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(C) The power output $P$ of a cell with $\operatorname{emf} E$ and internal resistance $r$ connected to an external load resistance $R$ is given by $P

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$E^{2} / 4 r$

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(C) The power output $P$ of a cell with $\operatorname{emf} E$ and internal resistance $r$ connected to an external load resistance $R$ is given by $P = I^{2} R$. Since $I = E / (R + r)$,we have $P = (E / (R + r))^{2} R = E^{2} R / (R + r)^{2}$. To find the maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $dP / dR = E^{2} [((R + r)^{2} - R(2(R + r))) / (R + r)^{4}] = 0$. This simplifies to $(R + r)^{2} - 2R(R + r) = 0$,which gives $R + r - 2R = 0$,or $R = r$. Substituting $R = r$ into the power formula,we get $P_{	ext{max}} = E^{2} r / (r + r)^{2} = E^{2} r / (2r)^{2} = E^{2} r / 4r^{2} = E^{2} / 4r$.

What is the maximum power output that can be obtained from a cell of $\operatorname{emf} E$ and internal resistance $r$?

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