Question

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?

Answer 1

Answer:

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:

$Q=Q_{max}[1-e^{-\frac{t}{RC}}]$         (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:

$Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC$

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