CHAPTER 2
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
=3Ω
Req = R1 +
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
=5Ω
Req =
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
+
= 52.1 Ω
Req =
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
= 1.5 kΩ
Req =
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 Ω
1 / R2 + 1 / R3 + 1 / R4
i1 =
20 V
20
=
= 1.04 A
R1 + Req 10 + 9.231
v eq = Req i1 = 9.600 V
i2 = v eq / R2 = 0.480 A
i4 = v eq / R4 = 0.240 A
22
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 Ω
v eq = 2 × Req = 20 V
Req 2 = R3 + R4 = 20 Ω Req =
i1 = v eq / Req 1 = 1 A
1 / Req 1
1
= 10 Ω
+ 1 / Req 2
i2 = v eq / 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 Ω Req 2 =
v 2 = i1Req 2 = 20 V
E2.3
(a) v 1 = v s
1
= 20 Ω
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
= 10 V . v 2 = v s
R2
R1 + R2 + R3 + R4
R1 + R2 + R3 + R4
Similarly, we find v 3 = 30 V and v 4 = 60 V .
23
= 20 V .
(b) First combine R2 and R3 in parallel: Req = 1 (1 / R2 + 1 R3 ) = 2.917 Ω.
Then we have v 1 = v s
v2 = vs
E2.4
R1
= 6.05 V . Similarly, we find
R1 + Req + R4
Req
= 5.88 V and v 4 = 8.07 V .
R1 + Req + R4
(a) First combine R1 and R2 in series: Req = R1 + R2 = 30 Ω. Then we have
Req
R3
30
15
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 Ω. Then we have i1 = is
=
= 1A.
R1 + Req 10 + 5
Similarly, i2 = 1 A and i3 = 1 A.
E2.5
E2.6
E2.7
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.
v1 − v3 v1 − v2
v − v1 v2 v2 − v3
+
= ia
Node 2: 2
+
+
=0
R1
R2
R3
R4
R2
v
v − v2 v3 − v1
Node 3: 3 + 3
+
+ ib = 0
R5
R4
R1
Node 1:
Using determinants we can solve for the unknown voltages as follows:
6 − 0. 2
1 0. 5
3 + 0. 2
v1 =
= 10.32 V
=
0.7 − 0.2 0.35 − 0.04
− 0 . 2 0. 5
0. 7
v2 =
6
− 0. 2 1
0 . 7 + 1. 2
=
= 6.129 V
0.7 − 0.2 0.35 − 0.04
− 0. 2 0. 5
Many other methods exist for solving linear equations.
24
E2.8
First write KCL equations at nodes 1 and 2:
Node 1:
v 1 − 10 v 1
v1 − v2
=0
2
5
10
v − 10 v 2 v 2 − v 1
Node 2: 2
+
+
=0
10
5
10
+
+
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
Node 2: 2
+ 10 + 2
=0
10
5
v − v1 v3 v3 − v2
Node 3: 3
+
+
=0
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
− 0.05v 1 − 0.20v 2 + 0.35v 3 = 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.)
25
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.
26
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:
v1
5
+
v 1 − 2i y
2
+3= 0
v2
5
+
v 2 − 2i y
10
27
=3
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
28
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
29
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
Simplifying and solving these equations we obtain i1 = -4/3 A and i2 = -1/3
A.
E2.20
(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 .
30
(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.
31
E2.22
Under open-circuit conditions, 5 A circulates clockwise through the
current source and the 10-Ω resistance. The voltage across the 10-Ω
resistance is 50 V. No current flows through the 40-Ω 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-Ω
10
= 1 A. Then the Thévenin resistance is
resistance, which is isc = 5
10 + 40
Rt = v oc / isc = 50 Ω.
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 :
32
(a) Rt = 10 +
(c) Rt =
E2.25
1
= 14 Ω
1 / 5 + 1 / 20
1
+
10 6 +
1
1
(b) Rt = 10 + 20 = 30 Ω
=5Ω
1
(1 / 5 + 1 / 20)
(a) Zero sources to determine Thévenin
resistance. Thus
1
Rt =
= 9.375 Ω.
1 / 15 + 1 / 25
Then find short-circuit current:
I n = isc = 10 / 15 + 1 = 1.67 A
33
(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 Ω.
E2.26
First we transform the 2-A source and the 5-Ω resistance into a voltage
source and a series resistance:
34
10 + 10
= 1.333 A.
15
From the original circuit, we have i1 = i2 − 2, from which we find
i1 = −0.667 A.
Then we have i2 =
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 Ω .
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 Ω
1 / 5 + 1 / 15
35
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 Ω
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 = −
36
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 = −5I 1 − 5 + 25
Then substituting I 1 = 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.
