A 14.0 F capacitor is charged to 190 C and then connected across the ends of a 4

ID: 1428747 • Letter: A

Question

A 14.0 F capacitor is charged to 190 C and then connected across the ends of a 4.80 mH inductor.

Part A

Find the maximum current in the inductor.

Part B

At the instant the current in the inductor is maximal, how much charge is on the capacitor plates?

Part C

Find the maximum potential across the capacitor.

Part D

At the instant the potential across the capacitor is maximum, what is the current in the inductor?

Part E

Find the maximum energy stored in the inductor.

Part F

At the instant the energy stored in the inductor is maximum, what is the current in the circuit?

imax = A

PART A ) Q^2/(2*C) = (L*Imax^2)/2

Imax = Q/(C*L)^0.5 = 190*10^-6/(14*10^-6*4.80*10^-3)^0.5 = 0.7335 A

PART B) q = 0 micro coulomb

PART C) Vmax = Q/C = 190*10^-6/(14*10^-6) = 13.57 v

PART D) I = 0 A

PART E) Emax = (L*Imax^2)/2 = 4.80*10^-3*(0.7335)^2/2 = 0.00129 J

PART F) i = Imax = 0.7335 A

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