Question

For a cell terminal potential difference is 2.2 volt when circuit is open and reduces to 1.8 volt when cell is connected to an external resistance of 5 ohm. What is the internal resistance of the cell?

Answer 1

$P.D.=2.2volts$ , this is the EMF of the cell.

Now when the circuit is closed and a current is flowing through the circuit then the $P.D.=1.8V$

Thus the total resistance of the circuit is $R+r=5+rohms$

So, the current in the circuit is $i=\left(\frac{2.2}{5+r}\right)$

hence the PD across the internal resistance.

Given that,

$r\left(\frac{2.2}{5+r}\right)=2.2-1.8$ as the drop in the potential across the battery is due the P.D. across the internal resistance.

So,
$r\left(\frac{2.2}{5+r}\right)=0.4$

$1.8r=2$

$r=1.11ohms$