A capacitor of capacitance $$C = 1 \, \mu$$F is suddenly connected to a battery of 100 V through a resistance $$R = 100 \, \Omega$$. The time taken for the capacitor to be charged to get 50 V is:
(Take ln 2 = 0.69)

A

$$1.44 \times 10^{-4}$$ s

B

$$3.33 \times 10^{-4}$$ s

C

$$0.69 \times 10^{-4}$$ s

D

$$0.30 \times 10^{-4}$$ s