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(D) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$.The power $P$ delivered to the external resistance $R$ is $P = i^2 R$.Substitutin

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$R = r$

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(D) The current $i$ in the circuit is given by $i = \frac{E}{r + R}$.The power $P$ delivered to the external resistance $R$ is $P = i^2 R$.Substituting the expression for $i$,we get $P = \left(\frac{E}{r + R}\right)^2 R = \frac{E^2 R}{(r + R)^2}$.To find the condition for maximum power,we differentiate $P$ with respect to $R$ and set it to zero:$\frac{dP}{dR} = E^2 \left[ \frac{(r + R)^2 \cdot 1 - R \cdot 2(r + R)}{(r + R)^4} \right] = 0$.This simplifies to $(r + R)^2 - 2R(r + R) = 0$.Dividing by $(r + R)$,we get $(r + R) - 2R = 0$,which implies $r - R = 0$ or $R = r$.Thus,the power delivered is maximum when the external resistance equals the internal resistance of the cell.

$A$ cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when:

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