A cell of constant e.m.f. is connected first | vedclass

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(A) The power consumed by an external resistance $R$ connected to a cell of $e.m.f.$ $E$ and internal resistance $r$ is given by $P = I^2 R = \left( \

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$\sqrt{R_1 R_2}$

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(A) The power consumed by an external resistance $R$ connected to a cell of $e.m.f.$ $E$ and internal resistance $r$ is given by $P = I^2 R = \left( \frac{E}{R + r} ight)^2 R$.Given that the power consumed is the same for $R_1$ and $R_2$:$\left( \frac{E}{R_1 + r} ight)^2 R_1 = \left( \frac{E}{R_2 + r} ight)^2 R_2$Canceling $E^2$ from both sides:$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$Cross-multiplying:$R_1(R_2 + r)^2 = R_2(R_1 + r)^2$$R_1(R_2^2 + r^2 + 2R_2 r) = R_2(R_1^2 + r^2 + 2R_1 r)$$R_1 R_2^2 + R_1 r^2 + 2R_1 R_2 r = R_2 R_1^2 + R_2 r^2 + 2R_1 R_2 r$Subtracting $2R_1 R_2 r$ from both sides:$R_1 R_2^2 + R_1 r^2 = R_2 R_1^2 + R_2 r^2$$R_1 r^2 - R_2 r^2 = R_2 R_1^2 - R_1 R_2^2$$r^2(R_1 - R_2) = R_1 R_2(R_1 - R_2)$Assuming $R_1 
eq R_2$,we divide by $(R_1 - R_2)$:$r^2 = R_1 R_2$$r = \sqrt{R_1 R_2}$

$A$ cell of constant $e.m.f.$ is connected first to an external resistance $R_1$ and then to an external resistance $R_2$. If the power consumed in both cases is the same,what is the internal resistance of the cell?

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