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 equal, then the internal resistance of the cell is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{R_{1}R_{2}} \)
With a cell of emf E and internal resistance r connected to an external resistance R, the current is \( E/(R+r) \), so the power delivered to R is
\( \displaystyle P = \frac{E^{2}R}{(R+r)^{2}} \)
Equal power in the two cases:
\( \displaystyle \frac{E^{2}R_1}{(R_1+r)^{2}} = \frac{E^{2}R_2}{(R_2+r)^{2}} \)
\( \displaystyle R_1(R_2+r)^{2} = R_2(R_1+r)^{2} \)
Expanding both sides, the terms \( 2R_1R_2r \) cancel:
\( \displaystyle R_1R_2^{2} + R_1r^{2} = R_2R_1^{2} + R_2r^{2} \)
\( \displaystyle R_1R_2(R_2-R_1) = r^{2}(R_2-R_1) \)
Since \( R_1 \ne R_2 \), divide by \( (R_2-R_1) \):
\( \displaystyle r^{2} = R_1R_2 \)
\( \displaystyle r = \sqrt{R_1R_2} \)
Note: The same question was also asked in CUET (UG) 2025 on 29-05-2025 (Slot 2), and a similar question in CUET (UG) 2023 on 02-07-2023 (Slot 3).