Multi-node systems & matrix form
With several unknown node voltages, write one Kirchhoff-current-law equation per node, solve the coupled system, then recognize the conductance matrix hidden inside the equations.
- Construct coupled nodal equations for two or more unknown nodes.
- Keep mutual-branch current signs consistent in both node equations.
- Solve a two-node system by elimination before introducing matrix notation.
- Read diagonal and off-diagonal conductance coefficients from circuit topology.
- Write the nodal system compactly as Gv=i without treating the matrix as a new physical law.
- Verify the solved node voltages by branch-current balances and power conservation.
What changes when a circuit has two unknown node voltages instead of one?
The physics does not change. We still write resistor currents from voltage differences and apply Kirchhoff's current law (KCL). The only new feature is coupling: a resistor joining two unknown nodes makes each node equation contain both voltages. A linear system appears naturally, and matrix notation is only a compact way to preserve that structure.
02.1Count unknown node voltages before writing equations
Figure 2.1 — Two unknown node voltages require two coupled equations
The source fixes VIN at 10 V and ground fixes the 0 V reference. VA and VOUT remain unknown. All four resistors are 1 kΩ so attention stays on equation structure rather than arithmetic.
The circuit has four electrical nodes: ground, the 10 V input node fixed by the ideal source, node A, and node B (labelled VOUT in the schematic). Ground and the input node are already known. Therefore only
are unknown. Two unknowns require two independent equations. Nodal analysis obtains them by applying KCL at A and B.
This unknown-count step should happen before algebra. It prevents unnecessary branch-current unknowns and reveals immediately whether the number of equations you later write is plausible.
02.2Write the equation at node A without guessing current directions
Define every resistor current as leaving node A. Three resistor branches touch A:
KCL gives
With R1=R2=R3=1 kΩ, multiply by 1 kΩ and collect terms:
The negative coefficient of VB is not a memorized matrix rule yet. It came directly from expanding the shared-branch current VA−VB.
02.3The same mutual branch appears with the opposite sign at node B
Figure 2.2 — The mutual branch couples the two node equations
The R3 reference current is defined from VA toward VOUT, so it is (VA−VB)/R3. In the second node equation, current leaving through that same branch is the opposite expression: (VB−VA)/R3.
At node B, again define every branch current as leaving B. The resistor to A now contributes
and the resistor to ground contributes VB/R4. Therefore
With R3=R4=1 kΩ:
The same physical resistor R3 contributes −VB to the A equation and −VA to the B equation after terms are collected. This reciprocal coupling will become the pair of off-diagonal matrix entries.
02.4Solve the coupled equations by elimination first
Before using matrices, solve the two equations exactly as simultaneous linear equations:
The second equation gives
Substitute into the first:
The arithmetic is intentionally simple. The important skill is that each equation came from one node and every shared-branch term was written from the local node outward.
02.5Now expose the conductance structure
Define Gk=1/Rk. The two general KCL equations can be rearranged as
Each diagonal coefficient is the sum of conductances incident on that unknown node. Each off-diagonal coefficient is minus the conductance directly connecting the two unknown nodes.
The right-hand side contains contributions from known-voltage neighbors and, in more general circuits, known current injections. Nothing new has been added to KCL; the terms have only been grouped by unknown voltage.
02.6Matrix notation is compressed bookkeeping: Gv = i
This is conventionally summarized as
where G is the nodal conductance matrix, v is the vector of unknown node voltages, and i is the vector of known current-equivalent excitations.
Units provide an immediate check:
For this example every resistor is 1 kΩ, so every conductance is 1 mS:
Substituting [4 V, 2 V] gives [10 mA, 0], exactly as required.
02.7Recover branch currents and check both nodes independently
With VA=4 V and VB=2 V:
At node A, 6 mA enters from the source side while 4 mA+2 mA leaves. At node B, 2 mA enters from A and 2 mA leaves to ground. Both KCL balances close independently.
This is an important diagnostic advantage of nodal analysis: after solving, each row can be rechecked locally at its physical node.
02.8Power gives a global check on the whole solution
The resistors absorb 60 mW in total. The 10 V source supplies 6 mA:
Local KCL checks and global power balance test different aspects of the solution. Agreement among both is much stronger evidence than repeating the same algebra twice.
02.9Topology is visible inside the matrix
In a passive reciprocal resistor network, two unknown nodes that are not directly connected by a resistor do not create a mutual conductance term. Their corresponding off-diagonal entry is zero. Large circuit matrices are therefore often sparse: most node pairs do not directly interact.
For a resistor joining nodes A and B with conductance G, its contribution is
A resistor from an unknown node to a known fixed-voltage node contributes its conductance to the unknown node's diagonal and moves the known-voltage product to the right-hand side.
This pattern is the conceptual bridge from hand analysis to automated circuit solvers: software assembles equations from connectivity and element laws instead of visually “solving the drawing.”
02.10Singular systems, mistakes and mastery practice
If an entire resistor subnetwork is floating with no voltage reference, its absolute potential is undetermined. Only differences inside the floating network may be fixed. Algebraically, the nodal matrix becomes singular. This is not merely a numerical inconvenience: it reports missing physical information about the voltage origin of that connected component.
- Repeat the all-equal-resistor circuit with a 15 V source. Find VA and VB without rebuilding the entire elimination from scratch.
- All four resistors are 2 kΩ and the source is 12 V. Find VA, VB and all branch currents.
- Write the conductance matrix, but do not solve, for R1=2 kΩ, R2=4 kΩ, R3=1 kΩ, R4=2 kΩ and VS=10 V.
- A student writes the A-row mutual coefficient as +G3 after moving all unknown-voltage terms to the left. Explain the error.
- Two unknown nodes are joined only to each other by one resistor, with no connection to any fixed voltage or reference. What does nodal analysis determine, and what remains arbitrary?
- Why is matrix sparsity a topological statement rather than merely a property of the numerical resistor values?
Solutions
1. The circuit is linear and unchanged except for source scale, so VA=0.4·15 V=6 V and VB=0.2·15 V=3 V.
2. Equal scaling of all four resistances leaves the voltage ratios unchanged: VA=4.8 V and VB=2.4 V. Currents are 3.6 mA through R1, 2.4 mA through R2, 1.2 mA through R3 and 1.2 mA through R4.
3. G1=0.5 mS, G2=0.25 mS, G3=1 mS, G4=0.5 mS. The matrix is [[1.75,−1],[−1,1.5]] mS and the right-hand side is [5,0] mA.
4. The shared current leaving A is G3(VA−VB), so collecting unknowns gives +G3 VA and −G3 VB. A positive off-diagonal would reverse that contribution.
5. The resistor fixes only VA−VB through its branch relation together with any imposed current; without an external reference, adding the same constant to both potentials changes no physical voltage difference. Absolute node voltages are undetermined.
6. An off-diagonal entry exists only when two unknown nodes are directly linked by an element contributing mutual current. Whether the numerical conductance is 1 mS or 1 µS changes the value, not the connectivity pattern of zeros.
Chapter 03 addresses the first major exception to direct resistor-current substitution: an ideal voltage source placed between two unknown nodes. The correct tool is a supernode, not an invented source resistance.