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JEE Main Physics Class 12 Current Electricity Part 1 Questions
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© examsnet.com
Question : 95
Total: 100
. The length of a wire of a potentiometer is 100 cm, and the e. m.f. of its standard cell is E volt. It is employed to measure the e.m.f. of a battery whose internal resistance is 0.5Ω. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is
where i is the current in the potentiometer wire
[2003]
30
E
100.5
30
E
(
100
−
0.5
)
30
(
E
−
0.5
i
)
100
30
E
100
Validate
Solution:
From the principle of potentiometer,
V
∝
l
If a cell of emF
E
is employed in the circuit between the ends of potentiometer wire of length
L
,
then
V
E
=
l
L
⇒
V
=
E
l
L
=
30
E
100
Note :
In this arrangement, the internal resistance of the battery
E
does not play any role as current is not passing through the battery.
© examsnet.com
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