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 :

$\omega=\dfrac{Cv^2}{2}\\\\v=\sqrt{\dfrac{2\omega}{C}}\\\\v=\sqrt{\dfrac{2\times 10cos^2 (377t)}{55\times 10^{-6}}}\\\\v=cos(337t)\sqrt{\dfrac{2\times 10}{55\times 10^{-6}}}\\\\v=603.02\ cos( 337t)$

Now , current i is given by :

$i=C\dfrac{dv}{dt}\\\\i=C\dfrac{d[603.02cos(337t)]}{dt}\\\\i=-55\times 10^{-6}\times 603.03\times 337\times sin(337t)\\\\i=-11.18\ sin(337t)$

( differentiation of cos x is - sin x )

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

Hence , this is the required solution .