A cell of constant emf is first connected to a resistance \( R_1 \) and then to \( R_2 \). If power delivered in both cases are same, then the internal resistance of the cell is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{R_1R_2} \)
Let the emf be E and the internal resistance r. With an external resistance R the current is \( I = \frac{E}{R + r} \), so the power delivered to the external resistance is
\( \displaystyle P = I^{2}R = \frac{E^{2}R}{(R + r)^{2}} \)
Equal power for \( R_1 \) and \( R_2 \):
\( \displaystyle \frac{R_1}{(R_1 + r)^{2}} = \frac{R_2}{(R_2 + r)^{2}} \)
\( R_1(R_2 + r)^{2} = R_2(R_1 + r)^{2} \)
\( R_1R_2^{2} + 2R_1R_2\,r + R_1r^{2} = R_2R_1^{2} + 2R_1R_2\,r + R_2r^{2} \)
The terms \( 2R_1R_2\,r \) cancel:
\( R_1R_2(R_2 - R_1) = r^{2}(R_2 - R_1) \)
Since \( R_1 \ne R_2 \), divide by \( R_2 - R_1 \):
\( r^{2} = R_1R_2 \Rightarrow r = \sqrt{R_1R_2} \)
Shortcut: Writing \( P = \frac{E^{2}}{R + \frac{r^{2}}{R} + 2r} \), the power has the same value for R and for \( \frac{r^{2}}{R} \). Two different resistances giving the same power must therefore satisfy \( R_1R_2 = r^{2} \), the internal resistance is their geometric mean.
Note: A similar question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1), and in CUET (UG) 2025 on 29-05-2025 (Slot 2).