Y-Bus Formation

The conversion. Rationalising rather than dividing complex numbers directly:

The sign reverses: a positive (inductive) reactance becomes a negative susceptance. Every series branch of a transmission network therefore contributes a negative imaginary part, which is why every \(Y_{ii}\) below turns out to be large and negative-imaginary.

Line 1–2 worked in full:

All seven:

Every line has the same \(x/r = 3\), so every admittance has the same angle, \(-71.57^\circ\), and they differ only in magnitude. That is deliberate in this standard test system and makes the arithmetic checkable: \(b = -3g\) in every row.

The ratio of the largest to the smallest is 8:1. Line 3–4 at \(0.01+j0.03\) is a short, heavy interconnection; lines 1–3 and 4–5 at \(0.08+j0.24\) are four times longer. This spread of admittances is what will make bus 3 and bus 4 behave almost as one bus in the load flow of Set 19.

A note on the charging column. The \(B/2\) values are already halved — the nominal-\(\pi\) of Set 11, with half the line's total susceptance at each end. They are pure susceptances, positive because capacitive, and they never appear in an off-diagonal term.

KCL at bus \(i\). The current injected into the bus from outside equals the sum of the currents leaving it through every connected branch:

The first sum runs over branches to other buses; the last term is the shunt to ground, whose voltage is \(V_i\) itself.

Collect the coefficients of each voltage:

Which is the by-inspection rule, obtained rather than asserted. Writing one such equation per bus and stacking them gives the matrix form.

What the symbols mean — and the two usual misreadings:

Why the reference is excluded. Include ground as an \((n+1)\)th node and the matrix becomes singular. Every row and every column then sums to zero, so the rows are linearly dependent:

Physically: raising every node by the same voltage drives no current anywhere, so the all-ones vector is in the null space. Deleting the reference row and column removes that degeneracy and makes \(\mathbf{Y}_{\text{bus}}\) invertible — which is what makes \(\mathbf{Z}_{\text{bus}} = \mathbf{Y}_{\text{bus}}^{-1}\) meaningful in Set 18.

The shunts are what break the degeneracy. With the reference deleted, a row sums to \(y_{i0}\) rather than to zero — the shunt admittance is the only path by which current can leave the network's set of buses. A network with no shunt element anywhere would still have a singular \(\mathbf{Y}_{\text{bus}}\) after deletion, and its \(\mathbf{Z}_{\text{bus}}\) would not exist.

The dimensions. For \(n\) buses excluding the reference, \(\mathbf{Y}_{\text{bus}}\) is \(n\times n\) regardless of how many branches the network has. A network of 2000 buses and 3000 lines gives a \(2000\times2000\) matrix — and, as Problem 17 shows, one that is 99.8% zeros.

Neglecting charging. The diagonal of bus 1 is the sum of the two admittances meeting there:

The off-diagonals are simply the negatives of the joining admittances:

The check: every row sums to exactly zero, because with no shunt element there is nowhere for current to go when all three buses are at the same potential.

Including charging. Only the diagonals change. Bus 1 carries half the charging of lines 1–2 and 1–3:

The off-diagonal terms are untouched. Charging is a shunt element and shunt elements never appear off the diagonal.

The row sums now equal the shunt at each bus:

Positive and purely imaginary, as line charging must be.

The size of the correction. Charging changes \(Y_{11}\) by 0.055 in 18.75 — three parts in a thousand. It is negligible for a fault study, where the inductive terms dominate entirely, and not negligible for a load flow, where it is the whole of the Ferranti effect and a substantial part of the network's reactive balance.

The rule needs no derivation now: \(Y_{ij} = -y_{ij}\) if a branch joins \(i\) and \(j\), and \(Y_{ij} = 0\) otherwise.

The zeros. Three bus pairs have no direct branch:

Six zero entries out of the twenty off-diagonal positions. The zeros are the network's topology, written down: \(\mathbf{Y}_{\text{bus}}\) is the graph's adjacency structure with admittances in place of ones.

The signs are worth pausing on. Every off-diagonal has a negative real part and a positive imaginary part — the exact opposite of the branch admittance, because of the minus sign in \(Y_{ij} = -y_{ij}\). A hand-formed matrix in which some off-diagonal has a positive real part contains an error, without exception.

Symmetry. \(Y_{ij} = Y_{ji}\) throughout, because a passive branch has the same admittance in both directions. This holds for every element in this network and fails only for the phase shifter of Problem 16 — the one device in a power system that is not reciprocal.

