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(B) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$,where $R$ is the external resistance.Power $P$ delivered to the external resistan

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$E^2 / 4r$

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(B) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$,where $R$ is the external resistance.Power $P$ delivered to the external resistance is $P = i^2 R = \frac{E^2 R}{(r + R)^2}$.To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = E^2 \left[ \frac{(r + R)^2 - R(2)(r + R)}{(r + R)^4} \right] = 0$.This simplifies to $(r + R) - 2R = 0$,which gives $R = r$.Substituting $R = r$ into the power equation: $P_{\max} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.

The maximum power drawn from a source with electromotive force $E$ and internal resistance $r$ is given by:

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