Problem set #5 EE 221, 09/26/2002 – 10/03/2002
Chapter 3, Problem 42.
In the circuit of Fig. 3.75, choose v1 to obtain a
current ix of 2 A.
Chapter 3, Solution 42.
We first simplify as shown, making use of the fact that we are told ix = 2 A to find the
voltage across the middle and right-most 1-W resistors as labeled.
By KVL, then, we find that v1 = 2 + 3 = 5 V.
Chapter 3, Problem 63.
What is the power dissipated by
(absorbed by) the 47-kΩ resistor in
Fig. 3.93?
Chapter 3, Solution 63.
The controlling voltage v1, needed to obtain the power into the 47-kΩ resistor, can be
found separately as that network does not depend on the left-hand network.
The right-most 2 kΩ resistor can be neglected.
By current division, then, in combination with Ohm’s law,
v1 = 3000[5×10-3 (2000)/ (2000 + 3000 + 7000)] = 2.5 V
Voltage division gives the voltage across the 47-kΩ resistor:
0.5v1
47
47 + 100 || 20
=
So that p47kΩ = (0.9928)2 / 47×103 = 18.12 µW
0.5(2.5)(47)
47 + 16.67
= 0.9228 V
1
Problem set #5 EE 221, 09/26/2002 – 10/03/2002
2
Chapter 4, Problem 3.
Use nodal analysis to find vP in the
circuit shown in Fig. 4.25.
Chapter 4, Solution 3.
The bottom node has the largest number of branch connections, so we choose that as
our reference node. This also makes vP easier to find, as it will be a nodal voltage. Working from left to right, we
name our nodes 1, P, 2, and 3.
NODE 1:
10 = v1/ 20 + (v1 – vP)/ 40
[1]
NODE P:
0 = (vP – v1)/ 40 + vP/ 100 + (vP – v2)/ 50
[2]
NODE 2:
-2.5 + 2 = (v2 – vP)/ 50 + (v2 – v3)/ 10
[3]
NODE 3:
5 – 2 = v3/ 200 + (v3 – v2)/ 10
[4]
Simplifying,
60v1 - 20vP
= 8000
-50v1 + 110 vP - 40v2
=0
- vP
+ 6v2 - 5v3
= -25
-200v2 + 210v3 = 6000
Solving,
[1]
[2]
[3]
[4]
vP = 171.6 V
Chapter 4, Problem 5.
For the circuit of Fig. 4.27,
(a) use nodal analysis to determine v1 and v2,
(b) Compute the power absorbed by the 6-Ω
resistor.
Chapter 4, Solution 5.
Designate the node between the 3-Ω and 6-Ω resistors as node X, and the right-hand
node of the 6-Ω resistor as node Y. The bottom node is chosen as the reference node.
(a) Writing the two nodal equations, then
NODE X:
–10 = (vX – 240)/ 3 + (vX – vY)/ 6
NODE Y:
0 = (vY – vX)/ 6 + vY/ 30 + (vY – 60)/ 12
Simplifying,
Solving,
-180 + 1440 = 9 vX – 3 vY [1]
10800 = - 360 vX + 612 vY
vX = 181.5 V
58.50 V
and
and
[1]
[2]
[2]
vY = 124.4 V
Thus,
v1 = 240 – vX =
v2 = vY – 60 =
64.40 V
(b)
The power absorbed by the 6-W resistor is (vX – vY)2 / 6 = 543.4 W
Problem set #5 EE 221, 09/26/2002 – 10/03/2002
3
Chapter 4, Problem 12.
With the help of nodal analysis on the
circuit of Fig. 4.34, find
(a) vA;
(b) the power dissipated in the 2.5-Ω resistor.
Chapter 4, Solution 12.
Choosing the bottom node as the reference terminal and naming the left node “1”, the
center node “2” and the right node “3”, we next form a supernode about nodes 1 and 3, encompassing the
dependent voltage source.
At the supernode, 5 – 8 = (v1 – v2)/ 2 + v3/ 2.5
At node 2,
8 = v2 / 5 + (v2 – v1)/ 2
[1]
[2]
Our supernode equation is v1 - v3 = 0.8 vA
[3]
Since vA = v2, we can rewrite [3] as v1 – v3 = 0.8v2
Simplifying and collecting terms,
0.5 v1 - 0.5 v2 + 0.4 v3 = -3
-0.5 v1 + 0.7 v2
= 8
v1 - 0.8 v2
- v3 = 0
(a) Solving for v2 = vA, we find that
[1]
[2]
[3]
vA = 25.91 V
(b) The power absorbed by the 2.5-W resistor is
(v3)2/ 2.5 = (-0.4546)2/ 2.5
= 82.66 mW.
4
Problem set #5 EE 221, 09/26/2002 – 10/03/2002
Chapter 4, Problem 22.
Calculate the power being dissipated in the
2-Ω resistor for the circuit of Fig. 4.44.
Chapter 4, Solution 22.
We define four clockwise mesh currents. The top mesh current is labeled i4. The
bottom left mesh current is labeled i1, the bottom right mesh current is labeled i3, and the remaining mesh current
is labeled i2. Define a voltage “v4A” across the 4-A current source with the “+” reference terminal on the left.
By inspection,
i3 = 5 A
and ia = i4.
MESH 1: -60 + 2i1 – 2i4 + 6i4 = 0
or
MESH 2: -6i4 + v4A + 4i2 – 4(5) = 0
or
MESH 4: 2i4 – 2i1 + 5i4 + 3i4 – 3(5) – v4A = 0
or
2i1
-2i1
+ 4i4
= 60
[1]
4i2 - 6i4 + v4A = 30
[2]
+ 10i4 - v4A = 15
[3]
At this point, we are short an equation. Returning to the circuit diagram, we note that
i2 – i4 = 4
[4]
Collecting these equations and writing in matrix form, we have
 2
 0

