What is the current in a circuit?

Question

Cons‎i‎der an LC circ‎uit with L = 50.0 [mH] and C = 5.00 [μF]. The ma‎ximum c‎rrent in t‎he induc‎tor is 0.60 [A]. What is the cu‎rr‎ent in the c‎ir‎cuit if the ca‎p‎acitor carri‎es a t‎‎hird of its maxim‎‎um cha‎rge?

Answer 1

For a charged capacitor connected to an inductor in a circuit (with voltage or current source absent), the current I and capacitor charge Q will oscillate when the circuit is closed. For circuit resistance R at 0, no energy is dissipated or "bled off" as heat in Joules and oscillations will continue.

Disregard any circuit resistance here and assume initial charge Q_max on the capacitor with circuit closed at t = 0.

In terms of energy, at full capacitor charge, total energy U in the circuit is stored in the electric field of the capacitor and is given as Q_max²/2C. At t = 0, current I equals 0 so there is no energy in the inductor.

As capacitor discharge begins, energy in the capacitor electric field decreases. At the same time, the current increases and some energy is stored in the magnetic field of the inductor. Then energy passes from the electric field of the capacitor to the magnetic field of the inductor.

At full capacitor discharge, the capacitor holds no energy and the current is maximum with all energy in the inductor.

The process then repeats in the reverse direction; energy will pass between inductor and capacitor indefinitely, tied to oscillations in the current and the charge.

While potential energy in the capacitor is Q_max²/2C, kinetic energy in the inductor is given by 0.5LI²; this inductor energy requires the presence of moving charges.

For T the time period of a complete oscillation cycle (capacitor charge goes from maximum to zero and then back to maximum), all energy is potential and in the capacitor at t = 0 & t = 0.5T. At t = 0.25T & t = 0.75T, all energy is kinetic and in the inductor and equal to 0.5LI².  

Then, with values given above, construct Q_max²/2C = 0.5LI_max².

Place values given in the problem statement to obtain:

Q_max²/2(5E-6 Farads) = 0.5(50E-3 Henries)(0.6 Ampères)², which gives Q_max as 3E-4 Coulombs.

Finally, for a third of the maximum charge, or 1E-4 Coulombs, write (0.0001)² [Coulombs]²/2(5E-6 Farads)

equals 0.5(50E-3 Henries)I² and gain I (at 1/3 the maximum charge) equal to 0.2 Ampères.