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EE302 Homework #3
Chapter 2, Problem 26.
For the circuit in Fig. 2.90, i
o
=2 A.
Calculate i
x
and the total power dissipated by the
circuit.
2
Ω
4
Ω
8
Ω
i
o
16
Ω
i
x
Chapter 2, Solution 26.
If
i
16
= i
o
= 2A, then
v = 16x2 = 32 V
8
4 A
8
v
i
==
,
42
8 A,
i
16
vv
i
=
=
2481
6
16 8 4 2
30 A
x
iiiii
=+++ = +++=
22222
16
2 8
4
4
8 2 16
960 W
Pi
R
x
x
x
x
=
=
+++
=
∑
or
30 32
960 W
x
v
x
=
=0.960 kW
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View Full Document Chapter 2, Problem 32.
Find
i
1
through
i
4
in the circuit in Fig. 2.96.
Chapter 2, Solution 32.
We first combine resistors in parallel.
=
30
20
=
50
30
x
20
12
Ω
=
40
10
=
50
40
x
10
8
Ω
Using current division principle,
A
12
)
20
(
20
12
i
i
,
A
8
)
20
(
12
8
8
i
i
4
3
2
1
=
=
+
=
+
=
+
=
=
)
8
(
50
20
i
1
3.20 A
=
=
)
8
(
50
30
i
2
4.80 A
=
=
)
12
(
50
10
i
3
2.40A
=
=
)
12
(
50
40
i
4
9.60 A
Chapter 2, Problem 34.
Using series/parallel resistance combination, find the equivalent resistance seen by the
source in the circuit of Fig. 2.98.
Find the overall dissipated power.
8
Ω
10
Ω
20
Ω
Chapter 2, Solution 34.
40//(10 + 20 + 10)= 20
Ω
,
40//(8+12 + 20) = 20
Ω
20 20
40
eq
R
=+=Ω
2
12
12/ 40,
3.6 W
40
eq
V
IP
V
I
R
==
=
=3.60W
+
_
40
Ω
40
Ω
12 V
20
Ω
12
Ω
10
Ω
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View Full Document Chapter 2, Problem 38.
Find R
eq
and
i
o
in the circuit of Fig. 2.102.
5
Ω
6
Ω
+
_
15
Ω
20
Ω
40 V
60
Ω
12
Ω
80
Ω
R
eq
i
o
Chapter 2, Solution 38.
20//80 = 80x20/100
= 16,
6//12 = 6x12/18 = 4
The circuit is reduced to that shown below.
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This note was uploaded on 09/18/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.
 Fall '06
 MCCANN

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