AC Mesh and Nodal Analysis

Write the mesh equations by exactly the rules of Set 4, with impedances in place of resistances:

The diagonal is the sum round each mesh; the off-diagonal is minus the shared impedance. Note \(j8 - j4 = j4\) — the reactances partly cancel in mesh 2.

In matrix form:

The determinant, remembering \((-j8)^2 = j^2 64 = -64\):

The sign trap: \(-(-j8)^2 = -(-64) = +64\). Two sign errors are available in that one step, and they cancel only if both are made.

Solve by Cramer's rule:

The shared-branch current is the difference of the mesh currents, as in Set 4:

Here both mesh currents happen to share a phase angle, so the subtraction is simple. In general it must be done rectangularly.

Check by substituting into mesh 2, the equation not used to obtain \(\mathbf{I}_1\):

More directly: \(-j8(10) + j4(20) = -j80 + j80 = 0\) once the common phase factor is taken out.

The construction rules are identical to Set 4:

What survives:

What does not survive is everything that depended on the entries being positive real:

In Problem 1, mesh 2's diagonal was \(j4\) — purely imaginary, with no resistance at all. Had the capacitor been \(-j8\) the diagonal would have been exactly zero, which cannot happen in a resistive network.

Three practical consequences:

The consequence for method. Two unknowns are comfortable by hand, three are tedious, and four are best left to a calculator or computer. The structure of the matrix is worth writing out regardless, because it is where the circuit's topology lives — and because a symmetry failure is an immediate signal that an equation has been written wrongly.

The difficulty is unchanged from Set 5. A current source has no defined voltage across it, so KVL cannot be written round either mesh individually.

Equation 1 — the constraint. The source fixes the difference of the mesh currents:

This replaces one KVL equation, so the count still balances.

Equation 2 — KVL round the supermesh, the path enclosing both meshes and avoiding the current source:

Substitute the constraint \(\mathbf{I}_2 = \mathbf{I}_1 + 4\):

And the second current:

The addition must be done rectangularly — adding 4 to a polar quantity is not a change of magnitude.

Check the constraint:

The imaginary parts cancel exactly, which they must — the constraint was real.

Write the equations, treating the dependent source exactly as Set 4 did — as a source, then substituting its controlling variable:

Collect terms:

The matrix is not symmetric: \(Z_{12} = -j4\) but \(Z_{21} = 3-j4\). The dependent source has added 3 to one off-diagonal only.

Solve. The determinant:

Check in mesh 2, which was not used for \(\mathbf{I}_1\):

Two equal magnitudes exactly 180° apart — they cancel, as required.

What the asymmetry means. Set 14 traced reciprocity to \(\mathbf{Z} = \mathbf{Z}^{\mathsf T}\), and Set 15, Problem 12 showed that active devices break it. Exactly the same happens here: this circuit is not reciprocal, so interchanging source and meter would give a different reading. The symmetry check of Problem 17 must therefore be applied only to circuits without dependent sources.

First, verify the matrix is well formed. It is symmetric — \(Z_{12} = Z_{21} = -j5\), \(Z_{13} = Z_{31} = -4\), \(Z_{23} = Z_{32} = -6\) — so no dependent sources are present and the setup is consistent with Problem 2.

Solve by elimination rather than by three 3×3 determinants, which is far less error-prone by hand. From row 3:

Substituting into rows 1 and 2 reduces the system to 2×2, which can then be handled as in Problem 1.

The solution:

Verify in row 2, which the elimination used only indirectly:

Real parts: \(-20.09+36.61-16.53 = -0.01\). Imaginary: \(-31.69+30.44+1.27 = 0.02\). Both zero to rounding.

Note the phase spread. The three currents lie at \(-32°\), \(+54°\) and \(-4°\) — nearly 90° apart in places. In a resistive version of this circuit all three would be in phase, and their relative sizes could be estimated by inspection. Here neither magnitude nor phase can be guessed, which is why the verification step is not optional.

On hand computation. Three complex unknowns is about the practical limit. Beyond it, set the matrix up carefully by inspection — that part is still worth doing by hand, since it encodes the topology — and hand the arithmetic to a calculator.

