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Q.

A capacitor of capacitance $10 μ F$ is connected to a resistance $2 Ω$ and battery of emf 5V of negligible internal resistance, after $20 μ s$ of completing the circuit power delivered by the battery is

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a

4.6 W

b

6.5 W

c

7.2 W

d

8.5 W

answer is A.

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Detailed Solution

$I = \frac{E}{R} e^{- \frac{t}{T}} = \frac{5}{2} e^{\frac{- 20}{20}} = \frac{5}{2 e}$

Power delivered by battery is

$P = E I = 5 \times \frac{5}{2 e} = \frac{25}{2 e} = 4.6 W$

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