Question

A 3.0 µF capacitor is initially connected to a 6.0 V battery. Once the capacitor is fully charged the battery is removed and a 4.0 Ω resistor is connected between the two terminals of the capacitor. Find the current flowing through the resistor 14.0 µsec after the capacitor begins to be discharged.

Answer 1

Answer:

$i=-66.6\times10^{-6}\;\;Amp$

Explanation:

Given,

$C=3\mu F\\V=6.0 volt\\R=4\Omega\\t=14\mu sec$

Now, charge in capacitor

$q_o=CV\\q_o=3\times10^{-6}\times6\\q_o=18\times10^{-6}\;\;C$

For discharging the RC circuit,

$q=q_oe^{-t/\tau}$

Differentiate with respect to the 't'

$\frac{dq}{dt}=-q_o\frac{t}{RC}e^{t/RC}\\i=-q_o\frac{t}{RC}e^{t/RC}\\i=-18\times10^{-6}\frac{14\times10^{-6}}{4\times3\times10^{-6}}e^{\frac{14\times10^{-6}}{4\times3\times10^{-6}}}\\i=-66.6\times10^{-6}\;\;Amp$