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Homework set # 7
Problems: 18.14, 15, 48
Solutions:
18.14
Going counterclockwise around the upper loop,
applying Kirchhoff’s loop rule, gives
1
15.0 V
7.00
5.00 2.00 A
0
I
or
1
15.0 V 10.0 V
0.714 A
7.00
I
From Kirchhoff’s junction rule,
12
2.00 A
0
II
so
21
2.00 A
2.00 A
0.714 A
1.29 A
Going around the lower loop in a clockwise direction gives
2
2.00
5.00
2.00 A
0
I
E
or
2.00
1.29 A
5.00
2.00 A
12.6 V
E
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View Full Document18.15
We name the currents
12
3
, , and
II
I
as shown.
Using Kirchhoff’s loop rule on the rightmost
loop gives
3
2
12.0 V 1.00+3.00
5.00 1.00
4.00 V
0
I
I
or
32
2.00
3.00
4.00 V
(1)
Applying the loop rule to the leftmost loop yields
21
4.00 V+ 1.00+5.00
8.00
0
or
4.00
3.00
2.00 V
(2)
From Kirchhoff’s junction rule,
123
+=
III
(3)
Solving equations (1), (2) and (3) simultaneously gives
3
=0.846 A,
=0.462 A, and
1.31 A
I
All currents are in the directions indicated by the arrows in the circuit diagram.
18.48
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 Spring '08
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