The emf of a battery is 2, V and its internal | vedclass

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(C) The power delivered to an external circuit by a battery with emf $E$ and internal resistance $r$ is given by $P = I^2 R$,where $I = \frac{E}{R+r}$

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$2$

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(C) The power delivered to an external circuit by a battery with emf $E$ and internal resistance $r$ is given by $P = I^2 R$,where $I = \frac{E}{R+r}$.Thus,$P = \left( \frac{E}{R+r} ight)^2 R$.For maximum power transfer,the external resistance $R$ must be equal to the internal resistance $r$ $(R = r)$.The formula for maximum power is $P_{	ext{max}} = \frac{E^2}{4r}$.Given $E = 2\, V$ and $r = 0.5\,\Omega$,we substitute these values:$P_{	ext{max}} = \frac{(2)^2}{4 	imes 0.5} = \frac{4}{2} = 2\, W$.

The emf of a battery is $2\, V$ and its internal resistance is $0.5 \,\Omega$. The maximum power which it can deliver to any external circuit will be ............. $watt$.

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