Reading the network back out. From the off-diagonals alone one can reconstruct the single-line diagram: bus 2 has four connections and is the hub; buses 1 and 5 have two each; the 3–4 branch is six times stiffer than any other. Nothing about generation, load or voltage appears — \(\mathbf{Y}_{\text{bus}}\) describes the network and only the network.

Bus 1 connects to buses 2 and 3, and carries the half-charging of both lines:

Bus 2 connects to four others — the hub of the network:

Buses 3 and 4 are identical in this system — each joins the other through the stiff 3–4 branch and has one further connection to bus 2, plus one long line:

Equal by coincidence of the data, not by symmetry of the network — bus 3 connects to 1, 2, 4 and bus 4 connects to 2, 3, 5, and the impedances happen to match.

Bus 5 has only two connections:

The complete matrix, in per unit on 100 MVA:

Two structural observations. Every diagonal is positive-real and negative-imaginary, and every diagonal is larger in magnitude than any individual off-diagonal in its row — necessarily, since it is a sum that includes them. The largest entry in the matrix is \(Y_{33} = Y_{44}\) at magnitude 40.79, dominated by the stiff 3–4 branch.

Row 1 summed term by term:

Exactly the half-charging of lines 1–2 and 1–3, \(0.030+0.025\).

All five rows:

Every real part is zero and every imaginary part matches the tabulated charging at that bus.

Why it works. Set every bus voltage to 1.0 and the nodal equation gives

With all buses at the same potential no current flows in any series branch, so the only current injected is that drawn by the shunts. The row sum is the shunt admittance, and the test is a physical statement rather than an algebraic trick.

What it catches. Every error in which a term is dropped, doubled, mis-signed or placed in the wrong column shows up:

A worked example of the failure it does catch. Suppose \(Y_{24}\) were entered as \(+1.6667-j5.0\) — the branch admittance rather than its negative. Row 2 would then sum to

A real part of 3.33 where zero was expected. The error is unmissable, and its size — twice the branch admittance — identifies which branch is at fault.

The one error it misses is a symmetric transposition: entering line 2–5's admittance at position (2,3) and line 2–3's at (2,5), with the same total. The row sums are unaffected because addition is commutative. Only a column-sum check on the matching column, or a comparison against the line list, will find it.

The per-unit admittance. A capacitor rated \(Q\) at nominal voltage has, in per unit on the same base,

Positive imaginary: a capacitor is a positive susceptance, the opposite sign to every series branch in the network.

The modification is one element. A shunt connects a bus to the reference, so it appears in that bus's diagonal alone:

Nothing else in the matrix changes — not \(Y_{25}\), not \(Y_{45}\), not any other diagonal.

The new row sum is \(j0.040 + j0.5 = j0.54\), and the check still passes because the capacitor is a shunt admittance and the test was written to include it.

The same slot serves every shunt element. Line charging, capacitor banks, shunt reactors, the transformer's magnetising branch, and the equivalent shunt of a constant-impedance load all add to the same diagonal:

The last is the only one with a real part, and it is also the only one that changes the diagonal's conductance. It is how a load is represented in a fault study, where the iteration of a load flow is unavailable.

The magnitude matters. \(j0.5\) against \(-j11.21\) is a 4.5% change in \(Y_{55}\) — twenty times larger than the line charging's effect, and large enough to move a load-flow solution appreciably. A shunt element is the cheapest lever a planner has on a bus voltage, and this single matrix entry is the whole of its representation.

The new branch admittance:

Four entries change, and no others:

The results:

The last is the interesting one: a position that was zero is now occupied. The network's graph has gained an edge.

The check. Row 1 now sums to

And \(0.055 + 0.020 = 0.075\) — correct. Row 4 similarly becomes \(j0.075\).

Why this is not the general case. The modification was purely additive because the new branch joins two buses that already exist. Three other cases behave differently:

Only the last requires the singular transformation of Problem 13; the rest are done by inspection.

The contrast with \(\mathbf{Z}_{\text{bus}}\) is the point. Adding one line changed four numbers here. The same addition changes every element of the impedance matrix, because every bus's driving-point and transfer impedance is affected by a new path anywhere in the network. That asymmetry is why \(\mathbf{Y}_{\text{bus}}\) is the matrix that gets modified and \(\mathbf{Z}_{\text{bus}}\) is the one that gets rebuilt — the subject of Set 18's building algorithm.

Removal is addition with the signs reversed. With \(y_{34} = 10.0 - j30.0\) and \(B/2 = 0.010\):

The diagonals collapse by 77%. Buses 3 and 4 go from being the two stiffest nodes in the network to being among the weakest:

Bus 5, untouched by the outage, is now stiffer than either of them.

