A battery with EMF E and internal resistance | vedclass

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

Vedclass · Accepted Answer

$R=r$

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(D) The current $I$ in the circuit is given by $I = \frac{E}{R+r}$.The power $P$ consumed by the external resistance $R$ 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: $\frac{dP}{dR} = 0$.$\frac{dP}{dR} = E^2 \left[ \frac{(R+r)^2(1) - R(2)(R+r)}{(R+r)^4} \right] = 0$.This implies $(R+r)^2 - 2R(R+r) = 0$.Dividing by $(R+r)$,we get $(R+r) - 2R = 0$,which simplifies to $r - R = 0$ or $R = r$.Thus,the power consumption is maximum when the external resistance equals the internal resistance of the battery.

$A$ battery with $EMF$ $E$ and internal resistance $r$ is connected across a resistance $R$. The power consumption in $R$ will be maximum when:

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