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

Vedclass · Accepted Answer

$0.25$

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

$A$ capacitor of capacitance $C$ is discharging 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. Then the ratio $t_1 / t_2$ is:

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