Problems
P2.1*
(a) Req = 20 Ω
P2.2
(a) Req = 20.32 Ω
(b) Req = 23 Ω
(b) Req = 33.44 Ω
37
P2.3
(a) Req = 151.5 Ω
P2.4*
We have 4 +
P2.5
We have
P2.6
We have Req =
(b) Req = 99.80 Ω
1
= 8 which yields Rx = 5 Ω.
1 / 20 + 1 / Rx
1
= 34 which yields Rx = 44.70 Ω.
1 / 142 + 1 / Rx
R (3R ) 3R
=
. Clearly, for Req to be an integer, R must be
R + 3R
4
an integer multiple of 4.
P2.7
Rab = 6 Ω
P2.8
Because the resistances are in parallel, the same voltage v appears across
both of them. The current through R1 is i1 = v/102. The current through
R2 is i2 = 4i1 = 4v/102. Finally we have R2 = v/i2 = v/(4v/102) = 25.5 Ω.
P2.9
Combining the resistances shown in Figure P2.9b, we have
Req
1
Req = 1 +
+1 =2+
1 + 1 Req
1 + Req
Req (1 + Req ) = 2(1 + Req ) + Req
(R )
eq
2
− 2Req − 2 = 0
Req = 2.732 Ω
(Req = −0.732 Ω is another root, but is not physically reasonable.)
P2.10*
Rab = 8.276 Ω
38
P2.11
Req =
1
1
1
1
+
+
+ ...
1000 1000 1000
=
1
n
=
1000
1000
n
P2.12*
P2.13
In the lowest power mode, the power is Plowest =
1202
= 83.33 W.
R1 + R2
For the highest power mode, the two elements should be in parallel with
an applied voltage of 240 V. The resulting power is
2402 2402
Phighest =
+
= 1000 + 500 = 1500 W.
R1
R2
Some other modes and resulting powers are:
R1 operated separately from 240 V yielding 1000 W
R2 operated separately from 240 V yielding 500 W
R1 in series with R2 operated from 240 V yielding 333.3 W
R1 operated separately from 120 V yielding 250 W
P2.14
For operation at the lowest power, we have
120 2
P = 300 =
R1 + R2
At the high power setting, we have
120 2 120 2
P = 1200 =
+
R1
R2
Solving these equations we find R1 = R2 = 24 Ω.
The intermediate power setting is obtained by operating one of the
elements from 120 V resulting in a power of 600 W.
39
P2.15
By symmetry, we find the currents in the resistors as shown below:
Then, the voltage between terminals a and b is
v ab = Req = 1 3 + 1 6 + 1 3 = 5 6
P2.16*
Rab = 11.63 Ω
P2.17
Rab = 19.72 Ω.
P2.18
(a) For a series combination Geq =
G1G2
1
=
1 / G1 + 1 / G2 G1 + G2
(b) For a parallel combination of conductances Geq = G1 + G2
P2.19
To supply the loads in such a way that turning one load on or off does not
affect the other loads, we must connect the loads in series with a switch
in parallel with each load:
40
To turn a load on, we open the corresponding switch, and to turn a load
off, we close the switch.
P2.20
We have Ra + Rb = Rab = 32, Rb + Rc = Rbc = 18 and Ra + Rc = Rca = 16.
These equations can be solved to find that Ra = 15 Ω, Rb = 17 Ω,
and Rc = 1 Ω. After shorting terminals b and c, the equivalent resistance
between terminal a and the shorted terminals is
1
Req = Ra +
= 15.94 Ω
1 / Rb + 1 / Rc
P2.21
The steps in solving a circuit by network reduction are:
1. Find a series or parallel combination of resistances.
2. Combine them.
3. Repeat until the network is reduced to a single resistance and a
single source (if possible).
The method does not always work because some networks cannot be
reduced sufficiently. Then, another method such as node voltages or
mesh currents must be used.
P2.22*
v 1 = 1.384 V v 2 = 0.06125 V
P2.23
Using Ohm's and Kirchhoff's laws, we work from right to left resulting in
P2.24
The equivalent resistance seen by the current source is
Req = 23.73 Ω . Then, we have v = 5Req = 118.67 V, i2 = 1.458 A, and
i1 = 2.875 A.
P2.25*
Combining resistors in series and parallel, we find that the equivalent
resistance seen by the current source is Req = 17.5 Ω. Thus,
41
v = 8 × 17.5 = 140 V. Also, i = 1 A.
P2.26*
i1 = 1 728A i2 = 0.5185 A
P2.27
The equivalent resistance seen by the voltage source is
Req = 20.19 Ω
Then, we have
13 V
i1 =
= 0.6439 A and i2 = i1
Req
P2.28
19
= 0.2308 A.
