For the circuit shown below, instantaneous current through inductor ' L ' and capacitor ' C ' is respectively. (a) (-e₀)/(ω L)cos ω t; e₀ω Ccos ω t (b) (-e₀)/(ω L)sin ω t; (e₀)/(ω C)cos ω t (c) (e₀ C)/(L)cos ω t; (e₀ L)/(C)sin ω t (d) (-e₀ C)/(L)sin ω t; (e₀ L)/(C)cos ω t

Answer: (a). Each element has the full source voltage e = e₀sin ω t. Inductor: iL = 1/L∫ e dt = -(e₀)/(ω L)cos ω t Capacitor: iC = Cde/dt = e₀ω Ccos ω t So the answer is (a).

For the circuit shown below, instantaneous current through inductor '… | MHT CET 2021 (22 Sep, Shift 2)

Q8MHT CET 2021 · 22 SEP · SHIFT 2Medium

Parallel AC Circuits

For the circuit shown below, instantaneous current through inductor ' $L$ ' and capacitor ' $C$ ' is respectively.

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Answer: (a)

Each element has the full source voltage $e = e_0\sin\omega t$ .
Inductor: $i_L = \frac{1}{L}\int e\,dt = -\frac{e_0}{\omega L}\cos\omega t$
Capacitor: $i_C = C\frac{de}{dt} = e_0\omega C\cos\omega t$
So the answer is (a).

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