The topology after the outage. With 3–4 gone, bus 3 connects only to 1 and 2, and bus 4 only to 2 and 5. Every path between them now goes through bus 2:

A path of impedance 0.12+j0.36 in place of one of 0.01+j0.03 — twelve times longer. Any power that was flowing 3–4 must now go the long way round, and the losses rise accordingly.

The check still passes: row 3 sums to \(j0.045\), which is \(0.025+0.020\) — the charging of the two lines that remain. The removed line's \(0.010\) has gone with it.

A trap worth naming. It is tempting to remove a line by setting \(Y_{34} = 0\) and stopping there. That leaves the diagonals containing the branch's admittance, so the matrix now describes a network in which the 3–4 branch has been replaced by two shunt admittances of \(10-j30\) at each end. The row sums would immediately expose it: row 3 would sum to \(10 - j29.955\) instead of \(j0.045\).

The new admittance:

Against \(5.0 - j15.0\) before. Note that the conductance has risen even though \(r\) is unchanged — because \(g = r/(r^{2}+x^{2})\) depends on the reactance too. This is the usual surprise in this calculation.

The change to apply:

Four entries, as in Problem 8 — a change of impedance is a removal followed by an addition, and the charging terms cancel:

The row sums are unchanged at \(j0.055\) and \(j0.085\), because only the series admittance moved and the series terms always cancel within a row. That is a useful discriminator: a change that alters the row sums has touched a shunt; one that does not has touched only a series branch.

The general rule, covering all three modifications of Problems 8 to 10 in one line. For a branch \(i\)–\(j\) whose admittance changes by \(\Delta y\) and whose half-charging changes by \(\Delta b\):

With \(\mathbf{e}_i\) the \(i\)th unit column. Addition is \(\Delta y = +y\), removal \(\Delta y = -y\), and a change of impedance is \(\Delta y = y^{\text{new}} - y^{\text{old}}\) — one formula for all three.

The rank-one structure is not decoration. \((\mathbf{e}_i-\mathbf{e}_j)(\mathbf{e}_i-\mathbf{e}_j)^{T}\) is a rank-one matrix, and a rank-one update to an inverted matrix is handled by the Sherman–Morrison formula in \(O(n^{2})\) instead of the \(O(n^{3})\) of a re-inversion. That single fact is the computational basis of the line-outage distribution factors used in every real-time contingency screening.

The convention. The element–bus incidence matrix \(\mathbf{A}\) has one row per element and one column per bus, with

The reference — ground — gets no column. An element joining a bus to ground therefore has a single \(+1\) in its row rather than a \(+1\) and a \(-1\).

The element list. Twelve elements: seven series branches and five aggregated shunts, the shunt at each bus being the sum of the half-charging of every line meeting there.

The matrix, \(12\times5\):

Every entry is 0 or \(\pm1\). \(\mathbf{A}\) contains no electrical information at all — only the topology. The impedances live entirely in the primitive matrix of the next problem, and this separation is what makes the formulation attractive to a program: topology and parameters are read from different files and combined by one matrix product.

What \(\mathbf{A}\) does. Multiplying it by the bus voltage vector gives the voltage across each element:

Row 1 gives \(V_1 - V_2\), row 8 gives \(V_1 - 0 = V_1\). And its transpose does the reverse: \(\mathbf{I}_{\text{bus}} = \mathbf{A}^{T}\mathbf{i}\) sums the element currents into each bus. Those two statements are Kirchhoff's voltage and current laws respectively, and they are all that the next two problems need.

The sign convention is arbitrary but must be consistent. Reversing a row's signs reverses that element's assumed current direction, which changes the sign of the element current but not the final \(\mathbf{Y}_{\text{bus}}\) — because \(\mathbf{A}\) appears twice in \(\mathbf{A}^{T}[\mathbf{y}]\mathbf{A}\) and the two sign changes cancel. What must not vary is the convention within a row.

The definition. \([\mathbf{y}]\) relates the current in each element to the voltage across it, before the elements are connected to one another:

For unconnected, uncoupled elements this is simply \(i_k = y_kv_k\) for each — so the matrix is diagonal, and each diagonal entry is one branch admittance.

For this network it is \(12\times12\) and diagonal:

The first seven are the series branches, the last five the shunts. Note that the shunt entries are pure positive imaginaries and the series entries have negative imaginary parts — the two families are immediately distinguishable.

