Let C be the capacitance of a capacitor disch | vedclass

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Question

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.For energy to reduce to half its initial value: $\frac{U_0}{2} = \frac{q^2}{2C}

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.For energy to reduce to half its initial value: $\frac{U_0}{2} = \frac{q^2}{2C} \Rightarrow \frac{q_0^2}{2} = q^2 \Rightarrow q = \frac{q_0}{\sqrt{2}}$.Using the discharging equation $q = q_0 e^{-t/RC}$,we have $\frac{q_0}{\sqrt{2}} = q_0 e^{-t_1/RC}$.Taking the natural logarithm: $-\frac{t_1}{RC} = \ln(1/\sqrt{2}) = -\frac{1}{2} \ln 2$,so $t_1 = \frac{RC \ln 2}{2}$.For charge to reduce to one-fourth its initial value: $\frac{q_0}{4} = q_0 e^{-t_2/RC}$.Taking the natural logarithm: $-\frac{t_2}{RC} = \ln(1/4) = -2 \ln 2$,so $t_2 = 2RC \ln 2$.The ratio $\frac{t_1}{t_2} = \frac{(RC \ln 2) / 2}{2RC \ln 2} = \frac{1}{4} = 0.25$.

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