A battery of e.m.f. E and internal resistance | vedclass

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Question

(B) The current $I$ in the circuit is given by $I = \frac{E}{R + r}$.The power $P$ delivered to the external resistor $R$ is $P = I^2 R = \left( \frac

Vedclass · Accepted Answer

Power delivered to the resistor is maximum when $R = r$.

Vedclass · Answer

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

$A$ battery of $e.m.f.$ $E$ and internal resistance $r$ is connected to a variable resistor $R$ as shown in the figure. Which one of the following is true?

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