Current through wire $X Y$ of circuit shown is

1 A

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4 A

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2 A

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3 A

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Solution

The correct option is B $2 A$
We know that potential difference in a loop equals to zero,

$L o o p 1$
$- i_{1} + 0 \times i_{x y} + 3 i_{2} = 0$ i.e. $i_{1} = 3 i_{2}$ .....(i)

$L o o p 2$
Also $- 2 (i_{1} - i_{x y} + 4 (i_{2} + i_{x y}) = 0$
i.e. $2 i_{1} - 4 i_{2} = 6 i_{x y}$ ....(ii)

$L o o p 3$
Also $V_{A B} - 1 \times i_{1} - 2 (i_{1} - i_{x y} = 0 \Rightarrow 50 = i_{1} + 2 (i_{1} - i_{x y})$
$= 3 i_{1} - 2 i_{x y}$ ....(iii)

Solving, (i), (ii) and (iii), $i_{x y} = 2 A$

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