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(C) The heat produced in a resistor is given by $Q = I^2 R t$. Since the battery has an electromotive force $E$ and internal resistance $r$,the curren

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$\sqrt {{R_1} {R_2}}$

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(C) The heat produced in a resistor is given by $Q = I^2 R t$. Since the battery has an electromotive force $E$ and internal resistance $r$,the current $I$ is given by $I = \frac{E}{R+r}$.Given that the heat $Q$ produced is the same for both resistors in the same time $t$,we have:$Q = \left( \frac{E}{R_1 + r} \right)^2 R_1 t = \left( \frac{E}{R_2 + r} \right)^2 R_2 t$Canceling $E^2$ and $t$ from both sides:$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$Taking the square root of both sides:$\frac{\sqrt{R_1}}{R_1 + r} = \frac{\sqrt{R_2}}{R_2 + r}$$\sqrt{R_1}(R_2 + r) = \sqrt{R_2}(R_1 + r)$$\sqrt{R_1} R_2 + r\sqrt{R_1} = \sqrt{R_2} R_1 + r\sqrt{R_2}$$r(\sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_2} R_1 - \sqrt{R_1} R_2$$r(\sqrt{R_1} - \sqrt{R_2}) = \sqrt{R_1 R_2}(\sqrt{R_1} - \sqrt{R_2})$$r = \sqrt{R_1 R_2}$

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