A capacitor of capacity `0.1 muF` connected in series to a resistor of `10 M Omega` is charged to a certain potential and then made to discharge through resistor. The time in which the potential will take to fall to half its original value is (Given, `log_(10) 2 = 0.3010`)

Options

A. `2s`
B. `0.693 s`
C. `0.5 s`
D. `1.0 s`

Answer

B

Explanation

By equation of charging,
`q = q_0 (1 - e^(-t//CR))`
According to question,
`(q)/(q_0) = 1/2 = 0.50`
`therefore 0.50 = 1 - e^(-t//CR)`
`e^(-t//CR) = 1 - 0.50 = 0.50`
`e^(-t//CR) = 2`
or `1/(CR) = log_e 2`
or `t/(CR) = 2.3026 log_10 2`
or `t = CR xx 2.3026 log_10 2`
or `t = 0.1 xx 10^(-6) xx 10 xx 10^6 xx 2.3026 log_10 2`
or `t = 2.3026 xx 0.3010 `
or `t = 0.693 s`