Question

An LC circuit like that in the figure below consists of a 3.10-H inductor and an 890-pF capacitor that initially carries a 105-µC charge. The switch is open for t < 0 and is then thrown closed at t = 0. Compute the following quantities at t = 2.00 ms.

(a) the energy stored in the capacitor

(b) the energy stored in the inductor

(c) the total energy in the circuit

Answer 1

Vc(0) = Q(0)/C = 105e-6/890e-12 = 117,978 V

(b) Ec(0) = ½CV² = ½QV = ½*105e-6*117,978 = 6.19 J (answer)

ω = 1/√(LC) = 19,038 rad/s

At t = 2ms = 0.002s

ωt = 19,038*0.002 = 38.08 rad

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

Vc(t) = 117,978*cosωt = 117,978*cos(38.08) = 109,687 V

½CV² = ½*890e-12*109,687² = 5.35 J at t = 0.002s <----- (a)

6.19 – 5.35 = 0.836 J <------ (c)