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In the circuit shown in figure, the resistance R can be varied. By changing R power dissipated in it is made maximum. If the emf of the battery is 10 volt and internal resistance is 1 ohm then find the maximum power dissipated in R.

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10 W

15 W

20 W

25 W

answer is D.

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Detailed Solution

When power dissipated in R is maximum, $R = r = 1 Ω$

$∴ i = \frac{ε}{R + r} = \frac{10}{1 + 1} A = 5 A$

$∴$ Maximum power in $R = {(i)}^{2} R = {(5)}^{2} \times 1 W = 25 W$

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