A capacitor of capacitance C = 1 , mu F is su | vedclass

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(D) The instantaneous voltage $V$ across a charging capacitor is given by the formula: $V = V_0 (1 - e^{-t/RC})$.Here,$V_0 = 100 \, V$,$V = 50 \, V$,$

Vedclass · Accepted Answer

$0.69$

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(D) The instantaneous voltage $V$ across a charging capacitor is given by the formula: $V = V_0 (1 - e^{-t/RC})$.Here,$V_0 = 100 \, V$,$V = 50 \, V$,$R = 100 \, \Omega$,and $C = 1 \, \mu F = 10^{-6} \, F$.The time constant $	au = RC = 100 	imes 10^{-6} = 10^{-4} \, s$.Substituting the values into the equation:$50 = 100 (1 - e^{-t/10^{-4}})$$0.5 = 1 - e^{-t/10^{-4}}$$e^{-t/10^{-4}} = 0.5$Taking the natural logarithm on both sides:$-t/10^{-4} = \ln(0.5) = -\ln(2)$$t/10^{-4} = \ln 2$Given $\ln 2 = 0.69$,we get:$t = 0.69 	imes 10^{-4} \, s$.Thus,the required value is $0.69$.

$A$ capacitor of capacitance $C = 1 \, \mu F$ is suddenly connected to a battery of $100 \, V$ through a resistance $R = 100 \, \Omega$. The time taken for the capacitor to be charged to $50 \, V$ is $.... \times 10^{-4} \, s$. (Take $\ln 2 = 0.69$)

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