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ggrree

  • 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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  1. ash2326
    • 4 years ago
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    Charge 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= \[\frac{Q_0-0.049 Q_0}{Q_0} \times 100\] \[=\frac{Q_0(0.951)}{Q_0} \times 100=95.1 \%\]

  2. Taufique
    • 4 years ago
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    Q=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=Qo-Q=0.951Qo hence percentage decrease=((decrease in charge)/Qo)*100=0.951*100=95.1%

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