Question

The AC voltage in a resonant LC circuit (the resistance, R is considered to be zero) is given as V(t)=120 cos (80 t). Assume that L=0.1 H and C=0.001 F. The value of the magnetic energy stored in the inductor (UL) when the charge on the plates of the capacitor is one fourth of the maximum charge value that the plate can hold is what ?

Answer 1

Charge 'Q' across the capacitor is given by

Q=C x V

Where, C is capacitance

V is voltage across capacitor

Threefore

Q(t) = 0.001 x 120cos 80t

= 0.12cos80t

Maximum charge capacitor can hold is '0. 12C'

At this instant total energy of system is equal to energy stored in capacitor because no current will flow through inductor so no magnetic energy

Therefore

Total enrgy= Q²/2C

=0.12²/(2 x 0.001)

=7.2J

When charge is 1/4 of maximum charge on capacitor

Energy in capacitor=(Q/4) ²/2C

Total energy=energy in capacitor + energy in inductor

7.2 = (0.12/4) ²/(2x0.001) + magnetic enrgy in inductor

Magnetic energy in inductor=6.75 J