Passage II
In fig., `i_(1)=10e^(-2t)A,i_(2)=4A`, and `V_(C)=3e^(-2t)V`.

The current `i_(L)` is

Options

A. `[2-2(1-e^(-2t))]`
B. `[2+2(1-e^(-2t))]A`
C. `[3-2(1-e^(-2t))]A`
D. `[2+3(1-e^(-2t))]A`

Answer

B

Explanation

Instantaneous current in the capacitor,
`q="CV"_(C)=(2)(3e^(-2t))=6e^(-2t)C," Current",i_(C)=("dq")/("dt")=-12e^(-2t)A`
This current flows from B to O