In the given at `t = 0`, switch `S` is closed.

The current through the `10 Omega` resistor at any instant `t(0 lt t lt oo)` will be

Options

A. `(1)/(6)e^(-(1000//3)t)`
B. `(5)/(6)e^(-(1000//3)t)`
C. `(1)/(6)e^((1000//3)t)`
D. `(6)/(5)e^((1000//3)t)`

Answer

B

Explanation

From loop, applying kirchhoff's law,

`12I - 10 I' = 10`
From loop (ii) (i)
`-L (dI')/(dt) + 10 I - 10 I' = 0`
`I - I' = (L)/(10) (dI')/(dt)` (ii)
Solving simultaneously (i) and (ii), we have
`I' = 5 - 5e^(-(5t)/(3L)` (iii)
and `I = 5 - (25)/(6) e^(-(5t)/(3L)` (iv)
`I - I' = (5)/(6)e^(-(1000t)/(3)`
`E_(L) = (1)/(2) L(I')^(2), E_(L) = (125)/(2) (1 - e^(-(1000t)/(L)))^(2) mJ`
Current in the inductor at `t = oo, I' = 5 A`
`E_(L)(t rarr oo)= (1)/(2) xx 5xx 10^(-3)(5)^(2)`
`E_(L) = 62.5 mJ`
`E_(C) (t rarr oo) = (1)/(2) CV^(2) = (1)/(2) xx 20xx 10^(-6) xx 100 = 1mJ`