# Question 1.SGPYQ.2: A segment of a circuit is shown in the figure given below. v......

A segment of a circuit is shown in the figure given below. $v_{ R }=5 V , v_{ C }=4 \sin 2 t . \text { The voltage } v_{ L }$ is given by

(a) 3 – 8cos2t      (b) 32sin2t

(c) 16sin2t           (d) 16cos2t

Step-by-Step
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Applying Kirchhoff’s current law, sum of incoming currents is equal to sum of outgoing currents.

$i_{ L }=i_{ C }+1+2 \Rightarrow i_{ L }=i_{ C }+3$

We know that

$i_{ C }=-C \frac{d v_{ C }}{d t}=-1 \cdot \frac{d}{d t}(4 \sin 2 t)=-8 \cos 2 t$

Therefore, $i_{ L }=-8 \cos 2 t+3$

\begin{aligned} v_{ L } & =L \frac{d i_{ L }}{d t}=2 \times \frac{d}{d t}(-8 \cos 2 t+3) \\ & =2[8 \cdot \sin 2 t \cdot 2+0]=32 \sin 2 t \end{aligned}

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