LetC 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. 2
B. 1
C. `(1)/(2)`
D. `(1)/(4)`

Answer

D

Explanation

As `U = (Q^(2))/(2C) = ((Q_(0) e^(-t//tau))^(2))/(2C)`
`= (Q_(0)^(2))/(2C) e ^(-2t//tau) = U_(0) e^(-2t//tau)`
As when `t = t_(1), U = U_(0)//2`, so `(U_(0))/(2) = U_(0) e^(-2t_(1)//tau)`
`(1)/(2) = e^(-(2t_(1))/(tau))` or `t_(1) = (tau)/(2) 1n 2`
`(Q_(0))/(4) = Q_(0) e^(-t_(2)//tau)`
`t_(2) = tau 1n 4 = 2 tau 1n 2`
`(t_(1))/(t_(2)) = (1)/(4)`