Answer to Question #159585 in Electric Circuits for Yolande

Question

A charged capacitor is connected to a resistor and switch all in series. The circuit has a time constant of 1.50 s. Soon after the switch is closed, the charge of the capacitor is 75.0% of its initial charge.
(i) Find the time interval required for the capacitor to reach this charge.
(ii) If R = 250 Kiloohm, what is the value of C?

Answer 1

(i)

$$Q=Q_0e^{-t/\tau},$$ $$0.75Q_0=Q_0e^{-t/\tau}$$ $$ln(0.75)=ln(e^{-t/\tau}),$$ $$ln(0.75)=-\dfrac{t}{\tau},$$ $$t=-\tau ln(0.75)=-1.5\ s\cdot ln(0.75)=0.43\ s.$$

(ii) From the definition of the time constant, we have:

$$\tau=RC,$$ $$C=\dfrac{\tau}{R}=\dfrac{1.5\ s}{250\cdot10^3\ \Omega}=6.0\cdot10^6\ F=6.0\ \mu F.$$