34 + 19
The currents through the 2-Ω resistance and the 6-Ω resistance are
zero because they are series with an open circuit. Thus, we can consider
the 16-Ω and the 9-Ω resistances to be in series. The current circulating
5
= 0.2 , and we have
clockwise in the left-hand loop is given by i1 =
9 + 16
v 1 = 9i1 = 1.8 V. The current circulating counterclockwise in the right
hand loop is 6 A. By Ohm's law, we have v 2 = 48 V. Then, using KVL we
have v ab = v 1 − v 2 = −46.2 V.
P2.29
The equivalent resistance seen by the current source is
1
Req = 4 +
= 10 Ω
1 / 10 + 1 /(5 + 10)
15
10
Then, we have v s = 2Req = 20 V, i2 = 2
= 1.2 A, i1 = 2
= 0. 8 A
10 + 15
10 + 15
and v 1 = −10i1 = −8 V.
P2.30
In a similar fashion to the solution for Problem P2.9, we can write the
following expression for the resistance seen by the 2-V source.
1
Req = 1 +
1 / Req + 1 / 2
42
The solutions to this equation are Req = 2 Ω and Req = -1 Ω. However, we
reason that the resistance must be positive and discard the negative
Req
i
2V
= 1 A, i2 = i1
= 1 = 0.5 A, and
root. Then, we have i1 =
2 + Req 2
Req
i3 =
P2.31
i1
2
= 0.5 A. Similarly, i4 =
i3
2
=
i1
2
2
= 0.25 A and i18 =
i1
29
= 1.953 mA.
20 V
= 4A
i1 = i 2 − 3 = 1 A
5Ω
Pcurrent −source = 3 A × 20 V = 60 W
Pvoltage −source = 20i1 = 20 W
i2 =
Power is delivered by both sources.
P2.32
With the switch open, the current flowing clockwise in the circuit is given
10R2
10
, and we have v 2 = R2i =
= 5. Solving we find R2 = 6 Ω.
by i =
6 + R2
6 + R2
With the switch closed, R2 and RL are in parallel with an equivalent
1
1
. The current through
resistance given by Req =
=
1 / R2 + 1 / RL 1 / 6 + 1 / RL
10Req
10
and we have v 2 = Req i =
Req is given by i =
= 4. Solving, we
6 + Req
6 + Req
find Req = 4 Ω. Then, we can write Req =
RL = 12 Ω.
P2.33*
Req =
1
= 3.75 Ω
1 5 + 1 15
i1 = v x 5 = 1.5 A
1
= 4. Solving, we find
1 / 6 + 1 / RL
v x = 2 A × Req = 7.5 V
i2 = v x 15 = 0.5 A
P4A = 4 × 7.5 = 30 W delivering
P2A = 2 × 7.5 = 15 W absorbing
P5 Ω = 7.52 5 = 11.25 W absorbing
P15 Ω = (7.5)2 15 = 3.75 W absorbing
43
P2.34
i =
P 26 W
=
= 4.333 A
v
6V
6
i = 4.333 =
Req
=
Req = R +
6
1.5R
1
= 1.5R
1 R +1 R
R = 0.9231 Ω
P2.35*
i1 = 3.857 A
i2 = 0.8571 A
P2.36*
v1 =
P2.37*
i1 =
P2.38
We have 120
P2.39
First, we combine the 30 Ω and 15 Ω resistances in parallel yielding an
equivalent resistance of 10 Ω, which is in parallel with Rx. Then, applying
the current division principle, we have
10
4
=1
10 + Rx
R1
× v s = 3.857 V
R1 + R2 + R3
R3
v3 =
× v s = 11.571 V
R1 + R2 + R3
R2
R1 + R2
i s = 0.8823 A
v2 =
i2 =
R1 + R2
(a) R1 + R2 =
i s = 2.1177 A
5
= 20, which yields Rx = 15 Ω.
10 + 5 + Rx
which yields Rx = 30 Ω.
P2.40
R1
R2
× v s = 2.571 V
R1 + R2 + R3
R2
12 V
= 120 Ω
0. 1 A
R1 + R2
Solving, we find R2 = 50 Ω and R1 = 70 Ω .
(b)
44
× 12 = 5
The equivalent resistance for the parallel combination of R2 and the load
is
1
Req =
= 40 Ω
1 50 + 1 200
Then using the voltage division principle, we have
vo =
P2.41*
v = 0.1 mA × Rw = 50 mV
Rg =
P2.42
Req
× 12 V = 4.364 V
R1 + Req
50 mV
= 25 mΩ
2 A − 0.1 mA
The circuit diagram is:
With iL = 0 and v L = 5 V , we must have
R2
R1 + R2
× 9 = 5 V . Rearranging,
this gives
R1
= 0. 8
(1)
R2
With iL = 25 mA and v L = 4.5 V , we have 9 − R1 (4.5 R2 + 25 mA) = 4.5 .
Rearranging, this gives
4.5
R1
+ R1 × 0.025 = 4.5 .
R2
(2)
Using Equation (1) to substitute into Equation (2) and solving, we obtain
R1 = 36 Ω and R2 = 45 Ω.
Maximum power is dissipated in R1 for iL = 25 mA , for which the voltage
4.52
= 562.5 mW . Thus, R1 must be
36
rated for at least 562.5 mW of power dissipation.
across R1 is 4.5 V. Thus, Pmax R 1 =
45
Maximum power is dissipated in R2 for iL = 0 , in which case the voltage
52
across R2 is 5 V. Thus, Pmax R 2 =
= 555.5 mW . (Standard resistors are
45
available in 1-W ratings and would be suitable for this circuit.)
P2.43*
Combining R2 and R3 , we have an equivalent resistance
1
= 10 Ω . Then, using the voltage-division principle, we
Req =
1 R2 + 1 R3
have v =
R2
Req
10
×v s =
× 10 = 3.333 V .
R1 + Req
20 + 10
i3 =
P2.45
We need to place a resistor in series with the load and the voltage source
as shown
R2 + R3
× is =
15
×8 = 6A
15 + 5
P2.44
Applying the voltage-division principle, we have 12
find R = 60 Ω.
P2.46
30
= 4. Solving, we
30 + R
We have P = 32 = I L2RL = I L2 8. Solving, we find that the current through
the load is I L = 2 A. Thus, we must place a resistor in parallel with the
current source and the load.
46
To divide the 4 A source current equally between R and RL, we need to
have R = 8 Ω.
P2.47*
At node 1 we have:
v1
v1 − v2
=1
10
v
v − v1
=2
At node 2 we have: 2 + 2
5
10
In standard form, the equations become
0.15v 1 − 0.1v 2 = 1
− 0.1v 1 + 0.3v 2 = 2
20
+
Solving, we find v 1 = 14.29 V and v 2 = 11.43 V .
v − v2
Then we have i1 = 1
= 0.2857 A.
10
P2.48*
Writing a KVL equation, we have v 1 − v 2 = 10 .
At the reference node, we write a KCL equation:
v1
Solving, we find v 1 = 6.667 and v 2 = −3.333 .
Then, writing KCL at node 1, we have is =
P2.49
v2 − v1
5
−
5
v1
5
+
v2
10
= −3.333 A .
Writing KCL equations at nodes 1, 2, and 3, we have
v1 v1 − v2 v1 − v3
+
+
=0
4
16
16
v2 − v1 v2 − v3
+
=1
16
16
v3 v3 − v2 v3 − v1
+
+
=0
5
16
16
Solving, we find
v 1 = 2.102
P2.50
v 2 = 10.237
v 3 = 2.373
Writing KCL equations at nodes 1, 2, and 3, we have
v1 v1 − v2
+
+1 = 0
17
16
v2 − v1 v2 − v3 v2
+
+
=0
16
9
6
v3 v3 − v2
+
=1
9
9
47
=1.
Solving, we find
v 1 = −8.273 V
P2.51
v 2 = −0.06 V
v 3 = 4.470 V
Writing KCL equations at nodes 1 and 2, we have
v1 v1 v1 −v 2
+
+
=3
21 28
9
v2 − v1 v 2
+
= −3
9
6
In standard form, we have:
0.1944v 1 − 0.1111v 2 = 3
− 0.1111v1 + 0.2778v2 = −3
Solving, we findv1 = 12.0 V and v2 = −6.00 V .
If the source is reversed, the algebraic signs of the node voltages are
reversed.
P2.52
To minimize the number of unknowns, we select the reference node at
one end of the voltage source. Then, we define the node voltages and
write a KCL equation at each node.
v 1 − 20
5
+
v1 − v2
2
=2
v2 − v1
2
+
v 2 − 20
10
Solving, we find v 1 = 18.24 V and v 2 = 13.53 V .
20 − v 2
= 0.647 A .
Then, we have i1 =
10
48
= −3
P2.53
We must not use all of the nodes (including those that are inside
supernodes) in writing KCL equations. Otherwise, dependent equations
result.
P2.54
The circuit with a 1-A source connected is:
v1 − v2
10
+
v1 − v3
20
=1
v2
+
v2 − v1
+
v2 − v3
v3
+
v3 − v1
+
v3 − v2
10
20
10
20
10
10
=0
=0
Solving, we find Req = v 1 = 13.33 .
