Voltage at the node where R2 is connected is Vo, consider voltage at other node is \$V_1\$ By KCL:

$$\frac{V_o}{R_2} + \frac{V_o-V_1}{R_1} + \frac{V_o-V_2-2V_o}{R_3} = 0$$

To find \$V_2\$ here it is simply \$R_4 \cdot I_1 = 48V\$.

Now put the values and you will get your answer.

However you can do same using the loop analysis and KVL.

The thing is keep the dependent source variable as it is and just solve as you normally do.

The currents split up as shown in the following schematic:

^{simulate this circuit – Schematic created using CircuitLab}

You could easily combine \$R_1\$ and \$R_2\$ into a single effective value and work out the voltage at their shared top node... and from that work out their currents. But you can also just as well see that \$R_1\$ is 3 times less conductive than \$R_2\$, so this means that if one part of the current flows through \$R_1\$ then it must be that three more parts of the current must flow through \$R_2\$. From this argument, you should consider dividing the total current of 12 A into four parts of 3 A each. And thus, \$i_1=3\:\textrm{A}\$ and \$i_2=9\:\textrm{A}\$.

Similar logic then applies to the other two resistors. Here, it's not hard to imagine a total of 7 parts of the -2 A net total current in \$R_3\$ and \$R_4\$, where now three parts go through \$R_3\$ and four parts through \$R_4\$. Each part is \$-\tfrac{2}{7}\:\textrm{A}\$, so this means \$i_3=-\tfrac{6}{7}\:\textrm{A}\$ and \$i_4=-\tfrac{8}{7}\:\textrm{A}\$.

Looking at the bottom node, we know the following must be true:

$$ -8\:\textrm{A} - 2\:\textrm{A} + 3\:\textrm{A}+9\:\textrm{A}+-\tfrac{6}{7}\:\textrm{A}+-\tfrac{8}{7}\:\textrm{A}=0$$

And it is.

## Best Answer

I assume the controlled source is a CCCS.

The left most clockwise mesh current, \$i_a\$ is by inspection \$i_a = 1A\$.

Clearly, the right most clockwise mesh current is just \$i_b = i_x\$. But \$i_b\$ cannot be found by KVL (as you've pointed out) so we need an auxiliary constraint to replace the KVL equation.

That constraint is given by the controlled source:

$$i_a - i_b = 3i_x = 3i_b \Rightarrow i_b = \frac{i_a}{4} = 0.25A = i_x$$

Remarkably, the solution does not depend on the resistor values.