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(A) The current $I$ flowing through the circuit is given by $I = \frac{V}{R + r}$.The power $P$ dissipated as heat in the external resistor $R$ is $P

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$r$

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(A) The current $I$ flowing through the circuit is given by $I = \frac{V}{R + r}$.The power $P$ dissipated as heat in the external resistor $R$ is $P = I^2 R = \left( \frac{V}{R + r} \right)^2 R$.To find the value of $R$ for which $P$ is maximum,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = V^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$.This simplifies to $(R+r) - 2R = 0$,which gives $R = r$.Thus,the power dissipated in the external resistor is maximum when the external resistance is equal to the internal resistance of the battery.

An external resistance $R$ is connected to a battery of $e.m.f.$ $V$ and internal resistance $r$. The joule heat produced in resistor $R$ is maximum when $R$ is equal to

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