The rule is unchanged. For \(\mathbf{Z}\mathbf{I} = \mathbf{V}\), replace column \(k\) by \(\mathbf{V}\) and divide:

Everything about the method survives; only the arithmetic inside changes.

The \(2\times2\) determinant:

with \(ad\) and \(bc\) complex products. Products are easiest in polar form, but the subtraction must then be done rectangularly — so both forms are needed within one calculation.

The traps, all arising from \(j^2 = -1\):

The third is worth dwelling on: an inductive reactance multiplied by a capacitive one gives a positive real number. That is how purely reactive networks produce real terms, as Set 20, Problem 9 found in the delta–wye conversion.

Demonstrating on Problem 1. Written out with every sign explicit:

Writing \(j^2\) explicitly before replacing it with \(-1\) is a small discipline that removes most of these errors.

A useful check on \(\Delta\). For a passive network the determinant can be anything, but it should not be zero unless the circuit is degenerate. If \(|\Delta|\) comes out very small compared with the products that formed it, either an error has been made or the circuit is near resonance — Problem 18 distinguishes the two.

For \(3\times3\) and larger, prefer Gaussian elimination to cofactor expansion. A \(3\times3\) complex determinant requires six triple products with signs; elimination requires fewer operations and, more importantly, produces intermediate results that can be checked as they appear.

Convert to admittances first, since nodal analysis adds admittances:

Note the sign flip: \(1/(-j4) = +j0.25\), because \(1/j = -j\). A capacitor has positive susceptance and negative reactance — the reversal Set 20, Problem 11 warned about.

Write the nodal equations by the rules of Set 6:

Diagonal: sum of admittances at the node. Off-diagonal: minus the connecting admittance. Symmetric, as expected.

The second diagonal is \(j0.15\) — purely imaginary, since node 2 touches no resistance. Compute the determinant:

Solve:

Node 2's voltage exceeds node 1's, even though node 2 is further from the source and has no source of its own. This is the magnification of Set 20, Problem 8 appearing again: the capacitor and inductor form a partially resonant pair, and the voltage across the inductor is boosted.

The ratio is exactly the divider ratio \(Y_{12}/(Y_{12}+Y_2)\) inverted — nothing mysterious, but impossible in a resistive network.

Check KCL at node 1:

The difficulty, unchanged from Set 7. A voltage source between two non-reference nodes carries an unknown current, so KCL cannot be written at either node alone. The remedy is to enclose both in a supernode.

Equation 1 — the constraint imposed by the source:

Equation 2 — KCL over the supernode. The source's current is internal to the boundary and cancels:

The two reactive admittances partly cancel, as they always do when an inductor and capacitor share a node.

Substitute \(\mathbf{V}_1 = \mathbf{V}_2 + 17.32 + j10\):

And node 1:

Check the constraint, which is the fastest verification available:

The general procedure, identical to Set 6:

The four source types and how each enters:

Current-controlled sources need the extra step of converting a branch current to node voltages — where the admittance of the controlling branch appears, complete with its \(j\).

A worked instance. Suppose a VCCS of \(0.5\mathbf{V}_x\) feeds node 2, with \(\mathbf{V}_x = \mathbf{V}_1 - \mathbf{V}_3\) the voltage across a capacitor. KCL at node 2 initially reads

which becomes, after substitution and collection:

The \(-0.5\) is added to \(Y_{21}\) but nothing is added to \(Y_{12}\) — hence the asymmetry.

Why the substitution must precede assembly. The matrix form \(\mathbf{Y}\mathbf{V} = \mathbf{I}\) requires every unknown on the left and every known on the right. A dependent source left on the right-hand side is an unknown masquerading as a source, and the system cannot be solved as written.

One AC-specific hazard. When a current-controlled source is controlled by the current through a reactive branch, the conversion introduces a complex admittance:

so a gain that looked real, say \(4\mathbf{I}_x\), contributes a complex coefficient \(4j\omega C\) to the matrix. Forgetting the \(j\) here is a genuinely AC error with no DC counterpart.

