The e.m.f of a battery A is balanced by a length 75 cm on a potentiometer wire. The emf of a standard cell 1.02 V is balanced by a length of 50 cm. What is the emf of A?

Given,

Let $E_1$ be the emf of a cell A

Balancing length for $A \; (l_1) = 75\;cm$

Emf of standard cell $(E_2) = 1.02\;V$

Balancing length for $E_2 \; (l_2) = 50\;cm $

We know,

$\frac{E_1}{E_2} = \frac{l_1}{l_2}$

$E_1 = \frac{75}{50} * 1.53\;V$

Hence, emf of the cell A is $11.53\;V$

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