# Current in the $3\: ohm$ resistor is

$\begin {array} {1 1} (A)\;1A & \quad (B)\;\large\frac{3}{7}A \\ (C)\;\large\frac{5}{7}A & \quad (D)\;\large\frac{10}{7}A \end {array}$

Firstly, short all the grounds. Then apply KVL to the first loop and then to the
second loop.
$6-2 i_1 -6 i_2 =0$
$2(i_1 - i_2) - 6 i_2 =0$
solving we get the value of $i_2$ the current the 3 ohm resistor to be 3/7 oms
Ans : (B)
edited Sep 11, 2014 by meena.p