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A capacitor of $0.1 μ F$ is charged from a 100V battery through a series resistance of $1000 Ω$ . Find i) the time for the capacitor to receive 63.2% of its final charge ii) the charge received in this time

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$1 \times 10^{- 4} s, 6.32 μ C$

$1 \times 10^{- 3} s, 20 μ C$

$1 \times 10^{- 2} s, 50 μ C$

$1 \times 10^{- 8} s, 70 μ C$

answer is A.

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Detailed Solution

$(1) λ = C R = 0.1 \times 10^{- 6} \times 1000$

$= 10^{- 4} \sec$

(2) final charge $Q = C V$ $= 0.1 \times 100 = 10 μ C$

Charge received during this time is

$0.632 \times 10 μ C = 6.32 μ C$

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