## 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?

1. ash2326

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. anonymous

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%