Question

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?

Answer 1

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= \[\frac{Q_0-0.049 Q_0}{Q_0} \times 100\]
\[=\frac{Q_0(0.951)}{Q_0} \times 100=95.1 \%\]

Answer 2

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%