Let C be the capacitance of a capacitor discharging through a resistor R. Suppose `t_1` is the time taken for the energy stored in the capacitor to reduce to half its initial value and `t_2` is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio `t_1//t_2` will be

Options

A. 1
B. `1/2`
C. `1//4`
D. 2

Answer

C

Explanation

Energy stored in capacitor,
`U=(1)/(2)(q^(2))/(C)=(1)/(2C)(q_(0)e^(-t//tau))^(2)=(q_(0)^(2))/(2C)e^(-2t//tau)` (where, `tau=CR`)
`U=U_(i)e^(-2t//tau),(1)/(2)U_(i)=U_(i)e^(-2t_(1)//tau), (1)/(2) =e^(-2t_(1)//tau) implies t_(1)=(tau)/(2)ln2`
Now, `q=q_(0)e^(-t//tau`
`(1)/(4)q_(0)=q_(0)e^(-t//2tau),t_(2)=tauln4=2tauln2` `therefore (t_(1))/(t_(2))=(1)/(4)`.