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(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.According to the problem,the potential falls to half its original val

Vedclass · Accepted Answer

$0.693$

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(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.According to the problem,the potential falls to half its original value,so $V = V_0/2$.Substituting this into the equation: $V_0/2 = V_0 e^{-t/RC}$.This simplifies to $1/2 = e^{-t/RC}$,or $e^{t/RC} = 2$.Taking the natural logarithm on both sides: $t/RC = \ln(2)$.We know that $\ln(2) = 2.3026 	imes \log_{10}(2)$.Given $\log_{10}(2) = 0.3010$,so $\ln(2) = 2.3026 	imes 0.3010 \approx 0.693$.Now,calculate the time constant $RC$:$R = 10 M\Omega = 10 	imes 10^6 \Omega = 10^7 \Omega$.$C = 0.1 \mu F = 0.1 	imes 10^{-6} F = 10^{-7} F$.$RC = 10^7 	imes 10^{-7} = 1 ~s$.Therefore,$t = RC 	imes 0.693 = 1 	imes 0.693 = 0.693 ~s$.

$A$ capacitor of capacity $0.1 \mu F$ connected in series to a resistor of $10 M \Omega$ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given,$\log _{10} 2=0.3010$ ) (in $~s$)

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