This preview shows pages 1–3. Sign up to view the full content.
HW 9 solutions
1
Chapter 4, Problem 50.
Obtain the Norton equivalent of the circuit in Fig. 4.116 to the left of terminals
a

b
. Use
the result to find current
i
Figure 4.116
Chapter 4, Solution 50.
From Fig. (a),
R
N
=
6 + 4
=
10 ohms
From Fig. (b),
2 + (12 – v)/6
=
v/4,
or
v
= 9.6 V
I
N
=
(12 – v)/6
=
0.4,
which leads to
I
N
=
0.4 A
Combining the Norton equivalent with the righthand side of the original circuit produces
the circuit in Fig. (c).
i
=
[10/(10 + 5)] (4 – 0.4)
=
2.4 A
6
Ω
4
Ω
(a)
12V
+
−
I
sc
= I
N
6
Ω
4
Ω
(b)
2A
i
10
Ω
(c)
0.4A
5
Ω
4A
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document HW 9 solutions
2
Chapter 4, Problem 54.
Find the Thèvenin equivalent between terminals
a

b
of the circuit in Fig. 4.120.
Figure 4.120
Chapter 4, Solution 54.
To find
V
Th
=V
x
,
consider the left loop.
x
o
x
o
V
i
V
i
2
1000
3
0
2
1000
3
+
=
→
=
+
+
−
(1)
For the right loop,
o
o
x
i
i
x
V
2000
40
50
−
=
−
=
(2)
Combining (1) and (2),
mA
1
3000
4000
1000
3
−
=
→
−
=
−
=
o
o
o
o
i
i
i
i
2
2
2000
=
→
=
−
=
Th
o
x
V
i
V
To find R
Th
,
insert a 1V source at terminals ab and remove the 3V independent
source, as shown below.
1 k
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 04/18/2008 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.
 Spring '06
 MCCANN

Click to edit the document details