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The emf 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?

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The emf 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?
Solution:
Given,
balancing length for battery A, \(l_A\) = 75 cm
balancing length for battery B, \(l_B\) = 50 cm
emf of battery A, \(\mathcal{E}_A\) = ?
emf of battery B, \(\mathcal{E}_B\) = 1.02 V
Now,
\[\begin{align*} \frac{\mathcal{E}_A}{\mathcal{E}_B}&=\frac{l_A}{l_B}\\ \mathcal{E}_A&=1.02 \times \frac{75}{50}\\ \therefore \mathcal{E}_A&= 1.53 \hspace{0.1cm} V\\ \end{align*}\] Hence the emf of battery A is 1.53 V.

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