Current through wire XY of circuit shown is-

Options

A. `1 A`
B. `4 A`
C. `2 A`
D. `3 A`

Answer

C

Explanation

Loop ABCFA
`-I_(2) xx 1 - 0 xx I_(XY) + 3I_(1) =0`
`rArr I_(2) = 3I_(1)`…….. (i)
Loop CDEFC
`-2(I_(2) - I_(XY)) + 4(I_(1) + I_(XY)) + 0 xx I_(XY) = 0` …….. (ii)
`rArr 4I_(1) - 6I_(1) + 6I_(XY) = 0`
Loop AFEGHA
`-3I_(1) -4(I_(1) + I_(XY)) + 50 = 0`
`rArr -7I_(1) - 4I_(XY) + 50 = 0`..... (iii)
From equation (ii)
`rArr 4I_(1) - 6I_(1) + 6I_(XY) = 0` ....... (iv)
Now, by elimination method
`-14I_(1) - 8I_(XY) + 100 = 0 to (iii) xx 2`
`-14 I_(1) + 42 I_(XY) = 0 to (iv) xx 7`
`I_(XY) = 100/50 = 2A`