Question

The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is $E$ volt. It is employed to measure the emf of a battery whose internal resistance is $0.5\Omega$. If the balance point is obtained at $l=30$ cm from the positive end, the emf of the battery is

• A. $\frac{30E}{100}$
• B. $\frac{30E}{100.5}$
• C. $\frac{30R}{(100-0.5)}$
• D. $\frac{30(E-0.5i)}{100}$

Answer 1

The correct option is A $\frac{30E}{100}$

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,
$\frac{V}{E}=\frac{l}{L}$
$V=\frac{l}{L}E=\frac{30}{100}E=\frac{30E}{100}$