Supernodes & voltage sources
An ideal voltage source between two unknown nodes fixes their voltage difference but does not provide a resistor-like branch-current equation. Enclose the source in a supernode, write one boundary KCL equation, then add the source-voltage constraint.
- Recognize why direct resistor-style nodal substitution fails for an ideal voltage source between two unknown nodes.
- Construct the supernode boundary and write Kirchhoff's current law using only currents crossing that boundary.
- Write the additional voltage constraint from the source polarity.
- Solve the supernode KCL equation together with the voltage-source constraint.
- Recover the current through the floating voltage source after solving the node voltages.
- Distinguish a floating voltage-source supernode from the simpler case of a voltage source tied directly to the reference node.
What should nodal analysis do with an ideal voltage source placed directly between two unknown nodes?
A resistor gives a branch current from the node-voltage difference. An ideal voltage source does something different: it fixes its voltage while its current is whatever the surrounding network requires. We therefore cannot invent a fictitious resistance. Instead, we combine the two source-connected nodes into a supernode, write one current balance around its external boundary, and add the source-voltage constraint.
03.1Why the ordinary resistor substitution stops working
Figure 3.1 — An ideal voltage source floats between two unknown nodes
VX connects VA directly to VB without touching ground. It fixes a voltage difference between the nodes, but its current is not given by a resistor-like ΔV/R relation.
The worked network contains a known 8 V input node, two unknown nodes VA and VB, three resistors, and a floating ideal 2 V source VX between VA and VB. For each resistor, current is available immediately from a voltage difference divided by resistance.
For VX, however, the element law is
It does not supply an equation of the form IX=(VA−VB)/R because an ideal voltage source has no finite internal resistance in this model. Its current is an additional internal quantity determined by the rest of the network.
03.2The supernode is an analysis boundary, not a new component
Imagine enclosing VA, VB and the ideal source VX inside one boundary. Currents through VX stay inside that boundary. The only currents that cross the boundary are those through R1, R2 and R3.
This can be derived rather than memorized. Let IX be positive from VA to VB. KCL at VA is
KCL at VB is
Add the two equations. The internal source current cancels:
That sum is the supernode KCL equation. The cancellation is the reason the method works: internal current need not be known to balance current crossing the external boundary.
03.3The source polarity supplies the missing independent equation
Figure 3.2 — Source polarity supplies the constraint equation
VX has its positive terminal at VA and negative terminal at VB. The voltage arrow therefore runs from negative reference VB toward positive VA, following ADR-0018.
One supernode KCL equation cannot determine both VA and VB. The second independent equation comes from the source itself. Figure 3.2 places the positive terminal at VA and the negative terminal at VB, so
For VX=2 V:
If the source polarity were reversed, the constraint would reverse sign. The supernode method does not remove sign discipline; it makes the source polarity an explicit algebraic constraint.
03.4Complete worked example: solve KCL plus the voltage constraint
Use R1=2 kΩ, R2=4 kΩ, R3=2 kΩ, VS=8 V and VX=2 V. The supernode KCL equation is
Multiply by 4 kΩ:
The source constraint is
Hence VB=VA−2 V. Substitute:
Both node voltages now follow without ever assigning an artificial resistance to the ideal source.
03.5Recover the current through the floating source after solving the nodes
Figure 3.3 — Recover the internal source current after solving the voltages
Once VA and VB are known, return to either individual node and use KCL to recover IX through the floating source from VA toward VB.
The supernode equation intentionally eliminated IX. If that current is required, recover it afterward from an individual-node KCL equation.
At VA, with IX defined from VA to VB:
Check node B independently:
The recovered source current is therefore 1 mA from VA to VB. Both individual KCL equations are now satisfied.
03.6Power balance checks the source signs as well as the resistor currents
The resistor powers are
IX=1 mA enters the positive terminal of the 2 V floating source, so the passive-sign convention gives
The grounded 8 V source supplies 2 mA from its positive terminal to R1, so its absorbed power is
Total absorbed power is 8+4+2+2=16 mW, exactly equal to the 16 mW supplied by VS. The floating voltage source is absorbing power in this operating point; an ideal voltage source is not inherently a power source—it can absorb or deliver depending on current direction.
03.7A voltage source connected to the reference node is the simpler special case
Suppose an ideal 5 V source connects directly between an unknown node and the reference node, with its positive terminal at the unknown node. Then the node voltage is known immediately:
If the source polarity is reversed, the same node is at −5 V relative to the reference. No supernode is needed because the source itself has removed that node voltage from the list of unknowns.
The supernode becomes necessary when the ideal voltage source constrains a difference between unknown node voltages rather than directly fixing one of them relative to the reference.
03.8Dependent voltage sources use the same structural idea
If a dependent ideal voltage source lies between two unknown nodes, the current through the source is still not given by a resistor law. The supernode KCL construction therefore remains useful.
What changes is the constraint. For example, a voltage-controlled source might impose
where μ is a dimensionless gain and Vx is a controlling voltage defined elsewhere in the circuit. The controlling relation must remain in the system; dependent sources are not deactivated or replaced arbitrarily.
03.9Common supernode errors and a robust workflow
Reliable sequence: identify the floating ideal voltage source → enclose its terminal nodes and source → write KCL using currents crossing the boundary → write the source constraint from polarity → solve → recover any internal source current afterward → check local KCL and power.
03.10Practice for mastery
- Repeat the worked network with VX=4 V while all other values remain unchanged. Find VA and VB.
- Keep VX=2 V but reverse its polarity. Find VA and VB.
- For the original worked network, calculate IX using node B rather than node A.
- An ideal 3 V source connects from ground (negative terminal) to node C (positive terminal). Is a supernode needed? What is VC?
- A student writes the supernode KCL and also includes +IX as a current leaving the supernode. Explain why this double-counts an internal current.
- A dependent source imposes VA−VB=2VC. What extra information is required before the nodal system can be solved?
Solutions
1. KCL remains 3VA+2VB=16 V, while VA−VB=4 V. Thus VB=0.8 V and VA=4.8 V.
2. Reversed polarity gives VA−VB=−2 V. Together with 3VA+2VB=16 V, this yields VA=2.4 V and VB=4.4 V.
3. At B, 2 V/2 kΩ−IX=0, so IX=1 mA from A to B.
4. No supernode is required. The source directly fixes VC=+3 V relative to ground.
5. IX flows from one point inside the chosen supernode boundary to another point inside it. It does not cross the boundary and cancels when the two individual KCL equations are added.
6. VC must be defined by the rest of the circuit—either as another node voltage or as a specified voltage difference. The dependent-source constraint must be accompanied by all equations needed to determine its controlling variable.
The next chapter changes coordinates: instead of choosing node voltages and using KCL, mesh analysis chooses loop currents in planar circuits and organizes Kirchhoff's voltage law.