Question

For the circuit given below, evaluate the current (I) passing through $2\Omega$.

Answer 1

According to the kirchhoff’s voltage law (KVL) - the sum of all voltages around a closed loop in any circuit must be equal to zero.
Let us assume the direction of current (I) is clockwise.

Using Kirchoff's law's sign conventions as shown below:
From loop CABDC
$-4-(I\times 2)-6-(I\times 4)=0$
$-4-(I\times 2)-6-(I\times 4)=0$
$-10-\left(6I\right)=0$
$6I=-10$
$⟹$$I=\frac{-10}{6}$
$⟹$$I=\frac{-5}{3}$
$⟹$$I=-1.66A$
As the direction current comes out to be negative.So, the direction of current will be opposite to the assumed direction.
Hence, 1.66 A current will flow in the $2\Omega$ resistance from B to A.