Question

A 17.0-μF capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280-mH inductor.

a) Calculate the energy stored in the inductor at t = 1.30 ms. (Express your answer to three significant figures and include the appropriate units.)

Answer 1

We know that the change of the charge on the Capacitor

q=Q cos(wt)

On differentiating

I=dq/dt =-Qw sin(wt). ...1

here Q=CV =(17F )(150V) =2550C =2.55mC =2.55×10^(-3)C, t=1.3×10^(-3)s=0.0013s

Put these values in equation 1

I=-(2.55×10^(-3)×14494.2759)sin(14494.2759×0.0013)

I=0.25861913 A

So

the energy store in the inductor

U=LI^2/2

U=(0.28×10^(-3)H)(0.25861913A)^2/2

U=0.0093637396×10^(-3) J

U=9.36×10^(-6) J

U=9.36 J