Two unequal resistances, R_{1} and R_{2}_{,} are connected across two identical batteries of emf ε and internal resistance r (see the figure). Can the thermal energies developed in R_{1} and R_{2} be equal in a given time? If yes, what will be the condition?

#### Solution

For the given time *t*, let the currents passing through the resistance *R*_{1}_{ }and *R*_{2} be *i*_{1} and *i*_{2}, respectively.

Applying Kirchoff's Voltage Law to circuit-1, we get:-

\[\epsilon - i_1 r - i_1 R_1 = 0\]

\[ \Rightarrow i_1 = \frac{\epsilon}{r + R_1}\]

Similarly, the current in the other circuit,

\[i_2 = \frac{\epsilon}{r + R_2}\]

The thermal energies through the resistances are given by

\[i_1^2 R_1 t = i_2^2 R_2 t\]

\[ \left( \frac{\epsilon}{r + R_1} \right)^2 R_1 t = \left( \frac{\epsilon}{r + R_2} \right)^2 R_2 t\]

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

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

\[\frac{r^2}{R_1} + R_1 = \frac{r^2}{R_2} + R_2 \]

\[ r^2 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = R_2 - R_1 \]

\[ r^2 \times \frac{R_2 - R_1}{R_1 R_2} = R_2 - R_1 \]

\[ r^2 = R_1 R_2 \]

\[ \Rightarrow r = \sqrt{R_1 R_2}\]