Question

The length of a wire of a potentiometer is $100 \,cm$, and the e.m.f. of its stand and 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
(where $i$ is the current in the potentiometer wire).

• A. $\frac{30\,E}{100}$
• B. $\frac{30\,E}{100-0.5}$
• C. $\frac{30\left(E-0.5 i \right)}{100}$
• D. $\frac{30\,E}{100 .5}$

Answer 1

Answer: (A)

Let $V$ be the potential across balance point and one end of wire. Hence acccording to the principle of potentiometer $V \propto l$
Also if a cell of emf $E$ is employed in the circuit between the ends of potentiometer wire of length $L$ then $E \propto L .$
Therefore, $\frac{ V }{ E }=\frac{1}{ L }$
$V =\frac{1}{ L } E =\frac{ 3 0 }{ 1 0 0 } E =\frac{ 3 0 E }{ 1 0 0 }$