DirectCurrent Circuits
115
18.21
First simplify the circuit
by combining the series
resistors. Then, apply
Kirchhoff’s junction rule
at point
a
to find
Next, we apply Kirchhoff’s loop rule to the rightmost loop to obtain
12
2.00 A
II
+=
( ) ( )
12.0
0
−
()
8.00 V
6.00
−+
=
or
( )
1
12.0
2.00
I
−
1
I
=
8.00 V
6.00
(
A
0
−
This yields
Finally, apply Kirchhoff’s loop rule to the leftmost loop to obtain
1
1.78 A
I
=
)( ) ( )
11
0 A
6.00
I
4.00
2.0
8.00 V
0
ε
+−
(
−
)
(
+
=
)
or
1
4.00
2.00
A
6.00
1.78
=+
A
8.00 V
10.7 V
−
=
18.22
From Kirchhoff’s point rule, note that
1
I
2
=
−
in the
circuit shown at the right. Going counterclockwise around
the upper pane of the circuit, Kirchhoff’s loop rule gives
( )( )
2
0
I
I
−=
22
50
30
90
−Ω−Ω+Ω
or
2
170
90
Ω
=
Ω
(1)
Now, going counterclockwise around the outer perimeter
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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Current, Work

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