Supernode Analysis

Neither node equation can be written alone. KCL at node \(A\) would require the current through the 6 V source, and an ideal voltage source passes whatever current the circuit demands — there is no expression for it in terms of node voltages.

Enclose both nodes and the source in a closed surface. By the generalised KCL of Set 3, everything crossing that boundary must sum to zero. The source's current is entirely internal and never appears:

One equation, two unknowns. The source supplies the second:

Substituting:

Now recover the source current. The supernode equation cannot give it — the current cancelled — so apply KCL to node \(B\) alone, where only the source branch and the 4 Ω meet:

Cross-check at node \(A\), which must give the same current:

Write both node equations honestly, keeping the unknown source current \(I\) as a symbol. Take \(I\) to flow from \(A\) to \(B\) inside the source, so it leaves node \(A\) and enters node \(B\):

Add them. The \(I\) terms cancel exactly:

This is precisely the supernode equation, obtained without drawing any surface or invoking any special technique.

The cancellation is guaranteed, not lucky. Any branch joining two nodes carries a current that leaves one and enters the other, so it enters the two KCL equations with opposite signs and always vanishes on addition. Drawing a surface around both nodes is just a graphical way of performing that addition.

Counting confirms nothing has been lost. Two nodes originally offered two KCL equations but carried three unknowns — \(V_A\), \(V_B\) and \(I\). Adding eliminates \(I\), leaving one equation and two unknowns; the source's own voltage restores the balance:

The surface is unchanged, because reversing a source does not move it. So the supernode equation is identical:

Only the constraint flips:

Substituting:

The source current, again from node \(B\) alone:

Check at node \(A\): \(6 = 6/2 + I\) gives \(I = 3\ \text{A}\;\checkmark\)

Side by side:

In the first case current enters the positive terminal, so the source absorbs; in the second it leaves the positive terminal, so it delivers.

The resistors, using \(P = V^2/R\) directly from the node voltages:

The 6 A source injects at node \(A\), which sits at 10 V, and returns to ground:

The 6 V source carries 1 A from \(A\) to \(B\), entering at the terminal marked \(+\). The passive sign convention is satisfied, so it absorbs:

The balance:

Nodes 1 and 2 are bridged by a voltage source, so enclose both. The supernode KCL, with every term a current leaving the surface:

Multiplying through by 300 clears every denominator:

With the constraint \(V_2 = V_1 + 6\):

The source current cannot come from the supernode equation, where it cancelled. Take node 1 alone, with \(I\) leaving towards node 2 through the source:

Cross-check at node 2, where the same current arrives:

Enclose both nodes. The 2 Ω runs from one enclosed node to the other, so its current crosses the boundary twice — once outward, once inward — and cancels identically. It does not appear:

With the constraint \(v_2 = v_1 + 4\):

What the 2 Ω does. Its voltage is forced to 4 V by the source it parallels, so it carries

That current is supplied by the voltage source and returns through it — a circulating current entirely inside the supernode, which the rest of the circuit never sees.

Check the balance across the boundary: the grounded resistors carry \(3.33/6 = 0.556\ \text{A}\) and \(7.33/3 = 2.444\ \text{A}\), totalling 3 A, which is exactly the net \(5 - 2 = 3\ \text{A}\) injected \(\checkmark\)

Nodes \(A\) and \(B\) form a supernode. The branch leaving \(B\) terminates on a known potential, so its current is perfectly ordinary — a fixed rail is just another known node voltage:

The supernode KCL:

With the constraint \(V_B = V_A + 4\):

Check the branch currents: the 2 Ω to ground carries 2.5 A, and the branch to the rail carries \((9-10)/2 = -0.5\ \text{A}\) — that is, 0.5 A flows from the rail into node \(B\). The balance is \(2.5 - 0.5 = 2\ \text{A}\;\checkmark\)

Write both node equations without eliminating anything, carrying \(I\) as a genuine unknown:

The source itself supplies the third equation:

In matrix form, with the unknown vector \((V_A, V_B, I)\):

The conductance block sits in the top-left corner, bordered by \(\pm1\) entries recording which nodes the source touches.

Solving: the second row gives \(I = V_B/4\), and the third gives \(V_A = V_B + 6\). Substituting both into the first:

Identical to Problem 1 — but the source current arrived as part of the solution rather than as a separate recovery step.

