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(D) The voltage across a discharging capacitor in a $C-R$ circuit is given by $V(t) = V_0 e^{-t/	au}$, where $	au = RC_{eq}$ is the time constant.Fo

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$2.5$

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(D) The voltage across a discharging capacitor in a $C-R$ circuit is given by $V(t) = V_0 e^{-t/	au}$, where $	au = RC_{eq}$ is the time constant.For the voltage to reduce to half its original value, $V_0/2 = V_0 e^{-t/	au}$, which implies $e^{-t/	au} = 1/2$, or $t = 	au \ln(2)$.For parallel combination, $C_{p} = C + C = 2C$. The time constant is $	au_p = R(2C) = 2RC$.Given $t_1 = 10\; s$, we have $10 = (2RC) \ln(2)$.For series combination, $C_{s} = (C \cdot C)/(C + C) = C/2$. The time constant is $	au_s = R(C/2) = RC/2$.Let the required time be $t_2$. Then $t_2 = (RC/2) \ln(2)$.Comparing the two expressions:$\frac{t_2}{t_1} = \frac{(RC/2) \ln(2)}{(2RC) \ln(2)} = \frac{1/2}{2} = \frac{1}{4}$.Therefore, $t_2 = t_1 / 4 = 10 / 4 = 2.5\; s$.

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