P2.55*
First, we can write: ix =
v1 − v2
.
5
Then, writing KCL equations at nodes 1 and 2, we have:
v1
+ ix = 1 and
v2
+ 0.5ix − ix = 0
10
20
Substituting for ix and simplifying, we have
0.3v 1 − 0.2v 2 = 1
− 0.1v 1 + 0.15v 2 = 0
Solving, we have v 1 = 6 and v 2 = 4 .
v − v2
= 0. 4 A .
Then, we have ix = 1
5
49
P2.56
First, we can write ix =
v1
10
. Then writing KVL, we have v 1 − 5ix − v 2 = 0 .
Writing KCL at the reference node, we have ix +
v2
= 8 . Using the first
20
equation to substitute for ix and simplifying, we have
0.5v 1 − v 2 = 0
2v 1 + v 2 = 160
Solving, we find v 1 = 64 , v 2 = 32 , and ix =
v1
= 6.4 A . Finally, the power
10
(v 1 − v 2 ) 2
delivered to the 16-Ω resistance is P =
= 64 W.
16
P2.57*
v x = v2 − v1
Writing KCL at nodes 1 and 2:
v1
+
v 1 − 2v x
+
v1 − v2
v2
+
v 2 − 2v x
+
v2 − v1
5
5
15
10
10
10
=1
=2
Substituting and simplifying, we have
15v 1 − 7v 2 = 30
and
v 1 + 2v 2 = 20 .
Solving, we find v 1 = 5.405 and v 2 = 7.297 .
P2.58
The circuit with a 1-A current source connected is
v x = v1 − v2
v1 v1 − v2
20
+
10
=1
50
v2 − v1
v2
vx
=0
10
5 5
Using the first equation to substitute for vx and simplifying, we have
0 . 15 v 1 − 0 . 1v 2 = 1
0.1v 1 + 0.1v 2 = 0
+
+
Solving we find v 1 = 4 . However, the equivalent resistance is equal in
value to v1 so we have Req = 4 Ω.
P2.59
The circuit with a 1-A current source connected is
ix = v 1 − v 2
v1 v1 − v2
+
=1
5
1
v2 − v1 v2
+ − 2ix = 0
1
2
Using the first equation to substitute for ix and simplifying, we have
1.2v 1 − v 2 = 1
− 3v 1 + 3.5v 2 = 0
Solving we find v 1 = 2.917 V. However, the equivalent resistance is equal
in value to v1 so we have Req = 2.917 Ω.
P2.60
First, we can write:
5i − v 2
ix = x
10
Simplifying, we find i x = −0.2v 2 .
Then write KCL at nodes 1 and 2:
v 1 − 5i x
v2
= −3
− ix = 1
5
20
Substituting for ix and simplifying, we have
v 1 − v 2 = −15
and
0.25v 2 = 1
which yield v 1 = −11 V and v 2 = 4 V .
51
P2.61*
Writing KVL equations around each mesh, we have
6i1 + 18(i1 − i2 ) = 30 and 18(i2 − i1 ) + 17i2 = 19
Solving we obtain i1 = 2.698 A and i2 = 1.930 A.
Then, the power delivered to the 18-Ω resistor is P = (i1 − i2 ) 2 18 = 10.60
W.
P2.62
Writing KVL equations around each mesh, we have
3i1 + 4(i1 − i3 ) + 107 = 0
20(i2 − i3 ) + 13i2 − 107 = 0
5i3 + 20(i3 − i2 ) + 4(i3 − i1 ) = 0
Solving, we obtain i1 = −15.141 A, i2 = 3.396 A, and i3 = 0.2539 A. Then,
the power delivered by the source is P = 107(i2 − i1 ) = 1983 W.
P2.63*
Writing and simplifying the mesh-current equations, we have:
28i1 − 10i2 = 12
− 10i1 + 40i2 − 30i3 = 0
− 30i2 + 60i3 = 0
Solving, we obtain
i1 = 0.500
i2 = 0.200
i3 = 0.100
Thus, v 2 = 5i3 = 0.500 V and the power delivered by the source is
P = 12i1 = 6 W.
P2.64
First, we select the mesh currents and then write three equations.
52
Mesh 1: 12i1 + 24(i1 − i3 ) = 0
Mesh 2: 12i2 + 6(i2 − i3 ) = 0
However by inspection, we have i3 = 2 . Solving, we obtain
i1 = 1.333 A and i2 = 0.6667 A.
P2.65
Writing and simplifying the
mesh equations yields:
14i1 − 8i2 = 10
− 8i1 + 16i2 = 0
Solving, we
find i1 = 1.000 and i2 = 0.500 .
