Explanation

(d) Let $E$ and $r$ be the emf and internal resistance of a battery respectively
[image:3_image144.emf]
In the first case current flowing in the circuit
$I_{1} = \frac{E}{R_{1}r}$
Or $E = I_{1}(R_{1} + r)$
[image:3_image145.emf]
In the second case current flowing in the circuit
$I_{2} = \frac{E}{R_{2} + r}$
Or $E = I_{2}\left( R_{2} + r \right)$
Equating equations (i) and (ii), we get
$I_{1}\left( R_{1} + r \right) = I_{2}\left( R_{2} + r \right) \Rightarrow I_{1}R_{1} + I_{1}r = I_{2}R_{2} + I_{2}r$
$I_{1}R_{1} - I_{2}R_{2} = \left( I_{2} - I_{1} \right)r \Rightarrow \left( I_{2} - I_{1} \right)r = I_{1}R_{1} - I_{2}R_{2}$
$r = \frac{I_{1}R_{1} - I_{2}R_{2}}{I_{2} - I_{1}}$