In the given circuit, the current through 2 `Omega` resistor is

Options

A. `9 A`
B. `0 . 9 A`
C. `(1)/(9) A`
D. `(1)/(0.9)A`

Answer

B

Explanation

The electric circuit as shown below.
At the steady state ,the capacitoer behaves as open circuit. Therefore, simple circuit will be
Applying KVL in loop (I)
`- 6 + 3 (I_(1) - I_(2)) +2.8I_(1) = 0 `
`rArr 5.8I_(2) - 31_(2) = 6" ". . . (i)`
Applying KVL in loop (II)
`2I_(2) +3(I_(2)-I_(2)) = 0 " ". . . (ii)`
From Eqs. (i) and (ii), we get
`rArr 5 . 8 ((5)/(3)I_(2)) - 3I_(2) = 6`
`rArr 6 . 66I_(2) = 6 rArr I_(2) = 0 . 9 A`
Hence, current through `2 Omega` resistor is 0 . 9A .