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In figure i(1) = 10 e^(-2t) A, i(2) = 4 ...

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

The potential difference across inductor `V_(L)` is :

A

`8 e^(-2t) V`

B

`9 e^(-2t)V`

C

`16 e^(-2t)V`

D

`18 e^(-2t)V`

Text Solution

Verified by Experts

The correct Answer is:
C
`V_(L) = L (di_(L))/(dt)`, `(4) (d)/(dt) (4 - 2 e^(-2t)) = 16 e^(-2t) V`
`V_(L)` decreases exponentially from `16 A` to 0 as shown in figure
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