Question

Making use of the passive sign convention, determine the current flowing through a 220 nF capacitor for t ≥ 0 if its voltage vC(t) =16.2e−9t V.

Answer 1

`i=-32076e^(-9t)\ nA`

Step-by-step explanation:

A capacitor is a device that stores electrical energy in an electric field. It is made up of two conductor plates separated by a dielectric. It stores electrical charges which then produces a potential difference between the plates.

The current flowing through a capacitor is given by:

`i=C(d)/(dt)v_c(t)\\ \\C=capacitance, i=current,v_c(t)=voltage\\\\given\ that\ C=220\ nF=220*10^(-9)F,v_c(t)=16.2e^(-9t)C\\\\i=220*10^(-9)F*(d)/(dt)(16.2e^(-9t)C) \\\\i=220*10^(-9)*-9*16.2e^(-9t)\\\\i=-32076e^(-9t)*10^(-9)\\\\i=-32076e^(-9t)\ nA`