in
v 1 v 1  10

 2ix  0
5
5
Then use ix  (10  v 1 ) / 5 to substitute and solve. We find v1 = 7.5 V.
10  v 1
Then we have ix 
 0.5 A.
5
(b) Choose the reference node and node voltages shown:
Then write KCL equations at nodes 1 and 2:
v 1 v 1  2i y

3  0
5
2
v 2 v 2  2i y

3
5
10
6
Finally use i y  v 2 / 5 to substitute and solve. This yields v 2  11.54 V and
i y  2.31 A.
E2.14
Refer to Figure 2.32b in the book. (a) Two mesh currents flow through
R2: i1 flows downward and i4 flows upward. Thus the current flowing in R2
referenced upward is i4 - i1. (b) Similarly, mesh current i1 flows to the
left through R4 and mesh current i2 flows to the right, so the total
current referenced to the right is i2 - i1. (c) Mesh current i3 flows
downward through R8 and mesh current i4 flows upward, so the total
current referenced downward is i3 - i4. (d) Finally, the total current
referenced upward through R8 is i4 - i3.
E2.15
Refer to Figure 2.32b in the book. Following each mesh current we have
R1i1  R2 (i1  i4 )  R4 (i1  i2 )  v A  0
R5i2  R4 (i2  i1 )  R6 (i2  i3 )  0
R7i3  R6 (i3  i2 )  R8 (i3  i4 )  0
R3i4  R2 (i4  i1 )  R8 (i4  i3 )  0
E2.16
We choose the mesh currents as shown:
Then the mesh equations are:
5i 1  10(i1  i2 )  100
and 10(i2  i1 )  7i2  3i2  0
Simplifying and solving these equations, we find that i1  10 A and
i2  5 A. The net current flowing downward through the 10-Ω resistance
is i1  i2  5 A.
To solve by node voltages, we select the reference node and node voltage
shown. (We do not need to assign a node voltage to the connection
7
between the 7-Ω resistance and the 3-Ω resistance because we can
treat the series combination as a single 10-Ω resistance.)
The node equation is (v 1  10) / 5  v 1 / 10  v 1 / 10  0 . Solving we find that
v1 = 50 V. Thus we again find that the current through the 10-Ω
resistance is i  v 1 / 10  5 A.
Combining resistances in series and parallel, we find that the resistance
“seen” by the voltage source is 10 Ω. Thus the current through the
source and 5-Ω resistance is (100 V)/(10 Ω) = 10 A. This current splits
equally between the 10-Ω resistance and the series combination of 7 Ω
and 3 Ω.
E2.17
First, we assign the mesh currents as shown.
Then we write KVL equations following each mesh current:
2(i1  i3 )  5(i1  i2 )  10
5i2  5(i2  i1 )  10(i2  i3 )  0
10i3  10(i3  i2 )  2(i3  i1 )  0
8
Simplifying and solving, we find that i1 = 2.194 A, i2 = 0.839 A, and i3 =
0.581 A. Thus the current in the 2-Ω resistance referenced to the right
is i1 - i3 = 2.194 - 0.581 = 1.613 A.
E2.18
Refer to Figure 2.37 in the book. In terms of the mesh currents the
current directed to the right in the 5-A current source is i1, however by
the definition of the current source, the current is 5 A directed to the
left. Thus we conclude that i1 = -5 A. Then we write a KVL equation
following i2, which results in 10(i2  i1 )  5i2  100.
E2.19
Refer to Figure 2.38 in the book. First, for the current source, we have
i2  i1  1
Then we write a KVL equation going around the perimeter of the entire
circuit:
5i1  10i2  20  10  0
E2.20
Simplifying and solving these equations we obtain i1 = -4/3 A andi2 = -1/3
A.
(a) As usual, we select the
mesh currents flowing
clockwise around the
meshes as shown.
Then for the current
source we have i2 =
-1 A. This is because
we defined the mesh
current i2 as the
current referenced downward through the current source. However,
we know that the current through this source is 1 A flowing upward.
Next we write a KVL equation around mesh 1: 10i1  10  5(i1  i2 )  0.
Solving we find that i1 = 1/3 A. Referring to Figure 2.29a in the book
we see that the value of the current ia referenced downward through
the 5 Ω resistance is to be found. In terms of the mesh currents we
have ia  i1  i2  4 / 3 A .
9
(b) As usual, we select
the mesh currents
flowing clockwise
around the meshes
as shown.
Then we write a
KVL equation for
each mesh.
 25  10(i1  i3 )  10(i1  i2 )  0
10(i2  i1 )  20(i2  i3 )  20i2  0
10(i3  i1 )  5i3  20(i3  i2 )  0
Simplifying and solving, we find i1 = 2.3276 A, i2 = 0.9483 A, and i3 =
1.2069 A. Finally, we have ib = i2 - i3 = -0.2586 A.
E2.21
(a) KVL mesh 1:
 10  5i1  5(i1  i2 )  0
For the current source:
i2  2ix
However, ix and i1 are
the same current, so we
also have i1 = ix.
Simplifying and solving, we find ix  i1  0.5 A.
(b) First for the current
source, we have: i1  3 A
Writing KVL around
meshes 2 and 3, we have:
2(i2  i1 )  2i y  5i2  0
10(i3  i1 )  5i3  2i y  0
However i3 and iy are the same current: i y  i3 . Simplifying and solving, we
find that i3  i y  2.31 A.
10
E2.22
Under open-circuit conditions, 5 A circulates clockwise through the
current source and the 10- resistance. The voltage across the 10-
resistance is 50 V. No current flows through the 40- resistance so the
open circuit voltage is Vt  50 V.
With the output shorted, the 5 A divides between the two resistances in
parallel. The short-circuit current is the current through the 40-
10
 1 A. Then the Thévenin resistance is
resistance, which is isc  5
10  40
Rt  v oc / isc  50 .
E2.23
Choose the reference node at the bottom of the circuit as shown:
Notice that the node voltage is the open-circuit voltage. Then write a
KCL equation:
v oc  20 v oc

