Nodal Analysis

Take the bottom rail as the reference. There are three nodes, so two equations are needed. At node 1, currents leaving must equal the 4 A injected:

At node 2, note that the bridging resistor's current is now written the other way round — always subtract the far node from the node you are standing on:

Solving. Multiply the first by 2 and add:

The bridging resistor carries

KCL check at node 1: \(6/2 + 1 = 3 + 1 = 4\ \text{A}\;\checkmark\). At node 2: \(4/2 = 2 = 1 + 1\ \text{A}\;\checkmark\)

KCL at node 1, with currents leaving taken as positive and the source as an inflow:

KCL at node 2:

From the first, \(V_2 = 2V_1 - 5\). Substituting:

The three resistor currents follow directly:

KCL check: node 1 gives \(4.75 + 0.25 = 5\ \text{A}\;\checkmark\) and node 2 gives \(2 + 0.25 = 2.25\ \text{A}\;\checkmark\)

With node 2 as reference, call the bottom rail \(V_0\) and node 1 \(V_1'\). KCL at node 1:

KCL at the bottom rail, which now carries both source returns:

Solving the pair gives

Compare with Problem 1. Every potential has fallen by exactly 4 V:

Now the branch currents, which are all differences and so are untouched by the shift:

Identical to Problem 1's 1 A, 3 A and 2 A.

The principle. Node voltages are potentials measured from an arbitrary datum; only their differences are physical. Moving the reference adds a constant to every potential, and every constant cancels in every branch equation.

The rules are the exact duals of the mesh ones. Self-conductance is the sum of everything meeting at the node; mutual conductance is minus what directly joins two nodes:

Nodes 1 and 3 are not directly joined, so that entry is zero and the matrix is tridiagonal.

The right-hand side is the net current injected at each node:

The third row gives \(1.5V_3 = V_2\), so \(V_2 = 1.5V_3\). The first row gives \(V_1 = (6 + V_2)/1.5\). Substituting both into the second row:

Hence

Check all three rows: \(1.5(8) - 6 = 6\;\checkmark\); \(-8 + 15 - 4 = 3\;\checkmark\); \(-6 + 6 = 0\;\checkmark\)

The voltage source has one terminal on the reference node, so it does not merely constrain \(V_1\) — it determines it:

No KCL equation is written at node 1, and none can be: the source's current is unknown. Two nodes have become one unknown.

KCL at node 2, the only equation needed:

Multiplying by 4:

Branch currents:

KCL at node 2: \(0.5 + 2 = 2.5\ \text{A}\;\checkmark\)

The 12 V source carries only the current in the first resistor, so

Balance: the 2 A source at 10 V supplies 20 W, total 26 W. The resistors absorb \((0.5)^2(4) + (2.5)^2(4) = 1 + 25 = 26\ \text{W}\;\checkmark\)

A short circuit is a zero-resistance branch, so both of its ends are the same node. Since one end is the reference, the shorted node sits at

This is the key simplification: the short does not merely have a small voltage across it, it has none at all, so it can be treated as a known potential.

KCL at node 1, whose three branches lead to \(E\), to ground and to the shorted node (also at 0 V):

KCL at node 2, fed by the 4 A source:

Now use the given short-circuit current. All the current arriving at the shorted node from nodes 1 and 2 must leave through the short:

Substituting \(i_R = 10\) and \(V_2 = 4\):

Hence

The negative sign says the source's actual polarity is opposite to the one assumed on the diagram.

KCL at node 1:

KCL at node 2, where the 4 A source is an outflow:

Solving by determinants:

The 5 Ω carries

flowing from node 1 to node 2.

Check node 2: \(1.818 = 4 + (-8.727/4) = 4 - 2.182 = 1.818\ \text{A}\;\checkmark\)

One node, one equation. Working in conductances avoids nested fractions:

Milliohms give conductances in hundreds of siemens — a useful sanity anchor for what follows.

The total injected current is

Hence the node voltage:

And the current in the 5 mΩ:

Check by current division — the branch currents must share the 19 mA in proportion to conductance:

The VCCS has transconductance 0.2 S, so its current is \(0.2V_1\) — dimensionally a conductance term, and it can be collected with the resistors:

Summing the conductances:

Hence

Checking the current split: the 1 Ω takes 9.23 mA, the 10 Ω takes 0.923 mA, and the VCCS takes \(0.2 \times 9.23\ \text{mV} = 1.85\ \text{mA}\). Total \(= 12.0\ \text{mA}\;\checkmark\)

Write KCL with the dependent source as a symbol first:

Then supply the constraint. The controlling current is the one in a 4 Ω resistor, so

Substituting:

Branch currents:

KCL check: \(1.25 + 1.25 + 2.5 = 5\ \text{A}\;\checkmark\)

The dependent source carries 2.5 A from the node down to ground, so it has 5 V across it with current entering the positive terminal:

Balance: the 5 A source supplies \(5 \times 5 = 25\ \text{W}\); the resistors take \(2 \times (1.25)^2(4) = 12.5\ \text{W}\); and \(12.5 + 12.5 = 25\ \text{W}\;\checkmark\)

KCL at node \(A\), with all currents leaving taken as positive:

Collecting conductances:

The three branch currents:

KCL check: \(2.4 = 1.6 + 0.8\;\checkmark\)

The dependent source drains 0.8 A from an 8 V node to ground, so current enters its positive terminal:

Power balance:

KCL at node \(v_1\):

KCL at node \(v_2\), with the dependent source injecting \(3i_1\) and the constraint \(i_1 = v_1/2\):

From the first equation, \(v_2 = 1.5v_1 - 15\). Substituting:

Hence \(i_1 = v_1/2 = -20\ \text{A}\) and the source current is \(3i_1 = -60\ \text{A}\). With \(v_2\) across it:

Verify with a full audit. The resistors absorb:

The 15 A source sits at \(-40\) V while pushing current out, so it absorbs \(40 \times 15 = 600\ \text{W}\). Total absorbed \(= 3900 + 600 = 4500\ \text{W}\;\checkmark\)

Neither node equation can be written alone. KCL at node 1 would require the current through the 6 V source, and that current cannot be expressed in terms of node voltages — an ideal voltage source will pass whatever the circuit demands.