Finally, the power delivered by
the source is P = 10i1 = 10 W.
P2.66
4iA + 28(iA − iB ) = 16
28(iB − iA ) + 7iB + 14iB = 0
Solving we find iA = 1 A and iB = 0.5714 A. Then we have i1 = iA = 1 A and
i2 = iA − iB = 0.4286 A.
P2.67
Mesh A: 10iA + 5iA + 10(iA − iB ) = 0
By inspection: iB = 2
53
Solving, we find iA = 0.8 A . Then we have i1 = iA = 0.8 A and
i2 = iB − iA = 1.2 A.
P2.68
First we select mesh-current variables as shown.
Then, we can write
(Rw + Rn + R1 )i1 − Rn i2 − R1i3 = 120
− Rn i1 + (Rw + Rn + R2 )i2 − R2i3 = 120
− R1i1 − R2i2 + (R1 + R2 + R3 )i3 = 0
Substituting values for the resistances and solving we find
i1 = i2 = 40.58 A and i3 = 28.99 A. Then, the voltages across R1 and R2 are
both 10(i1 − i3 ) = 115.9 V and the voltage across R3 is 8i3 =231.9 V. The
current through the neutral wire is i1 − i2 = 0.
54
P2.69
Writing and simplifying the mesh equations, we obtain:
− 20i1 + 40i2 = 0
40i1 − 20i2 = 10
Solving, we find i1 = 0.3333 and i2 = 0.1667 .
Thus, v = 20(i1 − i2 ) = 3.333 V .
The mesh currents and corresponding equations are:
P2.70
i1 = 8 A
15(i2 − i1 ) + 5i2 = 0
Solving, we find i2 = 6 A .
However, i3 shown in Figure P2.44 is the same as i2, so the answer is i3 = 6
A.
P2.71*
Because of the current sources, two of the mesh currents are known.
Writing a KVL equation around the middle loop we have
20(i1 − 1) + 10i1 + 5(i1 + 2) = 0
Solving, we find i1 = 0.2857 A.
55
P2.72
Current source in terms of mesh currents:
KVL for mesh 3:
− 15i1 − 15i2 + 45i3 = 0
− i1 + i 2 = 4
KVL around outside of network: 5i1 + 25i2 + 15i3 = 0
Solving, we find
i1 = −3.000 , i2 = 1.000 and i3 = −0.6667
Then, we have
v 3 = 25i2 = 25V
P2.73
20i1 − 10i2 − 10i3 = 1
− 10i1 + 40i2 − 10i3 = 0
− 10i1 − 10i2 + 40i3 = 0
Solving, we find
i1 = 0.075
Finally, we have
1V
= 13.33 Ω
Req =
i1
P2.74
Mesh 1: 3i1 + 7i1 + 30(i1 − i2 ) = 1
Mesh 2: 3i2 + 12(i2 − i3 ) + 4i2 + 30(i2 − i1 ) = 0
56
Mesh 3: 24i3 + 12(i3 − i2 ) = 0
Solving we find i1 = 0.05 A. Then Req = 1 / i1 = 20 Ω.
P2.75*
First, we write a node voltage
equation to solve for the opencircuit voltage:
v oc − 10 v oc
+
=1
10
5
Solving, we find v oc = 6.667 V .
Then zeroing the
sources, we have this
circuit:
Thus, Rt =
P2.76
1
= 3.333 Ω . The Thévenin and Norton equivalents are:
1 10 + 1 5
First, we solve the network with a short circuit:
57
1
= 14 Ω
1 20 + 1 5
= 24 Req = 1.714 A
Req = 10 +
i10
20
= 1.371
20 + 5
= i5 = 1.371
i5 = i10
isc
Zeroing the source, we have:
Combining resistances in series and parallel we find Rt = 6.563 Ω .
Then the Thévenin voltage is vt = isc Rt = 9.00 V .
P2.77*
The equivalent circuit of the battery with the resistance connected is
i = 6 100 = 0.06 A
Rt =
9−6
= 50 Ω
0.06
58
P2.78
With open-circuit conditions:
Solving, we find v ab = −5 V .
With the source zeroed:
Rt =
1
1 5 + 1 (5 + 10 )
= 3.75 Ω
The equivalent circuits are:
Notice the source polarity relative to terminals a and b.
P2.79
The 10-Ω resistor has no effect on the equivalent circuits because the
voltage across the 12-V source is independent of the resistor value.
P2.80
The Thévenin voltage is equal to the open-circuit voltage which is 12.6 V.
