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When two terminals of a battery having some internal resistance is short circuited, power dissipated in the battery is 10 W. When an external resistor is connected to the terminals of the same battery, maximum possible power that can be dissipated in the resistor is

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2.5 W

4 W

8 W

4.5 W

answer is A.

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Detailed Solution

If internal resistance of the battery is r and emf is E, then $\frac{E^{2}}{r} = 10 W$

If external resistor has a resistance R

Then power dissipated in it, $P' = {(\frac{E}{R + r})}^{2} xR$

P' is maximum when R = r

$∴ {(P')}_{\max} = {(\frac{E}{r + r})}^{2} x r = \frac{E^{2}}{4 r} = \frac{10}{4} W = 2.5 W$

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