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(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.We want to find the time $t$ when $V = V_0/2$.Substituting this into

Vedclass · Accepted Answer

$0.693$

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(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.We want to find the time $t$ when $V = V_0/2$.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: $t/RC = \ln(2)$.Using the conversion $\ln(2) = 2.3026 	imes \log_{10}(2)$, we get $t = RC 	imes 2.3026 	imes 0.3010$.Given $C = 0.1 \mu F = 0.1 	imes 10^{-6} F$ and $R = 10 M \Omega = 10 	imes 10^6 \Omega$.Calculating the time constant: $RC = (0.1 	imes 10^{-6}) 	imes (10 	imes 10^6) = 1 \ s$.Therefore, $t = 1 	imes 2.3026 	imes 0.3010 \approx 0.693 \ s$.

$A$ 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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