- 2

 0
0 4 0
4 - 6 1 
0 10 - 1 

1 -1 0 
i1 
60 
i 
 
 2  = 20
i4 
15 
 
 
 4
v 4 A 
Solving, i1 = 16.83 A, i2 = 10.58 A, i4 = 6.583 A and v4A = 17.17 V.
Thus, the power dissipated by the 2-Ω resistor is
(i1 – i4)2 • (2) = 210.0 W
Problem set #5 EE 221, 09/26/2002 – 10/03/2002
5
Chapter 4, Problem 26.
Determine each mesh current in the circuit of Fig. 4.47.
Chapter 4, Solution 26.
We define a clockwise mesh current i3 in the upper right mesh, a clockwise mesh
current i1 in the lower left mesh, and a clockwise mesh current i2 in the lower right mesh.
MESH 1:
-6 + 6 i1 - 2 = 0
[1]
MESH 2:
2 + 15 i2 – 12 i3 – 1.5 = 0 [2]
MESH 3:
i3 = 0.1 vx
[3]
Eq. [1] may be solved directly to obtain
i1 = 1.333 A.
It would help in the solution of Eqs. [2] and [3] if we could express the dependent source controlling variable vx in
terms of mesh currents. Referring to the circuit diagram, we see that vx = (1)( i1) = i1, so Eq. [3] reduces to
i3 = 0.1 vx = 0.1 i1 = 133.3 mA.
As a result, Eq. [1] reduces to
i2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA.
6
Problem set #5 EE 221, 09/26/2002 – 10/03/2002
Chapter 4, Problem 33.
Use the supermesh concept to determine the power
supplied by the 2.2-V source
of Fig. 4.54.
Chapter 4, Solution 33.
Define four mesh currents
i2
i1
i3
i4
By inspection, i1 = -4.5 A.
We form a supermesh with meshes 3 and 4 as defined above.
MESH 2:
SUPERMESH:
2.2 + 3 i2 + 4 i2 + 5 – 4 i3 = 0
[1]
3 i4 + 9 i4 – 9 i1 + 4 i3 – 4 i2 + 6 i3 + i3 – 3 = 0
Supermesh KCL equation:
[2]
i4 - i3 = 2
[3]
Simplifying and combining terms, we may rewrite these three equations as:
7 i2 – 4 i3
-4 i2 + 11 i3 + 12 i4
- i3
+ i4
= -7.2 [1]
= -37.5 [2]
=
2
[3]
Solving, we find that i2 = -2.839 A, i3 = -3.168 A, and i4 = -1.168 A.
The power supplied by the 2.2-V source is then 2.2 (i1 – i2) = -3.654 W.