Question

An RC circuit has a 55.0 mu F capacitor connected to a 780 ohm resistor. a). When the capacitor is charged, how long does it take for the capacitor to reach 80% of full capacity

Answer 1

Answer:

0.069 s

Explanation:

The proportion of the full charge of a capacitor at any time, t, is given by

$1-e^{-t/\tau}$ where τ is the time constant given by the product of the resistance and capacitance. Hence,

$\tau = 55.0\times10^{-6}\text{ F}\times780\ \Omega = 0.0429 \text{ s}$

At 80%,

$80\% =0.8= 1-e^{-t/0.0429}$

$e^{-t/0.0429} = 0.2$

$-t/0.0429 = \ln 0.2 = -1.609$$t = 0.069\text{ s}$