Comparing the two methods:

With \(B\) grounded, the 6 V source now has one terminal on the reference — so it no longer bridges two unknown nodes. It simply fixes:

No supernode, no constraint equation, and one unknown instead of two.

Only the old bottom rail, now called \(V_g\), remains unknown. KCL there — the 6 A source draws its return current from this node, and both resistors connect to it:

Multiplying by 4 and substituting \(V_A = 6\):

So \(V_A = 6\), \(V_B = 0\), \(V_g = -4\ \text{V}\). Shifting every potential up by 4 V to put the rail back at zero recovers Problem 1 exactly:

When the trick works. Only if the source has a terminal that can legitimately be made the reference — which requires that no other voltage source already needs that privilege. With a single bridging source it always works; with two or more, at most one can be eliminated this way, and the rest still need supernodes.

Seven nodes means six unknowns before any simplification:

The source tied to the reference fixes one node voltage outright — one unknown and one equation vanish together:

Each supernode consumes two nodes and returns one KCL equation instead of two, plus one constraint. With two disjoint supernodes, four of the remaining five nodes are absorbed:

The tally:

The general rule, for \(s\) bridging sources forming disjoint supernodes:

There are two bridging sources, and they touch no node in common — so they form two disjoint supernodes: \(\{1,2\}\) and \(\{3,4\}\). Each contributes one KCL equation and one constraint, giving four equations for four unknowns.

KCL on the first surface. Only two branches cross it — the 2 Ω to ground and the 3 Ω leaving node 2:

KCL on the second surface, which has no source feeding it:

Substituting the constraints \(V_2 = V_1 + 6\) and \(V_4 = V_3 + 4\). The second equation, multiplied by 6:

And the first, multiplied by 6:

From the first relation \(V_3 = (2V_1+8)/3\). Substituting:

Check both surfaces: \(8/2 + (14-8)/3 = 4 + 2 = 6\ \text{A}\;\checkmark\) and \((8-14)/3 + 12/6 = -2 + 2 = 0\;\checkmark\)

Node 2 is joined to node 1 by one source and to node 3 by another, so no surface can separate any pair of them. All three collapse into a single supernode, whose boundary crosses only the three grounded resistors:

The two constraints chain the voltages together, so everything is expressible in terms of \(V_1\):

Substituting:

Hence

Check: \(4/1 + 8/2 + 14/2 = 4 + 4 + 7 = 15\ \text{A}\;\checkmark\)

The general rule: a chain of \(s\) voltage sources linking \(s+1\) nodes produces

The supernode surface is drawn exactly as for an independent source — dependence is irrelevant to the boundary:

The constraint now involves an unknown on both sides. With \(v_x = V_1\):

Substituting:

Check: \(8/2 + 24/6 = 4 + 4 = 8\ \text{A}\;\checkmark\), and the source stands at \(2v_x = 16\ \text{V} = V_2 - V_1\;\checkmark\)

Recover the source current from node 2 alone, where only the source branch and the 6 Ω meet:

The source's positive terminal is at node 2, and current leaves that terminal, so it delivers:

Balance: the 8 A source supplies \(8 \times 8 = 64\ \text{W}\), the dependent source 64 W, total 128 W. The resistors absorb \(64/2 + 576/6 = 32 + 96 = 128\ \text{W}\;\checkmark\)

Node 1 is fixed by its source to ground:

The dependent source bridges nodes 2 and 3, so those two form a supernode. Its boundary is crossed by the 5 Ω into node 2, the 50 Ω and 100 Ω to ground, and the 4 A source:

The constraint, with the controlling current written in node voltages:

Substituting into the supernode equation and multiplying by 100:

Hence

And the required current:

Verify the constraint: \(10i_\phi = 20\ \text{V}\) and \(v_3 - v_2 = 80 - 60 = 20\ \text{V}\;\checkmark\)

No supernode is needed. The dependent source has a resistor in series with it, so the current in that branch follows from Ohm's law like any other:

KCL at node 1:

Multiplying by 20: \(10(V_1-20) + V_1 + 4(V_1-V_2) = 0\), that is \(15V_1 - 4V_2 = 200\).

KCL at node 2:

Substituting \(i_\phi = (V_1-V_2)/5\) and multiplying by 10:

so \(V_2 = 0.625\,V_1\).