The remedy is the generalised KCL of Set 3, Problem 14: draw a closed surface enclosing both nodes and the source, and apply KCL to that surface. The source's current is internal to it and never appears:

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

Substituting:

Check: the two resistors carry 1 A and 4 A, totalling the 5 A injected \(\checkmark\). The source itself carries 4 A into node 2 and out of node 1.

KCL at node \(v_1\), which is an ordinary node:

Nodes \(v_2\) and \(v_3\) are bridged by a voltage source, so enclose both in a supernode and write one KCL for the surface:

And the constraint:

Substituting the constraint into the supernode equation and collecting:

Solving simultaneously with the node-1 equation (also written in terms of \(v_2\)) gives

Verify node 1 directly:

The 11 A injected is exactly balanced by 11 A arriving from the two branches — both of which flow into node 1, since \(v_2\) and \(v_3\) both sit above it.

Enclose both nodes in a supernode. The 10 Ω runs from one enclosed node to the other, so its current is entirely internal to the surface — it crosses the boundary twice, once in each direction, and cancels:

With the constraint \(v_2 = v_1 + 2\):

The role of the 10 Ω. It has no effect whatever on the node voltages, because those are fixed by the supernode equation and the constraint, neither of which contains it. Its voltage is forced to be 2 V by the source it parallels, so it carries

That current is supplied by the voltage source and returns through it — a circulating current within the supernode that the rest of the circuit never sees.

Check: the two grounded resistors carry \(-3.667\ \text{A}\) and \(-1.333\ \text{A}\), summing to \(-5\ \text{A}\), which balances the net \(2 - 7 = -5\ \text{A}\) injected \(\checkmark\)

Ground node \(b\) and use the source to fix \(V_a = 8\ \text{V}\). That leaves only two unknowns, \(V_c\) and \(V_d\) — fewer than the three meshes the same circuit needed in Set 4.

KCL at node \(c\):

KCL at node \(d\):

Multiplying through by 36 to clear fractions:

Solving: from the first, \(V_d = 6V_c - 24\). Substituting:

The source current is what leaves node \(a\):

And the bridge arm:

Three shunt nodes, three equations. KCL at node 1:

At node 2 and node 3:

The third equation gives \(V_2 = 1.5\,V_3\) immediately. Multiplying the second by 36 to clear fractions:

Substituting \(V_2 = 1.5V_3\): \(-9V_1 + 34.5V_3 - 12V_3 = 0\), so \(V_1 = 2.5\,V_3\).

Multiplying the first equation by 60:

Solving the system gives

Verify each equation directly:

These are exactly the node voltages found by rung-by-rung reduction in Set 2, and the branch currents follow: 0.8 A in the 15 Ω, 0.4 A in the 18 Ω, 0.8 A in the 6 Ω, with 2 A from the source.

Sources first. Each delivers its current at the potential of the node it feeds:

Now the resistors, each using the difference across it:

The balance closes exactly:

Note how convenient the node-voltage form is here: \(P = V^2/R\) needs only the two node potentials, with no branch current computed at all.

The two methods are duals: every statement about one becomes a true statement about the other under a fixed dictionary.

Symmetry. \(G_{jk}\) is minus the conductance of the branch directly joining nodes \(j\) and \(k\) — a quantity that does not depend on the order in which the two nodes are named. Hence \(G_{jk} = G_{kj}\).

Positive diagonal. \(G_{jj}\) is the sum of the conductances of every branch meeting at node \(j\), each of which is positive.

Diagonal dominance. The off-diagonal magnitudes \(|G_{jk}|\) are precisely the conductances of the branches from \(j\) to other non-reference nodes. But \(G_{jj}\) includes those plus any branch from \(j\) to the reference node:

with equality only if node \(j\) has no direct path to the reference.

A symmetric, diagonally dominant matrix with positive diagonal is positive definite and therefore non-singular. So a purely resistive network always has a unique nodal solution — the same conclusion reached for mesh analysis in Set 4, by the same argument in the dual language.

Let \(\{V_k\}\) solve the network with node \(r\) as reference, so \(V_r = 0\). Propose the shifted set

where \(s\) is the new reference, which now correctly satisfies \(V_s' = 0\).

Every branch current is a difference, so the shift cancels:

Since every branch current and every element voltage is unchanged, every element power \(p = vi\) is unchanged. And because the branch currents are unchanged, KCL is still satisfied at every node, so the shifted set is a valid solution — and the unique one, by Problem 19.

What does change. The matrix \(\mathbf{G}\) is built from branch conductances alone, so its entries are the same numbers — but its size and indexing change, because a different node has been dropped from the system. The row and column belonging to the old reference reappear, and those of the new reference vanish.

The injection vector changes too. A current source returning to the old reference contributed nothing to any equation; after the shift it may inject into a node that is now in the system. Both effects are bookkeeping, not physics.

Problem 3 is this proof in miniature: shifting from the rail to node 2 moved every potential down by 4 V and left all three branch currents at 3 A, 1 A and 2 A.