Question

A 55-μF capacitor has energy ω (t) = 10 cos2 377t J and consider a positive v(t). Determine the current through the capacitor.

Answer 1

Given :

Capacitor , C = 55 μF .

Energy is given by :

`\omega(t)=10cos^2 (377t)\ J` .

To Find :

The current through the capacitor.

Solution :

Energy in capacitor is given by :

`\omega=(Cv^2)/(2)\\\\v=\sqrt{(2\omega)/(C)}\\\\v=\sqrt{(2* 10cos^2 (377t))/(55* 10^(-6))}\\\\v=cos(337t)\sqrt{(2* 10)/(55* 10^(-6))}\\\\v=603.02\ cos( 337t)`

Now , current i is given by :

`i=C(dv)/(dt)\\\\i=C(d[603.02cos(337t)])/(dt)\\\\i=-55* 10^(-6)* 603.03* 337* sin(337t)\\\\i=-11.18\ sin(337t)`

( differentiation of cos x is - sin x )

Therefore , the current through the capacitor is -11.18 sin ( 377t).

Hence , this is the required solution .