Question

For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___

(a) cos Ωt

(b) sin Ωt

(c) tan Ωt

(d) cos Ωt.sin Ωt

This question was addressed to me during an online exam.

The query is from Diode Circuits-2 topic in section Diode Circuits of Power Electronics

Answer 1

Correct choice is (b) sin Ωt

The best I can explain: When switch is closed,

Ldi/dt + 1/C ∫ idt  =  Vs

Laplace of the above gives,

                                L[sI(s)] + 1/C [I(s)/s] = Vs/s

                                       I(s) = Vs/(LΩ) * (Ω/Ω^2 + s^2)

Taking the inverse lapace gives,

                                       I(t) = Vs * √ (L/C) * sin Ωt.