Check the structure before solving. The matrix is symmetric, so no dependent sources are present. Note also that \(Y_{13} = +j0.25\) is positive imaginary — an off-diagonal is minus the connecting admittance, and minus an inductor's \(-j\) admittance is positive.

Off-diagonals are not automatically negative in AC — only their admittance is negated, and the sign of the result depends on the element.

Solve by elimination. The solution:

Verify at node 3, the equation least involved in the elimination:

Real parts sum to \(-0.002\), imaginary to \(-0.001\) — zero to the precision carried.

Reading the result. The voltages fall away from the driven node — 12.0, 4.9, 2.9 V — and their phases spread over nearly 70°. The phase spread is the AC-specific feature: in a resistive network every node voltage would be in phase with the source, and only the magnitudes would differ.

A note on effort. The setup took one pass of inspection; the solution took several minutes of complex arithmetic. That ratio is characteristic, and it is why the structural checks — symmetry, and the sign of each off-diagonal — are worth applying before any arithmetic begins. An error caught in the matrix costs seconds; the same error caught after solving costs the whole calculation.

The primary criterion is unchanged: count the unknowns.

for \(b\) branches and \(n\) nodes — Set 8's topology result, which knows nothing about impedance. Choose whichever is smaller, since a \(2\times2\) complex system is far less work than a \(3\times3\).

Secondary criteria:

What is new in AC. A third criterion appears that has no DC counterpart — which elements are present:

A capacitor-heavy circuit is easier by nodal analysis; an inductor-heavy one by mesh. Each avoids taking reciprocals of imaginary numbers, where the \(1/j = -j\) sign flip lives.

Applying this to earlier problems:

The last two show the criteria can conflict: a current source normally suggests nodal analysis, but one sitting between two meshes is handled perfectly well by a supermesh if mesh analysis is otherwise preferable.

A practical tiebreaker. When the counts are equal, prefer nodal. It applies to non-planar circuits, its unknowns are directly measurable with a voltmeter referred to ground, and node voltages are what most subsequent calculations need.

Why this needs its own problem. In Part 1 a wrong answer often announced itself — a negative resistance, a current larger than the source could supply, a node voltage outside the supply rails. Complex answers offer none of that. A result wrong by a factor of \(j\), or with two terms transposed, looks entirely reasonable.

Check 1 — matrix symmetry, before solving. Costs seconds:

Must hold if there are no dependent sources. Catches: mis-copied shared impedances, wrong signs on off-diagonals, omitted branches. Will not catch an error made symmetrically in both entries.

Check 2 — substitute into an unused equation. The single most valuable check:

Use the row not used to isolate the answer. Both real and imaginary parts must vanish, which is two independent conditions from one substitution. Catches nearly all arithmetic errors in the solving stage.

Check 3 — Kirchhoff on the original circuit. Reconstruct a branch voltage or current from the answer and verify a loop or node directly:

Catches errors in setting up the matrix itself, which Check 2 cannot — a consistently wrong system is solved consistently.

Check 4 — physical plausibility. Weaker in AC, but three things still hold:

Note what is not on the list: any expectation that internal voltages be smaller than the source. Set 20, Problem 8 disposed of that.

Worked example on Problem 1's answer. Check 2, using mesh 2:

Check 4: the net reactance in mesh 1 is inductive, and \(\mathbf{I}_1\) at \(+53°\)... which leads. Investigating: mesh 1's own impedance is \(6+j8\), but the coupled mesh 2 is net capacitive, and the combination seen by the source is

— capacitive overall, so a leading current is correct. The check did its job: it flagged something worth examining, and the examination confirmed the answer.

Work from the load backwards, exactly as Set 2 did for resistive ladders. The last two elements are in series:

That combines with the shunt capacitor:

Add the series inductor:

A clean result. Note that the network is capacitive overall despite the series inductor — the shunt capacitor dominates.

The source current:

Work forwards to the load. The voltage at the shunt node:

99.85 V from a 60 V source — the magnification of Set 20, Problem 8 again, now inside a ladder. The series inductor and shunt capacitor partly resonate.

