A parallel plate capacitor of `1 muF` capacity is discharging thorugh a resistor. If its energy reduces to half in one second. The value of resistance will be

Options

A. `(2)/(In(2))M Omega`
B. `(4)/(In(2))M Omega`
C. `(10)/(In(2))M Omega`
D. `(16)/(In(2))M Omega`

Answer

A

Explanation

`q = q_(0)q^(-t//tau)`
when energy is `50%`, then `q = (q_(0))/(sqrt(2))`
`:. (q_(0))/(sqrt(2)) = q_(0)e^(-t//tau)`
`e^(-t//tau) = sqrt(2)`
`(t)/(tau) = In (sqrt(2)) rArr tau = (t)/(In (sqrt(2)))`
or `RC = (1)/(In(sqrt(2)))` (`:' tau = RC)`
`R = (1)/(C In(sqrt(2))) = (1)/(10^(-6)In(sqrt(2))) = (10^(6))/(In(sqrt(2)))Omega = (2)/(In(sqrt(2)))M Omega`