DirectCurrent Circuits
133
Substitution of equation (4) into (3) yields
1
1
140
310
450
I
I
+=
or
2
11
310
450
140
0
II
−
Solving this quadratic equation gives two possible values for the current
. These are
. Then, from
1
I
1
A
and
0.452
=
1
1.0
A
=
2
1
20 W
=
R
I
, we find two possible values for the
resistance
R
. These are
20
or
98
RR
=Ω
=
Ω
18.57
When connected in series, the equivalent resistance is
12
eq
n
RRR
Rn
R
=
++
⋅
⋅
⋅
. Thus,
the current is
( )
( )
se
q
IV
R V
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This document was uploaded on 10/20/2011.
 Spring '09
 Physics, Current, Work

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