A resistor R and a 2 mu F capacitor in serie | vedclass

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Question

(B) The voltage across a charging capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200 \ V$,$V(t) = 120 \ V$,$C = 2

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$2.7 	imes 10^6 \ \Omega$

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(B) The voltage across a charging capacitor in an $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200 \ V$,$V(t) = 120 \ V$,$C = 2 	imes 10^{-6} \ F$,and $t = 5 \ s$.Substituting the values: $120 = 200(1 - e^{-5/(R 	imes 2 	imes 10^{-6})})$.$0.6 = 1 - e^{-5/(R 	imes 2 	imes 10^{-6})} \Rightarrow e^{-5/(R 	imes 2 	imes 10^{-6})} = 0.4$.Taking the natural logarithm on both sides: $-5/(R 	imes 2 	imes 10^{-6}) = \ln(0.4) = -\ln(2.5)$.$5/(R 	imes 2 	imes 10^{-6}) = \ln(2.5) = 2.303 	imes \log_{10}(2.5)$.Given $\log_{10}(2.5) = 0.4$,so $\ln(2.5) = 2.303 	imes 0.4 = 0.9212$.$R = 5 / (2 	imes 10^{-6} 	imes 0.9212) \approx 2.71 	imes 10^6 \ \Omega$.

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