A cell sends a current through a resistance R | vedclass

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(C) Let $\varepsilon$ be the $EMF$ and $r$ be the internal resistance of the cell.For resistance $R_1$,the current is $I_1 = \frac{\varepsilon}{R_1 +

Vedclass · Accepted Answer

$\sqrt{R_1 R_2}$

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(C) Let $\varepsilon$ be the $EMF$ and $r$ be the internal resistance of the cell.For resistance $R_1$,the current is $I_1 = \frac{\varepsilon}{R_1 + r}$. The heat produced is $H_1 = I_1^2 R_1 t = \frac{\varepsilon^2 R_1 t}{(R_1 + r)^2}$.For resistance $R_2$,the current is $I_2 = \frac{\varepsilon}{R_2 + r}$. The heat produced is $H_2 = I_2^2 R_2 t = \frac{\varepsilon^2 R_2 t}{(R_2 + r)^2}$.Given $H_1 = H_2$,we have:$\frac{\varepsilon^2 R_1 t}{(R_1 + r)^2} = \frac{\varepsilon^2 R_2 t}{(R_2 + r)^2}$$\frac{R_1}{(R_1 + r)^2} = \frac{R_2}{(R_2 + r)^2}$$R_1(R_2 + r)^2 = R_2(R_1 + r)^2$$R_1(R_2^2 + 2R_2 r + r^2) = R_2(R_1^2 + 2R_1 r + r^2)$$R_1 R_2^2 + 2R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2R_1 R_2 r + R_2 r^2$$R_1 R_2^2 + R_1 r^2 = R_2 R_1^2 + R_2 r^2$$r^2(R_1 - R_2) = R_1 R_2(R_1 - R_2)$Assuming $R_1 
eq R_2$,we get $r^2 = R_1 R_2$,so $r = \sqrt{R_1 R_2}$.

$A$ cell sends a current through a resistance $R_1$ for time $t$. Now,the same cell sends a current through a resistance $R_2$ for the same time $t$. If the heat produced in both cases is the same,then the internal resistance of the cell is:

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