A battery of emf ε and internal resistance r is connected to a resistor of resistance R₁ and QJ of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance R₂, the same quantity of heat is produced in the same time t. Then, the value of r is :

$\sqrt{R_1 R_2}$

In the first case,

Current in the circuit, $I_1=\frac{\varepsilon}{R_1+r}$

Amount of heat produced in resistor $R_1$ in time t is

$Q_1=I_1^2 R_1 t=\frac{\varepsilon^2}{\left(R_1+r\right)^2} R_1 t$                (i)

In the second case,

Current in $R_1$ the circuit, $I_2=\frac{\varepsilon}{r+R_2}$

Amount of heat produced in resistor $R_2$ in the same time t is

$Q_2=I_2^2 R_2 t=\frac{\varepsilon^2}{\left(r+R_2\right)^2} R_2 t$                  (ii)

As per question, $Q_1=Q_2=Q$

From (i) and (ii), we get

$\frac{\varepsilon^2 R_1 t}{\left(r+R_1\right)^2}=\frac{\varepsilon^2 R_2 t}{\left(r+R_2\right)^2}$

$\frac{R_1}{\left(r+R_1\right)^2}=\frac{R_2}{\left(r+R_2\right)^2}$

On solving, we get

$r=\sqrt{R_1 R_2}$