In an experiment with a C-R circuit,two ident | vedclass

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(D) The charging voltage of a capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC_{eq}})$.For parallel combination,equivalent capacitanc

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$2.5$

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(D) The charging voltage of a capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC_{eq}})$.For parallel combination,equivalent capacitance $C_p = C + C = 2C$. The time constant is $	au_p = R(2C) = 2RC$.Given $V(t_p) = V_0/2$ at $t_p = 10 \ s$,we have $V_0/2 = V_0(1 - e^{-t_p/2RC})$,which implies $1/2 = e^{-t_p/2RC}$,so $t_p = 2RC \ln(2) = 10 \ s$.For series combination,equivalent capacitance $C_s = C/2$. The time constant is $	au_s = R(C/2) = RC/2$.We need to find $t_s$ such that $V(t_s) = V_0/2$,so $1/2 = e^{-t_s/(RC/2)}$,which implies $t_s = (RC/2) \ln(2)$.Comparing the two expressions,$t_s = t_p / 4 = 10 / 4 = 2.5 \ s$.

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