A 2 mu F capacitor and an R resistor are conn | vedclass

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Question

(A) The voltage across the capacitor in a charging $RC$ circuit is given by $V(t) = V_0(1 - e^{-t/RC})$.Here,$V_0 = 200\ V$,$V(t) = 120\ V$,$t = 5\ s$

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$2.7 	imes 10^6\ \Omega$

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

$A$ $2\ \mu F$ capacitor and an $R$ resistor are connected in series to a $200\ V$ $DC$ supply. $A$ neon bulb is connected across the capacitor,which glows at $120\ V$. Calculate the value of $R$ so that the bulb glows for $5\ s$ after the switch is closed. (Given: $\log_{10} 2.5 = 0.4$)

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