Question

A cell of e.m.f. $E$ is connected to a resistance $R_1$ for time $t$ and the amount of heat generated in it is $H$. If the resistance $R_1$ is replaced by another resistance $R_2$ and is connected to the cell for the same time $t$, the amount of heat generated in $R_2$ is $4H$. Then the internal resistance of the cell is

• A. $\frac{2R_1+R_2}{2}$
• B. $\sqrt{R_1{R}_2}\frac{2\sqrt{R_2}-\sqrt{R_1}}{\sqrt{R_2}-2\sqrt{R_1}}$
• C. $\sqrt{R_1{R}_2}\frac{\sqrt{R_2}-2\sqrt{R_1}}{2\sqrt{R_2}-\sqrt{R_1}}$
• D. $\sqrt{R_1{R}_2}\frac{\sqrt{R_2}-\sqrt{R_1}}{\sqrt{R_2}-\sqrt{R_1}}$

Answer 1

Answer: (B)

We have, $H=l^{2}R$
According to given condition, $l_1^2{R}_1=H$
and $l_2^2{R}_2=4H$
$\Rightarrow \frac{E^2}{{\left(R_1+r\right)}^2}{R}_1=H$
and $\frac{E^2}{{\left(R_2+r\right)}^2}{R}_2=4H$
$\therefore \frac{R_2}{{\left(R_2+r\right)}^2}=\frac{4R_1}{{\left(R_1+r\right)}^2}$
$\Rightarrow \sqrt{R_2}\left(R_1+r\right)=2\sqrt{R_1}\left(R_2+r\right)$
On solving, $r=\frac{\sqrt{R_1{R}_2}\left[\sqrt{R_1}-2\sqrt{R_2}\right]}{2\sqrt{R_1}-\sqrt{R_2}}$