2
5
20
Solving we find that voc = 24 V which agrees with the value found in
Example 2.15.
E2.23
To zero the sources, the voltage sources become short circuits and the
current sources become open circuits. The resulting circuits are :
11
(a) Rt  10 
(c) Rt 
E2.25
1

10
1
 14 
1 / 5  1 / 20
1
6
1
(b) Rt  10  20  30 
5
1
(1 / 5  1 / 20)
(a) Zero sources to determine Thévenin
resistance. Thus
1
Rt 
 9.375 .
1 / 15  1 / 25
Then find short-circuit current:
I n  isc  10 / 15  1  1.67 A
12
(b) We cannot find the Thévenin resistance by zeroing the sources
because we have a controlled source. Thus we find the open-circuit
voltage and the short-circuit current.
v oc  2v x v oc

2
v oc  3v x
10
30
Solving we find Vt  v oc  30 V.
Now we find the short-circuit current:
2v x  v x  0

vx  0
Therefore isc  2 A. Then we have Rt  v oc / isc  15 .
E2.26
First we transform the 2-A source and the 5-Ω resistance into a voltage
source and a series resistance:
13
10  10
 1.333 A.
15
From the original circuit, we have i1  i2  2, from which we find
Then we have i2 
i1  0.667 A.
The other approach is to start from the original circuit and transform
the 10-Ω resistance and the 10-V voltage source into a current source
and parallel resistance:
1
 3.333  .
1 / 5  1 / 10
The current flowing upward through this resistance is 1 A. Thus the
voltage across Req referenced positive at the bottom is
3.333 V and i1  3.333 / 5  0.667 A. Then from the original circuit we
Then we combine the resistances in parallel. Req 
have i2  2  i1  1.333 A, as before.
E2.27
Refer to Figure 2.60b. We have i1  15 / 15  1 A.
Refer to Figure 2.60c. Using the current division principle, we have
5
i2  2 
 0.667 A. (The minus sign is because of the reference
5  10
direction of i2.) Finally, by superposition we have iT  i1  i2  0.333 A.
E2.28
With only the first source active we have:
Then we combine resistances in series and parallel:
1
Req  10 
 13.75 
1 / 5  1 / 15
14
Thus, i1  20 / 13.75  1.455 A, and v 1  3.75i1  5.45 V.
With only the second source active, we have:
Then we combine resistances in series and parallel:
1
Req 2  15 
 18.33 
1 / 5  1 / 10
Thus, is  10 / 18.33  0.546 A, and v 2  3.33is  1.818 V. Then we have
i2  (v 2 ) / 10  0.1818 A
Finally we have vT  v 1  v 2  5.45  1.818  7.27 V and
iT  i1  i2  1.455  0.1818  1.27 A.
E2.29
First, we replace the controlled source by an independent source denoted
as I1. Then, we activate one source at a time and solve for vx.
10
1

I
10  20 1 / 30  1 /(20  10) 1
 5I 1
vx  
15
v x  30
vx  3
10
 5
10  20  30
1
 25
1 / 10  1 /(20  30)
Adding the contributions from all three sources, we have
v x  5I1  5  25
Then substituting I1  0.6v x , we have
v x  5(0.6v x )  5  25
which yields v x  5 V. Then, applying Ohm's law and KCL in the original
circuit, we readily find that i y  0.5 A.
Answers for Selected Problems
P2.1*
(a) Req  20 
P2.4*
Rx  5 .
(b) Req  23 
16
P2.10*
Rab  10 
P2.12*
P2.16*
Rab  9.6 
P2.22*
v1  3 V
P2.25*
v  140 V; i  1 A
P2.26*
i1  1 A
P2.33*
i2  0.5 A
i1  1.5 A
P4A  30 W delivering
P2A  15 W absorbing
P5  11.25 W absorbing
P15  3.75 W absorbing
P2.35*
i1  2.5 A
i2  0.8333 A
P2.36*
v1  5 V
v2  7 V
P2.37*
i1  1 A
i2  2 A
P2.41*
Rg  25 m
P2.43*
v  3.333 V
v 2  0.5 V
i2  0.5 A
v 3  13 V
17
P2.47*
v 1  14.29 V
v 2  11.43 V
i1  0.2857 A
P2.48*
v 1  6.667 V
v 2  3.333 V
is  3.333 A
P2.55*
v1  6 V
v2  4 V
ix  0.4 A
P2.57*
v 1  5.405 V
v 2  7.297 V
P2.61*
i1  2.364 A
i2  1.818 A
P2.63*
v 2  0.500 V
P 6 W
P2.71*
i1  0.2857 A
P  4.471 W
P2.75*
P2.77*
Rt  50 
P2.86*
Rt  0
Pmax  80 W
P2.89*
iv  2 A
ic  2 A
P2.92*
is  3.333 A
P2.98*
i x  0.4 A
P2.102* R3  5932 
i  iv  ic  4 A
v1  6.000 V
idetector  31.65  10 9 A
18