Question

there in no current part of this circuit for time `t lt o`. Switch `S` is closed at `t = 0`.

The current through the inductor after a long time will be
A. zero
B. infinite
C. `(6)/(13) A`
D. None of these

Answer 1

Correct Answer - C
Initally there is no current in the inductor.
So in initially, `V_(A) - V_(B) = ((6)/(1 + 5)) xx 1 = 1 V` (i)
`V_(A) - V_(C) = ((6)/(2 + 4)) xx 2 = 2 V` (ii)
from (i) and (ii), `V_(B) - V_(C) = 2 - 1 = 1 V`
`rArr V_(B) - V_(C) = L(di)/(dt)`
`rArr 1 = 0.1 (di)/(dt) rArr (di)/(dt) = 10 As^(-1)`
Current through `6 W` resistor will remain constant because it is independently connected to `6 V`. After a long time, inductor will behave like a simple wire.
`1 Omega` and `2 Omega` are in parallel, their equivalent is `(2)/(3) Omega`
`5 Omega` and `4 Omega` are in parallel, their equivalent is `(20)/(9) Omega`.
`V_(1) = ((2)/(3) xx6)/((2)/(3) + (20)/(9)) = (18)/(13) V,V_(2) = V - V_(1) = 6 - (18)/(13) = (60)/(13)V`
`I_(1) = (V_(1))/(2) = (18)/(13 xx 2) = (9)/(13)A, I_(2) = (V_(2))/(4) = (60)/(13 xx 4) = (15)/(13) A`
Current through inductor: `I = I_(2) - I_(1) = (15)/(13) - (9)/(13) = (6)/(13)A`