Question

In a potentiometer arrangement, a cell of emf $1.25V$ gives a balance point at $35cm$ length of the wire. If the cell is replaced by another cell, the balance point shifts to $63cm$, then emf of the second cell is.

Answer 1

Step 1: Given data

Given that,

$E_1=1.25V$

$L_1=35cm$

$L_2=63cm$

Step 2: Formula used

The condition where the galvanometer shows a null point is,

$\frac{E_1}{E_2}=\frac{L_1}{L_2}$

$\therefore E_2=\frac{E_1\times L_2}{L_1}$

Step 3: Calculate the EMF

Upon substituting the values we get,

$E_2=\frac{1.25\times 63}{35}$

$=2.25V$

Hence, the emf of the second cell is $2.25V$.