Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a given cell.

Principle of working: The voltage drop along the wire is directly proportional to the length of the wire. The potentiometer works without drawing any current from the voltage source.

We can also use a potentiometer to measure internal resistance of a cell. For this the cell (emf E) whose internal resistance ( r ) is to be determined is connected across a resistance box through a

Key K_{2}, as shown in the figure.

With key K_{2} open, balance is obtained at length L1 (AN1).

Then,

E = ΦL_{1}

When key K2 is closed, the cell sends a current (I ) through the resistance box ®. If V is the terminal potential difference of the cell and balance is obtained at length L_{2} (AN_{2}), then

V = ΦL_{2}

So, we have E/V = L_{1}/L_{2}

But, E= I (r + R) and V = IR. This gives

E/V = (r + R)/R

On comparing the above equations, we get,

(R + r)/R = L_{1}/L_{2}

Using the above equation, we can find the internal resistance of a given cell.