Substituting into the node-1 equation:

Hence

Check node 2 numerically: \((10-16)/5 + 10/10 + (10 - 9.6)/2 = -1.2 + 1 + 0.2 = 0\;\checkmark\)

KCL at the base node, with \(i_B\) leaving into the transistor:

At the emitter node, the current arriving is the base current plus the collector current — the standard result that the emitter carries \((1+\beta)i_B\):

The junction model supplies the constraint linking the two nodes:

Substituting \(v_b\) into the base equation. Writing \(A = (1+\beta)R_E\) for brevity:

Collecting the \(i_B\) terms on one side:

Hence the closed form:

Reading the result. The numerator is the current the divider would push into the base if the emitter were grounded. The denominator is the price of the emitter resistor: the factor \((1+\beta)R_E\) is \(R_E\) as seen from the base, magnified by the transistor's current gain.

The source does not connect nodes 1 and 2 directly — there is a resistor in the same branch. The branch current is therefore perfectly ordinary:

Ohm's law applies to the resistor, and the source contributes a known offset. No unknown current, so no supernode.

KCL at node 1:

KCL at node 2:

From the second, \(V_1 = 2V_2 + 6\). Substituting into the first:

The branch carries \(10 - 6 - 2 = 2\ \text{A}\). Check node 1: \(10/5 + 2 = 4\ \text{A}\;\checkmark\)

Confirmation by source transformation. The 6 V source in series with 1 Ω is equivalent to a 6 A source in parallel with 1 Ω, injecting into node 1. The node equations become

Substituting \(V_1 = 10\), \(V_2 = 2\): the first gives \(2 + 8 = 10\;\checkmark\) and the second \(-8 + 2 + 6 = 0\;\checkmark\)

Every statement about one becomes a true statement about the other under a fixed dictionary:

The asymmetry. A supermesh exists only if the circuit is planar, since meshes themselves require a planar drawing. A supernode requires nothing: nodes exist in every circuit however it is drawn.

The consequence is practical. Given an unfamiliar circuit with sources of both kinds, nodal analysis is always available and mesh analysis may not be. The duality of the two techniques is exact at the level of the algebra, but not at the level of what they can be applied to.

Write the KCL equation at each node \(k\) of the chain, keeping every unknown source current as a symbol. Let \(\Sigma_k\) denote the sum of all resistive and current-source terms at node \(k\), and let \(I_j\) be the current through the \(j\)th voltage source, taken to flow from node \(j\) to node \(j+1\):

Each source current appears exactly twice with opposite signs, because it leaves one node and enters the next.

Add all \(s+1\) equations. The source currents telescope:

Leaving

This sum contains only resistive branches and current sources. Any branch internal to the chain has also cancelled, for exactly the same reason — which is why the resistor of Problem 6 disappeared. What survives is precisely the set of branches crossing the outer boundary: the supernode equation.

Counting. Before: \(s+1\) KCL equations carrying \(s+1\) node voltages plus \(s\) unknown source currents — \(2s+1\) unknowns for \(s+1\) equations, underdetermined by \(s\). After: one KCL equation, with the \(s\) source voltages supplying \(s\) constraints:

The derivation never referred to the shape of the surface — only to the cancellation. So any closed surface enclosing exactly these \(s+1\) nodes gives the same equation, which is why the boundary can be drawn wherever is most convenient.

The supernode requires a judgement. Before any equation can be written, something must examine the topology and decide which nodes are bridged by voltage sources, then merge them and generate a constraint. That is a graph-analysis pass, and it must be redone whenever the netlist changes.

MNA requires none. Each element stamps a fixed pattern into the matrix, independently of everything else. A resistor between nodes \(j\) and \(k\) adds \(1/R\) to two diagonal entries and subtracts it from two off-diagonal ones. A voltage source appends a row, a column and a new unknown:

Here \(\mathbf{G}\) is the ordinary conductance matrix and \(\mathbf{B}\) holds only \(0\) and \(\pm1\), recording which nodes each source touches. Problem 8 is the \(2\times2\) case of this bordered structure.

Three further advantages.

The cost. The system grows by one unknown per voltage source, so MNA solves a larger problem than the supernode method would. For hand calculation on two or three nodes that is a real penalty, which is exactly why the supernode remains the right technique for a person and the wrong one for a machine.

Note also that the \(\mathbf{0}\) block in the bottom-right destroys positive definiteness — the matrix is symmetric but indefinite. Simulators therefore use LU factorisation with pivoting rather than the Cholesky methods that a pure conductance matrix would allow.