Supernodes

The Problem

This is for dealing with a voltage source connecting two unknown node voltages (example: $2V$ source in schematic below). Since a voltage source may have any current flowing through it, there is no device equation to plug into the KCL equations.

The unknown current through the voltage source can be written as:

$$\begin{array}{rl}\text{Node A:}& I_{2V}=\frac{V_A-10}{4}+\frac{V_A}{2}+\frac{V_A-V_B}{1}\\ \text{Node B:}& 0=I_{2V}+\frac{V_B}{3}+\frac{V_B-V_A}{1}\end{array}$$

However, these currents are coupled by $I_{2V}$, which means it can be eliminated.

Supernode Usage

We can skip the step of defining the unknown current by taking a broader interpretation of KCL: The sum of currents into and out of any boundary in a circuit must equal zero. We can define a supernode that encompasses both nodes A and B. There is now only one nodal equation, with the 2 V source supplying the missing equation.

$$\begin{array}{rl}\text{Supernode:}& 0=\frac{V_A-10}{4}+\frac{V_A}{2}+\frac{V_A-V_B}{1}+\frac{V_B}{3}+\frac{V_B-V_A}{1}\\ \text{Voltage Source:}& V_B-V_A=2\end{array}$$

The currents entering the $1\Omega$ resistor cancel each other, so the supernode can also be expanded to include the resistor and eliminate the redundant currents:

$$\begin{array}{rl}\text{Supernode: }& 0=\frac{V_A-10}{4}+\frac{V_A}{2}+\frac{V_B}{3}\\ \text{Voltage Source:}& V_B-V_A=2\end{array}$$