Then the load current and voltage:

The load voltage also exceeds the source. A resistive ladder can only attenuate; a reactive one can amplify voltage — though not power, as Set 23 will confirm.

Check by KVL round the input loop:

Balance means zero detector current, which requires the two detector nodes to be at the same potential. With no detector current, each arm pair acts as an undisturbed divider:

Cross-multiplying and cancelling gives the classical condition:

Formally identical to Set 2's \(R_1R_4 = R_2R_3\) — but now a complex equation.

That is the crucial difference. A complex equation is two real equations:

Magnitudes and angles must balance simultaneously. A DC bridge has one condition and one adjustment; an AC bridge has two of each.

Solving for the unknown arm:

A pure capacitive reactance of 600 Ω. At 1 kHz that is

Verify both conditions:

Why AC balancing is harder in practice. Two independent adjustments are needed, and they usually interact — changing a resistance to correct the magnitude disturbs the angle, and vice versa. Balancing therefore proceeds iteratively, adjusting each in turn until the detector nulls. Three further complications:

Why bridges are used at all, despite this. A null measurement depends only on the ratio of known components, not on the detector's calibration or the source's amplitude. That makes bridge measurements of impedance among the most accurate available — which is why they survived long after direct-reading instruments became common.

The balance condition with the unknown opposite the parallel \(RC\) arm:

Placing the unknown opposite the \(RC\) arm is the whole trick: it makes \(\mathbf{Z}_4\) appear as a product rather than a quotient, so its admittance enters directly.

Rearranging in terms of admittance:

The parallel combination's admittance is a simple sum — which is why the arm is built as a parallel \(RC\) rather than a series one.

Equate real and imaginary parts separately, the two conditions of Problem 14:

The key advantage: \(\omega\) cancels from the inductance result.

So the measurement does not require the source frequency to be known accurately, and it does not need to be stable. Compare Problem 14's general bridge, where balance held at one frequency only.

The second advantage is that a capacitor is used as the standard. Precision capacitors are far easier to make than precision inductors: they have negligible loss, no stray magnetic coupling, and stable values. Measuring an inductance against a capacitance standard is therefore more accurate than against another inductor.

The limitation. The two balance conditions interact through \(R_4\) and \(C_4\), and convergence is poor when the unknown coil's \(Q\) is high:

A high-\(Q\) coil needs a large \(R_4\), which becomes impractical above \(Q \approx 10\). The Hay bridge — the same idea with a series \(RC\) arm — is used instead for high-\(Q\) coils, at the cost of reintroducing a frequency dependence.

Two separate analyses, as Set 20, Problem 18 requires — the impedance differs at each frequency, so no single phasor exists.

aAt \(\omega = 500\ \text{rad/s}\):

Capacitive, so the current leads.

bAt \(\omega = 1000\ \text{rad/s}\):

Inductive, so the current lags.

Add the time functions — never the phasors:

The symmetry explained. The reactances swapped exactly between the two frequencies, and the reason is the resonant frequency:

The two driving frequencies straddle it, and in fact \(500 \times 1000 = 500\,000 = \omega_0^2\) — they are geometrically symmetric about resonance. Whenever \(\omega_a\omega_b = \omega_0^2\), the net reactance at one is exactly the negative of the net reactance at the other, so the impedance magnitudes match and the angles are equal and opposite.

The RMS current, where mean squares add because different frequencies are orthogonal:

The argument is Set 14's, unchanged. Solving \(\mathbf{Z}\mathbf{I} = \mathbf{V}\) gives

and the inverse of a symmetric matrix is symmetric. So the transfer impedance from mesh \(j\) to mesh \(k\) equals that from \(k\) to \(j\):

Now a complex equality, so both magnitude and phase are reciprocal.

Verifying on Problem 5. A source of \(100\angle0°\) in mesh 1 produced \(\mathbf{I}_3 = 2.764\angle{-4.38°}\ \text{A}\). Moving the same source to mesh 3 and measuring in mesh 1 gives, from the transfer impedance,

Identical in both magnitude and angle — the phase is reciprocal too, which is a stronger statement than the DC version and is easy to overlook.

