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A cell of constant e.m.f. first connected to a resistance $R_1$ and then connected to a resistance $R_2$. If power delivered in both cases is then the internal resistance of the cell is
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The correct answer is:
$\sqrt{R_1 R_2}$
Power dissipated $=I^2 R=\left(\frac{E}{R+r}\right)^2 R$
$\begin{aligned} & \therefore\left(\frac{E}{R+r}\right)^2 R_1=\left(\frac{E}{R_2+r}\right)^2 R_2 \\ & \Rightarrow R_1\left(R_2^2+r_2+2 R_2 r\right)=R_2\left(R_1^2+r^2+2 R_1 r\right) \\ & \Rightarrow R_2^2 R_1+R_1 r^2+2 R_2 r=R_1^2 R_2+R_2 r^2+2 R_1 R_2 r \\ & \Rightarrow\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) R_1 R_2 \\ & \Rightarrow r=\sqrt{R_1 R_2}\end{aligned}$
$\begin{aligned} & \therefore\left(\frac{E}{R+r}\right)^2 R_1=\left(\frac{E}{R_2+r}\right)^2 R_2 \\ & \Rightarrow R_1\left(R_2^2+r_2+2 R_2 r\right)=R_2\left(R_1^2+r^2+2 R_1 r\right) \\ & \Rightarrow R_2^2 R_1+R_1 r^2+2 R_2 r=R_1^2 R_2+R_2 r^2+2 R_1 R_2 r \\ & \Rightarrow\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) r^2=\left(R_1-R_2\right) R_1 R_2 \\ & \Rightarrow r=\sqrt{R_1 R_2}\end{aligned}$
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