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DirectCurrent Circuits
113
18.17
We name the currents
as shown.
Using Kirchhoff’s loop rule on the rightmost
loop gives
12
3
, , and
II
I
( )
()
3
2
5.00
1.00
I
I
−+
12.0 V 1.00+3.00
4.00 V
0
+
−
=
(
or
)
32
4.00 V
=
2.00
3.00
+
(
(1)
Applying the loop rule to the leftmost loop yields
) ( )
21
4.00 V+ 1.00
0
(
+5.00
8.00
+−
=
or
) ( )
2.00 V
=
4.00
3.00
−
123
+=
I
(2)
From Kirchhoff’s junction rule,
(3)
Solving equations (1), (2) and (3) simultaneously gives
3
=0.846 A,
1.31 A
I
=
=0.462 A, and
All currents are in the directions indicated by the arrows in the circuit diagram.
±²³³´
±
±
µ
µ¶²³
·
¸²³³
·
µ²³³
±
±
¶
±
±
¹²³³
±
µ²³³
±
º²³³
±
18.18
Observe that the center branch of this circuit, that is the branch containing points
a
and
b
, is not a continuous conducting path, so no current can flow in this branch. The only
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 Spring '09
 Physics, Current, Work

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