In figure `i_(1) = 10 e^(-2t) A, i_(2) = 4 A, v_(C ) = 3 e^(-2t) V`

The current `i_(L)` is

Options

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

Answer

B

Explanation

In current in the capacitor,
`q = CV_(C ) = (2) (3 e^(-2t)) = 6 e^(-2t) A`
Current, `i_(c ) = (dq)/(dt) = - 12 e^(-2t) A`
Current flows from `B` to `O`.
From `KVL`, we have `i_(L) = i_(1) + i_(2) + i_(c )`
`10 e^(-2t) + 4 - 12 e^(-2t) = (4 - 2 e^(-2t)) A`
`= [2 + 2 (1 - e^(-2t))] A`