A cell of constant emf is first connected to a resistance \( R_1 \) and then connected to a resistance \( R_2 \). If power delivered in both cases is same, then the internal resistance of the cell is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{R_1 R_2} \)
A cell of emf \( \varepsilon \) and internal resistance \( r \) sends a current \( \varepsilon/(R + r) \) through an external resistance \( R \), so the power delivered to \( R \) is
\( \displaystyle P = \frac{\varepsilon^{2} R}{(R + r)^{2}} \)
Equal power in the two cases:
\( \displaystyle \frac{\varepsilon^{2} R_1}{(R_1 + r)^{2}} = \frac{\varepsilon^{2} R_2}{(R_2 + r)^{2}} \;\Rightarrow\; R_1 (R_2 + r)^{2} = R_2 (R_1 + r)^{2} \)
Expanding, the terms \( 2R_1R_2\,r \) cancel on both sides:
\( \displaystyle R_1 R_2^{2} + R_1 r^{2} = R_2 R_1^{2} + R_2 r^{2} \)
\( \displaystyle r^{2}(R_1 - R_2) = R_1 R_2 (R_1 - R_2) \;\Rightarrow\; r^{2} = R_1 R_2 \)
\( \displaystyle r = \sqrt{R_1 R_2} \)
Note: The same question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1), and a similar question in CUET (UG) 2023 on 02-07-2023 (Slot 3).