A capacitor of capacitance C is being dischar | vedclass

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Question

(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Since the charge $q$ decays as $q(t) = q_0 e^{-t/RC}$,the energy decays as $U(t)

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(C) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Since the charge $q$ decays as $q(t) = q_0 e^{-t/RC}$,the energy decays as $U(t) = \frac{q_0^2}{2C} e^{-2t/RC} = U_0 e^{-2t/RC}$,where $RC = 	au$ is the time constant.For $t_1$,the energy reduces to half: $U(t_1) = U_0/2$.$U_0/2 = U_0 e^{-2t_1/	au} \implies 1/2 = e^{-2t_1/	au} \implies \ln(1/2) = -2t_1/	au \implies -\ln(2) = -2t_1/	au \implies t_1 = \frac{	au \ln(2)}{2}$.For $t_2$,the charge reduces to one-fourth: $q(t_2) = q_0/4$.$q_0/4 = q_0 e^{-t_2/	au} \implies 1/4 = e^{-t_2/	au} \implies \ln(1/4) = -t_2/	au \implies -2\ln(2) = -t_2/	au \implies t_2 = 2	au \ln(2)$.Taking the ratio $t_1/t_2 = \frac{	au \ln(2) / 2}{2	au \ln(2)} = \frac{1/2}{2} = 1/4 = 0.25$.

$A$ capacitor of capacitance $C$ is being discharged through a resistor $R$. Let $t_1$ be the time taken for the energy stored in the capacitor to reduce to half of its initial value,and $t_2$ be the time taken for the charge to reduce to one-fourth of its initial value. Find the ratio $t_1/t_2$.

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