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Chapter 4, Problem 24.
Use source transformation to find the voltage V
x
in the circuit of Fig. 4.92.
Figure 4.92
For Prob. 4.24.
Solution.
Transform the two current sources in parallel with the resistors into their voltage source
equivalents yield,
a 30V source in series with a 10Ω resistor and a 20V
x
V sources in series
with a 10Ω resistor.
We now have the following circuit,
We now write the following mesh equation and constraint equation which will lead to a
solution for V
x
,
28I – 70 + 20V
x
= 0 or 28I + 20V
x
= 70, but V
x
= 8I which leads to
28I + 160I = 70 or I = 0.3723 A or V
x
=
2.978 V
.
1
+
_
8
Ω
10
Ω
10
Ω
3 A
40 V
+
–
V
x
2 V
x
I
8
Ω
10
Ω
10
Ω
+
_
40 V
+
–
V
x
+
–
20V
x
–
+
30 V
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Use source transformation to find i
o
in the circuit of Fig. 4.94.
Figure 4.94
For Prob. 4.26.
Solution.
Transforming the current sources gives the circuit below.
–12 + 11i
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This note was uploaded on 02/13/2010 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.
 Spring '06
 MCCANN

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