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

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}