First slide

potentiometer

Question

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 $0.5 Ω$ . If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is

Easy

A

$\frac{30 E}{100}$
B

$\frac{30 E}{100.5}$
C

$\frac{30 E}{(100 - 0.5)}$
D

$\frac{30 (E - 0.5 i)}{100}$ , where i is the current in the potentiometer

Solution

From the principle of potentiometet, $V \propto l$
$\Rightarrow \frac{V}{E} = \frac{l}{L};$ where V = emf of battery E = emf of standard cell, L = Length of potentiometer wire
$V = \frac{El}{L} = \frac{30 E}{100}$

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