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

- 1/4
- 1/2
- 1
- 2

**Answer**

U = q^{2}/2C

U = U_{max}/2

q = Q_{0}/v_{2}

q = Q_{0} e^{-t/RC}

ln(q/Q_{0}) = -t/RC

t = RC ln(Q_{0}/q)

At t_{1} ; q = Q_{0}/√2 and t_{1} = (RC/2) ln2

At t_{2} ; q = Q_{0}/4 and t_{2} = 2 RC ln2

t_{1}/t_{2} = 1/4

**The correct option is A.**