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. lt is employed to measure the e.m.f. of a battery whose internal resistance is 0.5 $\Omega$. lf the balance point is obtained at $l=30$ cm from the positive end, the e.m.f. of the battery is

• A. $\frac{30E}{100.5}$
• B. $\frac{30E}{100-0.5}$
• C. $\frac{30(E-0.5i)}{100}$, where i is the current in the potentiometer wire.
• D. $\frac{30E}{100}$

Answer 1

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

Since galvanometer is used to find the null point when the battery of unknown EMF is connected to the potentiometer wire at a particular point(null position), no current flows through the unknown battery. Hence, there is no drop across the internal resistance of the battery.

Therefore, the potential across the given segment of wire is the absolute EMF of the unknown battery $=\frac{30}{100}E$
Hence, correct answer is option D.