Question

Let c be the capacitance of a capacitor discharging through a resistor r. Suppose, t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one-fourth its initial value. Then, the ratio t1/t2 will be

Answer 1

U = q2/2C

U = Umax/2

q = Q0/v2

q = Q0 e-t/RC

ln(q/Q0) = -t/RC

t = RC ln(Q0/q)

At t1 ; q = Q0/√2 and t1 = (RC/2) ln2

At t2 ; q = Q0/4 and t2 = 2 RC ln2

t1/t2 = 1/4

The solution of this is given in the following image.