The equivalent circuit with the 0.1-Ω load connected is:
59
We have 12.6 /(Rt + 0.1) = 100 from which we find Rt = 0.026 Ω. The
Thévenin and Norton equivalent circuits are:
Because no energy is converted from chemical form to heat in a battery
under open-circuit conditions, the Thévenin equivalent seems more
realistic from an energy conversion standpoint.
P2.81
The Thévenin voltage is equal to the open-circuit voltage which is 19 V.
Consider the circuit with the load attached. We have
7
iL =
= 2.333 mA . Thus, the Thévenin resistance is
3000
19 − 7
Rt =
= 5.143 kΩ.
2.333 mA
P2.82
The equivalent circuit with a load
attached is:
For a load of 7 Ω, we have
i L = 7 / 7 = 1 A , and we can write
v L = Vt − Rt iL . Substituting values this becomes
7 =Vt − Rt
(1)
Similarly, for the 10-Ω load we obtain
8 = Vt − 0.8Rt
(2)
Solving Equations (1) and (2), we find Vt = 12 V and Rt = 5 Ω.
P2.83
With open-circuit conditions:
12 − v oc
v oc
− i x + 0.5i x = 0
29
8
Vt = v oc = 7.733 V .
ix =
With short-circuit conditions:
60
Solving, we find
i x = 12 / 8 = 1.5 A
i sc = i x − 0.5i x = 0.75 A
Then, we have Rt = v oc i sc = 5.155 Ω .
P2.84
As is Problem P2.75, we find the Thévenin equivalent:
Then maximum power is obtained for a load resistance equal to the
Thévenin resistance.
Pmax =
P2.85
(vt 2)2
Rt
= 3.333 W
As in Problem P2.78, we find the Thévenin equivalent:
Then, maximum power is obtained for a load resistance equal to the
Thévenin resistance.
Pmax =
P2.86*
(vt 2)2
Rt
= 1.667 W
To maximize the power to RL , we must maximize the voltage across it.
Thus we need to have Rt = 0 . The maximum power is
Pmax =
202
= 80 W
5
61
P2.87
The circuit is
By the current division principle:
iL = I n
Rt
RL + Rt
The power delivered to the load is
PL = (iL ) RL = (I n )
2
2
(Rt )2 RL
(RL + Rt )2
Taking the derivative and setting it equal to zero, we have
2
2
2
dPL
2 (Rt ) (Rt + RL ) − 2(Rt ) RL (Rt + RL )
= 0 = (I n )
dRL
(Rt + RL )4
which yields RL = Rt .
The maximum power is PL max = (I n )2 Rt 4 .
P2.88
For maximum power conditions, we have RL = Rt . The power taken from
the voltage source is
Ps =
(Vt )2
Rt + RL
=
(Vt )2
2Rt
Then half of Vt appears across the load and the power delivered to the
load is
PL =
(0.5Vt )2
Rt
Thus the percentage of the power taken from the source that is
delivered to the load is
PL
× 100% = 50%
Ps
On the other hand, for RL = 9Rt , we have
η=
62
Ps =
PL =
η=
(Vt )2
Rt + RL
=
(Vt )2
10Rt
(0.9Vt )2
9Rt
PL
× 100% = 90%
Ps
Thus, design for maximum power transfer is relatively inefficient. Thus,
systems in which power efficiency is important are almost never designed
for maximum power transfer.
P2.89*
First, we zero the current source and find the current due to the voltage
source.
iv = 30 15 = 2 A
Then, we zero the voltage source and use the current-division principle to
find the current due to the current source.
10
= 2A
5 + 10
Finally, the total current is the sum of the contributions from each
source.
i = iv + ic = 4 A
ic = 3
63
P2.90
Zero the 2 A source and use the current-division principle:
i1,1A = 1
20
= 0.5714 A
15 + 20
Then zero the 1 A source and use the current-division principle:
i1,2A = −2
Finally,
5
= −0.2857 A
5 + 30
i1 = i1,1A + i1,2A = 0.2857 A
P2.91
The circuits with only one source active at a time are:
i1.4A = 4 ×
= 3A
15
15 + 5
i1,2A = −2 ×
= −1.5
15
15 + 5
Finally, we add the components to find the current with both sources
active.
64
i1 = i1, 4A + i1,2A = 1.5 A
P2.92*
The circuits with only one source active at a time are:
1
= 3.75 Ω
1 5 + 1 15
10 V
=−
= −2.667 A
Req =
is ,v
is ,c = −1
10
= −0.667 A
10 + 5
Req
Then the total current due to both sources is is = is ,v + is ,c = −3.333 A .
