A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zero charge. How long after the switch is closed will it take the capacitor to reach 2/3 its maximum charge?

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

        (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax    

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC