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anonymous
 4 years ago
The time constant of an RC circuit is tau = RC. At time t = 3 tau, the charge on the discharging capacitor in an RC circuit will have decreased to approximately what percentage of its initial value?
anonymous
 4 years ago
The time constant of an RC circuit is tau = RC. At time t = 3 tau, the charge on the discharging capacitor in an RC circuit will have decreased to approximately what percentage of its initial value?

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ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1Charge of a discharging capacitor is given as \[\huge Q= Q_{0} e^{\frac{t}{\tau}}\] Q0 is the initial charge on the capacitor \[\tau\ is\ the\ time\ constant= RC\] at t= 3\(\tau\) \[\huge Q= Q_{0} e^{\frac{3 \tau}{\tau}}\] \[\huge Q= Q_{0} e^{3}=0.049 Q_0\] Initial Charge was \(Q_0\) So Percentage Decrease= \[\frac{Q_00.049 Q_0}{Q_0} \times 100\] \[=\frac{Q_0(0.951)}{Q_0} \times 100=95.1 \%\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Q=Qo*e^(t/(time constant)) put the value of each and get... Q=Qo*e^3 Q=Qo/e^3 Q=0.049Qo decrease in charge=QoQ=0.951Qo hence percentage decrease=((decrease in charge)/Qo)*100=0.951*100=95.1%
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