P2.93
The circuit, assuming that v 2 = 1 V is:
i2 = (v 2 5) = 0.2 A
v1 = 30i2 = 6 V
i10 = v 1 10 = 0.6 A
i30 = v 1 30 = 0.2 A
is = i2 + i10 + i30 = 1 A
v s = 12is + v 1 + 6is = 24 V
We have established that for v s = 24 V , we have v 2 = 1 V . Thus, for
v s = 12 V , we have:
v2 = 1 ×
12
= 0. 5 V
24
65
P2.94
We start by assuming i = 1 A and work back through the circuit to
determine the value of vs. This results in:
However, the circuit actually has vs = 70 V, so the actual value ofi is
70
× (1 A) = 0.5 A.
140
P2.95 We start by assuming i2 = 1 A and
work back through the circuit to
determine the value of vs. The
results are shown on the circuit
diagram.
However, the circuit actually has
vs = 10 V, so the actual value ofi2 is
P2.96
10
× (1 A) = 0.5 A.
20
(a) With only the 2-A source activated, we have
i2 = 2 and v 2 = 2(i2 )3 = 16 V
(b) With only the 1-A source activated, we have
i1 = 1 A and v 1 = 2(i1 )3 = 2 V
(c) With both sources activated, we have
i = 3 A and v = 2(i )3 = 54 V
Superposition does not apply because device A has a nonlinear
relationship between v and i.
66
P2.97
1. Replace the controlled source with an independent source of unknown
value.
2. Use superposition to determine the total response for the controlling
current or voltage in terms of the unknown source value and other circuit
parameters.
3. Subsititute the expression for the value of contolled source for the
unknown dependent source and solve for the controlling current or
voltage. Then, compute the value for unknown independent source.
4. Use superposition to solve for any other currents or voltages of
interest.
P2.98*
Replacing the dependent current source with an independent current
source and turning on one source at a time, we have
i xa = 1
10
= 0.2857
10 + 25
v 1a = 25i xa = 7.143
20
= 0.5714I
v 1b = −10I = −5.714I
20 + 15
Adding the results from the two analyses, we have
ix = ixa + ixb = 0.2857 + 0.5714I
i xb = I
Now, we substitute I = 0.5i x which yields
i x = 0.2857 + 0.5714(0.5ix )
67
Solving, we find i x = 0.4000 A. Then, we have I = 0.5i x = 0.2 and
v 1 = v 1a + v 1b = 7.143 − 5.714I = 6.000 V.
P2.99
Following the same procedure used in P2.98, we eventually have
ix = 1.333 + 0.6667I
Then, substituting I = 0.5i x and solving, we find i x = 2 A and I = 1 A.
Also, we have
v ab = 20
P2.100
10
1
−I
= 10 V
10 + 5
1 / 5 + 1 / 10
First, we replace the controlled source with an independent voltage
source of unknown value V. Then, we use superposition to find
expressions for the controlling current:
6
4
vx = 4
+V
= 1.333 + 0.2222V
6 + 12
18
Then, we substitute V = 2v x and solve, resulting in v x = 2.400 V and
V = 4.800 V. Finally, we use superpositon to solve for the current of
interest:
i1 =
P2.101
12 V
−
= 0.4 A
18 18
From Equation 2.82, we have
R2
1 kΩ
R3 =
× 3419 = 341.9 Ω
R1
10 kΩ
R
100 kΩ
× 3419 = 34.19 kΩ
(b) Rx = 2 R3 =
R1
10 kΩ
(a)
Rx =
P2.102* (a) Rearranging Equation 2.82, we have
R1
10 4
R3 = Rx = 4 × 5932 = 5932 Ω
R2
10
(b) The circuit is:
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The Thévenin resistance is
Rt =
1
1
+
= 7447 Ω
1 R3 + 1 R1 1 R2 + 1 Rx
The Thévenin voltage is
vt = v s
R3
R1 + R3
− vs
Rx
Rx + R2
= 0.3939 mV
Thus, the equivalent circuit is:
idetector =
Vt
= 31.65 × 10 −9 A
Rt + Rdetector
Thus, the detector must be sensitive to very small currents if the bridge
is to be accurately balanced.
P2.103
If R1 and R3 are too small, large currents are drawn from the source. If
the source were a battery, it would need to be replaced frequently.
Large power dissipation could occur, leading to heating of the components
and inaccuracy due to changes in resistance values with temperature.
If R1 and R3 are too large, we would have very small detector current
when the bridge is not balanced, and it would be difficult to balance the
bridge accurately.
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P2.104
With the source replaced by a short circuit and the detector removed,
the Wheatstone bridge circuit becomes
The Thévenin resistance seen looking back into the detector terminals is
1
1
Rt =
+
1 R3 + 1 R1 1 R2 + 1 Rx
The Thévenin voltage is zero when the bridge is balanced.
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71