$\begin {array} {1 1} (A)\;3A & \quad (B)\;3.5A \\ (C)\;4.5A & \quad (D)\;Zero \end {array}$

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Assume the values of current in the main branch as well as in the loop.

Write equations of KCL and KVL for the loop and the branch.

Let the current in the 2 ohm resistor be $i_1 + i_2$

then using KVL equation in closed loop

$ 9 - i_2 - 2(i_1 +i_2 ) = 0 $ ........................(1)

$4 i_1 +2 (i_1 +i_2 ) -3 +4 i_1 = 16 $ ................(2)

solving we get $i_1 =1.5$ and $i_2 = 2$

Thus current through 2 ohm resistor is 3.5 ohms

Ans : (B)

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