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EAMCET Engineering 2001 Solved Paper
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© examsnet.com
Question : 21
Total: 200
When a resistor of
11
Ω
is connected in series with an electric cell, the current flowing in it is 0.5 A. Instead, when a resistor of
5
Ω
is connected to the same electric cell in series, the current increases by
0.4
A
. The internal resistance of the cell is
1.5
Ω
2
Ω
2.5
Ω
3.5
Ω
Validate
Solution:
Current taken from the cell,
i
=
E
R
+
r
where
R
=
external resistance and
r
=
internal resistance
Ist Case
i
1
=
E
R
1
+
r
0.5
=
E
11
+
r
IInd Case
0.5
+
0.4
=
E
5
+
r
⇒
0.9
=
E
5
+
r
Dividing Eq. (ii) by (i),
0.5
0.9
=
E
(
11
+
r
)
E
(
5
+
r
)
5
9
=
5
+
r
11
+
r
⇒
55
+
5
r
=
45
+
9
r
10
=
4
r
⇒
r
=
2.5
Ω
© examsnet.com
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