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) → $\sqrt{R_1R_2}$
Let emf of the cell = $E$, internal resistance = $r$
Power across $R = \frac{E^2 R}{(R+r)^2}$
Given: Power across $R_1$ = Power across $R_2$
$\frac{E^2 R_1}{(R_1 + r)^2} = \frac{E^2 R_2}{(R_2 + r)^2}$
Cancel $E^2$:
$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$
Cross-multiply:
$R_1(R_2 + r)^2 = R_2(R_1 + r)^2$
Expanding:
$R_1(R_2^2 + 2R_2r + r^2) = R_2(R_1^2 + 2R_1r + r^2)$
$R_1R_2^2 + 2R_1R_2r + R_1r^2 = R_2R_1^2 + 2R_1R_2r + R_2r^2$
Cancel $2R_1R_2r$:
$R_1R_2^2 + R_1r^2 = R_2R_1^2 + R_2r^2$
Rearrange:
$R_1R_2^2 - R_2R_1^2 = R_2r^2 - R_1r^2$
$R_1R_2(R_2 - R_1) = r^2(R_2 - R_1)$
If $R_1 \neq R_2$, divide by $(R_2 - R_1)$:
$R_1R_2 = r^2$
$r = \sqrt{R_1R_2}$
Final Answer: $r = \sqrt{R_1R_2}$