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AIEEE 2008 Solved Paper
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© examsnet.com
Question : 58
Total: 104
A
5
V
battery with internal resistance
2
Ω
and a
2
V
battery with internal resistance
1
Ω
are connected to a
10
Ω
resistor as shown in the figure.
The current in the
10
Ω
resistor is
0.27
A
P
2
to
P
1
0.03
A
P
1
to
P
2
0.03
A
P
2
to
P
1
0.27
A
P
1
to
P
2
Validate
Solution:
Applying Kirchhoff's loop law in
A
B
P
2
P
1
A
we get
−
2
i
+
5
−
10
i
1
=
0
.
.
.
(
i
)
Again applying kirchoffs loop law in
P
2
C
D
P
1
P
2
we get,
10
i
1
+
2
−
i
+
i
1
=
0
From
(
i
)
and
(
i
i
)
11
i
1
+
2
−
[
5
−
10
i
1
2
]
=
0
⇒
i
1
=
1
32
A from
P
2
to
P
1
© examsnet.com
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