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(D) For a circuit with an electromotive force $E$ and internal resistance $r$,the power $P$ delivered to an external resistance $R$ is given by $P = I

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(D) For a circuit with an electromotive force $E$ and internal resistance $r$,the power $P$ delivered to an external resistance $R$ is given by $P = I^2 R = \left(\frac{E}{R+r}ight)^2 R$.To find the condition for maximum power,we differentiate $P$ with respect to $R$ and set it to zero: $\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2 - R \cdot 2(R+r)}{(R+r)^4} ight] = 0$.This gives $R+r - 2R = 0$,so $R = r$. Thus,statement $(i)$ is correct.Substituting $R = r$ into the power formula: $P = \left(\frac{E}{R+R}ight)^2 R = \left(\frac{E}{2R}ight)^2 R = \frac{E^2}{4R^2} \cdot R = \frac{E^2}{4R}$. Thus,statement $(ii)$ is correct.The total power supplied by the source (input power) is $P_{	ext{in}} = E \cdot I = E \cdot \left(\frac{E}{R+r}ight)$. Since $R=r$,$P_{	ext{in}} = E \cdot \left(\frac{E}{2R}ight) = \frac{E^2}{2R}$. Thus,statement $(iii)$ is correct.The efficiency $\eta$ is defined as the ratio of output power to input power: $\eta = \frac{P_{	ext{out}}}{P_{	ext{in}}} 	imes 100 = \frac{E^2/4R}{E^2/2R} 	imes 100 = \frac{1}{2} 	imes 100 = 50\%$. Thus,statement $(iv)$ is correct.Since all statements are correct,the correct option is $D$.

If the power in the external resistance $R$ is maximum,then which of the following statements are correct?
$(i) R = r$
$(ii)$ Power in $R$ is $\frac{E^2}{4R}$
$(iii)$ Input power is $\frac{E^2}{2R}$
$(iv)$ Efficiency is $50\%$

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If the power in the external resistance $R$ is maximum,then which of the following statements are correct?$(i) R = r$$(ii)$ Power in $R$ is $\frac{E^2}{4R}$$(iii)$ Input power is $\frac{E^2}{2R}$$(iv)$ Efficiency is $50\%$