Please show me step by step. The answers are correct!!! VA Ch32 x C e switch in

Question

Please show me step by step. The answers are correct!!!

VA Ch32 x C e switch in The Figure x C ponses/view key?dep 15617/86 16.4 45.5 mA (d) the total energy the circuit possesses at t 3.00 s 107 108 Need Help? Read it l watch it L -3 points ser-SE9 32P054. My Notes o Ask Your Teacher An LC circuit like that in the figure below consists of a 3.30-H inductor and an 831-pF capacitor that initially carries a 108-uc charge. The switch is open for t 0 and is then thrown closed at t 0. Compute the following quantities a t ms. (a) the energy stored in the capacitor No Response) 1.5/ (b) e lolal energy in lhe circuil No Response)! 6.99 (c) lhe energy stored in the inductor Re Need Help? Read it B20 PM a Type here to search 4/17/2017

Explanation / Answer

here,

inductor, L = 3.30 H

capacitnace, C = 834*10^-12 F

charge, qo = 108*10^-6 C

resonancee frequency, wo = 1/sqrt(L*C)
resonancee frequency, wo = 1/sqrt(3.30*834*10^-12)
resonancee frequency, wo = 19061.63 Hz

initial potential, vo = qo / C
initial potential, vo = 108*10^-6 / (834*10^-12)
initial potential, vo = 129496.403 V

Total energy in cicuit, E = 0.5 * C * vo^2
Total energy in cicuit, E = 0.5 * (834*10^-12) * (129496.403)^2
Total energy in cicuit, E = 6.993 J

At t = 5 ms = 0.005s
wt = 19061.63 * 0.005
wt = 95.308 rad
wt = 95.308 * 180/pi ( converting to degreess)
wt = 15 * 360 + 60.746 degrees

There is no energy or power consumed since there is no resistance.

Vc(t) = Vo*cost
Vc(t) = 129496.403*cos(60.746)
Vc(t) = 63282.581 V

Energy stored in capacitor, Ec = 0.5 * C* V^2
Energy stored in capacitor, Ec = 0.5*(834*10^-12)*(63282.581)^2
Energy stored in capacitor, Ec = 1.67 J

Energy stored in inductor, El = E - Ec
Energy stored in inductor, El = 6.993 - 1.67
Energy stored in inductor, El = 5.32 J

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