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ε
−
−
−
=
iR
i R
ir
21
1
0.
We apply the loop rule to the righthand loop to obtain
iR
i i R
V
11
1
0
−−
=
bg
.
The second equation yields
i
RR
R
i
V
V
=
+
1
1
We substitute this into the first equation to obtain
−
++
+=
Rr
R
i
V
V
1
0
b
g
This has the solution
i
R
R
R
V
VV
1
1
=
+
b
gb
g
.
The reading on the voltmeter is
()
(
)
( ) ( ) ( )
3
1
33
1
3.0V
5.0 10
250
300
100
250
5.0 10
250
5.0 10
1.12V.
V
RrRR
R
R
×Ω
Ω
==
+
Ω
+
Ω
Ω
+×Ω
+
Ω
=
The current in the absence of the voltmeter can be obtained by taking the limit as
R
V
becomes infinitely large. Then
R
RRr
1
12
30
250
250
300
100
115
=
=
=
.
.
V
V.
b
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 Spring '08
 Any
 Physics, Current

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