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55. (a) At
t
= 0 the capacitor is completely uncharged and the current in the capacitor branch is as it
would be if the capacitor were replaced by a wire. Let
i
1
be the current in
R
1
and take it to be
positive if it is to the right. Let
i
2
be the current in
R
2
and take it to be positive if it is downward.
Let
i
3
be the current in
R
3
and take it to be positive if it is downward. The junction rule produces
i
1
=
i
2
+
i
3
,
the loop rule applied to the lefthand loop produces
E−
i
1
R
1
−
i
2
R
2
=0
,
and the loop rule applied to the righthand loop produces
i
2
R
2
−
i
3
R
3
=0
.
Since the resistances are all the same we can simplify the mathematics by replacing
R
1
,
R
2
,and
R
3
with
R
. The solution to the three simultaneous equations is
i
1
=
2
E
3
R
=
2(1
.
2
×
10
3
V)
3(0
.
73
×
10
6
Ω)
=1
.
1
×
10
−
3
A
and
i
2
=
i
3
=
E
3
R
=
1
.
2
×
10
3
V
3(0
.
73
×
10
6
Ω)
=5
.
5
×
10
−
4
A
.
At
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge, Current

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