# Nodal analysis - what to do when a supernode spans entire circuit

I'm back again with a nodal analysis question. I'm getting better, but I've come to a problem again.

Consider the following circuit with the bottom node as ground: I specifically don't know how to deal with the two lower left branches. When I apply KVL I get equations of the form v1 = v1 and v2 = v2, which is not helpful. One of them is also a dependent source, so I can't just use source transformations as far as I know.

I know that this problem ought to be a lot easier with mesh current analysis. Unfortunately, I'm doing this problem for educational purposes and we're expected to be able to solve this problem with nodal analysis too.

• How are you using KVL? Can you give a direct example of how you're getting these "V1 = V1" equations? Why do you say that the supernode spans the entire circuit? Apr 7, 2015 at 11:02

You don't need to do anything weird to handle the lower left branches. Treat the dependent voltage source the same as you would an independent source. If $V_1$ is the node between the 10 and 20 ohm resistors, $V_2$ is the node between the 40 ohm resistor and the 120 V source, and $V_3$ is the node between the dependent current source and the 80 ohm resistor:

$$\frac{V_1 - 100\ \mathrm V}{10\ \Omega} + \frac{V_1 - 4V_O}{20\ \Omega} + \frac{V_1 - V_2}{40\ \Omega} = 0$$

You've introduced a new variable ($V_O$), so you need another equation to solve the circuit. This equation should define $V_O$ in terms of the node voltages. In this case, $V_O$ is simply equal to one of the node voltages:

$$V_O = V_3$$

You can use a similar technique for the dependent current source.

Let X be node voltage at junction of 10/20/40 resistors, then

(100-X)/10 + (4Vo-X)/20 = [X-(Vo-120)]/40 = Io