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) → $\sqrt{R_1\,R_2}$
Let 'r' be the internal resistance of the cell,
$I=\frac{E}{R+r}$ [Using Ohm's law]
and, $P=I^2R$
$⇒P_1\left(\frac{E}{R_1+r}\right)^2R,P_2\left(\frac{E}{R_2+r}\right)^2R$
$\left(\frac{E}{R_1+r}\right)^2R_1=\left(\frac{E}{R_2+r}\right)^2R_2$
$R_1({R_1}^2+r^2+2R_2r)=R_2({R_1}^2r^2+r^2+2R_1r)$
$R_1R_2(R_2-R_1)=(R_2-R_1)r^2$
$⇒r=\sqrt{R_1R_2}$