Answer:
[tex]i=-66.6\times10^{-6}\;\;Amp[/tex]
Explanation:
Given,
[tex]C=3\mu F\\V=6.0 volt\\R=4\Omega\\t=14\mu sec\\[/tex]
Now, charge in capacitor
[tex]q_o=CV\\q_o=3\times10^{-6}\times6\\q_o=18\times10^{-6}\;\;C[/tex]
For discharging the RC circuit,
[tex]q=q_oe^{-t/\tau}[/tex]
Differentiate with respect to the 't'
[tex]\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[/tex]