When it fails. Exactly when the matrix is asymmetric:

The mutual-inductance entry is worth noting in advance: coupled coils add off-diagonal terms \(\pm j\omega M\) to both \(Z_{jk}\) and \(Z_{kj}\) equally, so symmetry survives and Set 26's circuits remain reciprocal.

The practical use is as a check on the matrix, before any arithmetic. If a circuit contains only passive bilateral elements and the assembled matrix is asymmetric, an equation is wrong — most often a shared branch entered in one row and omitted from the other, or entered with the wrong sign.

A second use is labour-saving. Reciprocity means a transfer function need be computed once and can then be read in either direction, which halves the work when a network's behaviour is wanted from both ports — as it will be in Set 35's two-port parameters, where \(z_{12} = z_{21}\) is precisely this result.

Consider a single mesh containing \(R\), \(L\) and \(C\) in series. Its "matrix" is the scalar

At \(\omega = \omega_0\) the imaginary part vanishes and \(\Delta = R\). If \(R\) is small, so is \(\Delta\) — while the individual terms \(\omega L\) and \(1/\omega C\) may each be large.

That is the definition of ill-conditioning: a small result from the near-cancellation of large terms. With \(R = 10\ \Omega\) and reactances of 31.6 Ω each (Set 20, Problem 13):

A 1% error in either reactance produces a 3% error in \(\Delta\) and hence in every current — errors are amplified by the cancellation.

The consequences:

Distinguishing ill-conditioning from an error. Both produce a suspiciously small determinant, so the test is whether the smallness is explicable:

The third row is the useful one. For a passive network the determinant's real part cannot vanish — there is always some resistance somewhere. A determinant that is exactly zero, or has negative real part, is an error rather than a physical near-cancellation.

An exactly singular matrix means no unique solution exists. Physically this happens only in idealised circuits: a lossless \(LC\) loop driven exactly at \(\omega_0\) would demand infinite current. Real circuits always have resistance, so the singularity is an artefact of neglecting it — a reminder that the ideal \(L\) and \(C\) of Sets 16 and 17 are approximations.

The complete catalogue for mesh analysis:

And for nodal analysis, the exact dual:

The two tables are duals under \(V\leftrightarrow I\), mesh\(\leftrightarrow\)node, which is Set 17's duality operating on the analysis methods themselves.

The row worth exploiting is the third in each table. A voltage source connected from a node to the reference makes that node voltage known, removing an unknown entirely:

Choosing the reference node adjacent to a voltage source is therefore worth doing deliberately. Placing it at the source's negative terminal can convert a supernode problem into a trivial one.

The elimination. A voltage source in series with an impedance, or a current source in parallel with one, can be converted by source transformation:

Set 12's method, valid unchanged for impedances. This can turn a supermesh into an ordinary mesh, or a supernode into an ordinary node, and is often the quickest simplification available — Set 22 develops it.

A worked judgement. Problem 8's supernode arose from a voltage source between two non-reference nodes. Had that source had any impedance in series with it, transformation would have converted it to a current source in parallel — removing the supernode and leaving two ordinary node equations. The supernode was necessary only because the source was ideal.

Transferred with no modification whatever:

Nine major results, none requiring re-derivation. The proofs in Part 1 used only Kirchhoff's laws and linearity, both of which survive the phasor transform.

Needing modification:

Genuinely new:

and nothing else. A resistive network's matrix cannot be near-singular from cancellation, because resistances only accumulate. Reactances of opposite sign can cancel, and that possibility is the one structurally new feature of AC analysis.

The honest summary of this set's difficulty: it is arithmetic, not conceptual. Anyone who could solve Sets 4 to 7 can set up every problem here. What takes the time is complex multiplication, and what causes errors is \(j^2 = -1\) appearing where a sign was not expected.

What Set 22 will do is complete the transfer, taking Sets 9 to 14 — Thévenin, Norton, superposition, source transformation, maximum power and the additional theorems — into the frequency domain. One result there genuinely changes: the maximum power condition becomes the conjugate match \(\mathbf{Z}_L = \mathbf{Z}_{Th}^*\), previewed in Set 20's Challenge C2. Everything else transfers as here.