Figure: Find the given current through 1$\omega$ resistor.

Solution:

Using nodal analysis

$\because$ Mothod is not specified

KCL at node (A)

$\begin{aligned} I_x+3 I_x &=a+b \\\\ 4 I_x &=a+b \\\\ 4\left[\frac{12-V_A}{6}\right] &=\frac{V_A}{3}+\frac{V_A}{1} \\\\ 48-4 V_A &=2 V_A+6 V_A \\\\ 48 &=12 V_A \\\\ V_A &=4 \text { Volts } \\ \end{aligned} \\ $

$\therefore$ Current through $1 \Omega$ resistor is,

$\begin{aligned} \\ b=\frac{V_A}{1} &=4 \text { Amp } \\\\ b &=4 \text { Amp } \\ \end{aligned} \\ $