Respuesta :
Given :
Capacitor , C = 55 μF .
Energy is given by :
[tex]\omega(t)=10cos^2 (377t)\ J[/tex] .
To Find :
The current through the capacitor.
Solution :
Energy in capacitor is given by :
[tex]\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)[/tex]
Now , current i is given by :
[tex]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)[/tex]
( differentiation of cos x is - sin x )
Therefore , the current through the capacitor is -11.18 sin ( 377t).
Hence , this is the required solution .
