Question

A cell of constant emf 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 :

• A. $\sqrt{R_1 R_2} $
• B. $\sqrt{\frac{R_1}{R_2}}$
• C. $\frac{R_1 - R_2}{2}$
• D. $\frac{R_1 + R_2}{2}$

Answer 1

The Correct Option is A

Current given by cell
$I=\frac{E}{R +r}$
Power delivered in first case
$P_{1}=I^{2} R_{1}$
$=\left(\frac{E}{R_{1}+r}\right)^{2} R_{1}$
Power delivered in second case
$P_{2}=I^{2} R_{2}$
$=\left(\frac{E}{R_{2}+r}\right)^{2} R_{2}$
Power delivered is same in the both the cases
$\left(\frac{E}{R_{1}+r}\right)^{2} R_{1}=\left(\frac{E}{R_{2}+r}\right)^{2} R_{2}$
$\frac{R_{1}}{\left(R_{1}+r\right)^{2}}=\frac{R_{2}}{\left(R_{2}+r\right)^{2}}$
$R_{1}\left(R_{2}^{2}+r^{2}+2 R_{2} r\right)$
$=R_{2}\left(R_{1}^{2}+r^{3}+2 R_{1} r\right)$
$R_{1} R_{2}^{2}+R_{1} r^{2}+2 R_{1} R_{2} r$
$=R_{2} R_{1}^{2}+R_{2} r^{2}+2 R_{1} R_{2} r$
$R_{1} R_{2}^{2}-R_{2} R_{1}^{2}=R_{2} r^{2}-R_{1} r^{2}$
$R_{1} R_{2}\left(R_{2}-R_{1}\right)=r^{2}\left(R_{2}-R_{1}\right)$
$r=\sqrt{R_{1} R_{2}}$