It is not the network. \([\mathbf{y}]\) knows the admittance of every element and nothing about how they are joined. \(\mathbf{A}\) knows how they are joined and nothing about their admittances. Neither alone describes the system; their product does.

When it stops being diagonal. Mutual coupling between two elements — most often between two circuits of the same double-circuit tower, or between the zero-sequence networks of parallel lines sharing a right of way — puts off-diagonal terms in \([\mathbf{y}]\):

Obtained by inverting the primitive impedance matrix \([\mathbf{z}]\), which is where the mutual impedance is naturally written. That inversion is of a small block, not of the whole matrix — coupled elements come in twos and threes.

And that is when the formation by inspection fails. The by-inspection rule assumes \(Y_{ij}\) depends only on the branch joining \(i\) and \(j\). With mutual coupling, a current in one branch induces a voltage in another that may not share a bus with it at all, and the resulting \(\mathbf{Y}_{\text{bus}}\) can have non-zero entries between buses with no direct connection. Only \(\mathbf{A}^{T}[\mathbf{y}]\mathbf{A}\) handles it.

Where this matters in practice. Positive-sequence mutual coupling between transmission circuits is small — a few per cent — and is normally neglected. Zero-sequence coupling between parallel circuits is not: it can reach 60% of the self-impedance, and neglecting it produces earth-fault currents and relay reach settings that are substantially wrong. Set 22 returns to this.

The three statements needed:

Substitute forwards:

Three lines. The transformation is called singular because \(\mathbf{A}\) is rectangular — 12 by 5 here — and so has no inverse; the map from twelve element quantities to five bus quantities loses information and cannot be undone.

The element-by-element form. Writing out the product:

With \([\mathbf{y}]\) diagonal this collapses to \(Y_{ij} = \sum_k A_{ki}y_kA_{kj}\) — a single sum over elements.

Verifying \(Y_{11}\). Column 1 of \(\mathbf{A}\) is non-zero for elements 1 (branch 1–2), 2 (branch 1–3) and 8 (shunt at bus 1), with \(A_{k1} = +1\) in each. So \(A_{k1}^{2} = 1\) and

Matching Problem 5. The transformation has reproduced the by-inspection rule: \(A_{ki}^{2} = 1\) for every element touching bus \(i\), so the diagonal is a plain sum.

Verifying \(Y_{12}\). An off-diagonal needs elements with a non-zero entry in both columns 1 and 2 — only element 1, the branch 1–2, with \(A_{11} = +1\) and \(A_{12} = -1\):

The minus sign in \(Y_{ij} = -y_{ij}\) is the product of the \(+1\) and the \(-1\) — it was never a convention, but a consequence of the incidence signs.

And \(Y_{14} = 0\) for the same reason: no element has non-zero entries in both column 1 and column 4, so the sum is empty. The zeros of \(\mathbf{Y}_{\text{bus}}\) are the bus pairs sharing no element.

Why bother, when inspection is faster. Three reasons, none of them about speed:

The model. An ideal transformer of ratio \(a{:}1\) in series with the admittance \(y\), with a fictitious node \(t\) between them:

The current relation carries the conjugate. For a real tap it makes no difference; for the complex tap of Problem 16 it makes all the difference, so it is written in from the start.

The current through the admittance:

Refer it to bus \(i\):

Using \(aa^{*} = |a|^{2}\).

And at bus \(j\), where the current leaving is \(-I_t\):

The four contributions:

For a real tap, \(a = a^{*}\) and the matrix is symmetric: \(Y_{ii} = y/a^{2}\), \(Y_{ij} = Y_{ji} = -y/a\), \(Y_{jj} = y\). Note the asymmetry between the two diagonals — the tapped side carries the \(a^{2}\), the other does not.

The equivalent \(\pi\). For a real tap, match a \(\pi\) circuit of series \(y_s\) and shunts \(y_{si}\), \(y_{sj}\) against those four entries:

The two shunts have opposite signs, since \((1-a)\) and \((a-1)\) differ by a minus. A tap above nominal puts a capacitive shunt on the tapped side and an inductive one on the other; below nominal, the reverse. Neither corresponds to any physical component — they are the bookkeeping that lets an ideal transformer be represented in an admittance matrix that has no place for one.

At \(a = 1.05\), with \(a^{2} = 1.1025\):

The equivalent \(\pi\):

A capacitive \(+j0.227\) at the tapped bus and an inductive \(-j0.238\) at the other. Check: \(-j4.762 + j0.227 = -j4.535 = Y_{22}\) and \(-j4.762 - j0.238 = -j5.0 = Y_{44}\). Both confirm.

At \(a = 0.975\), with \(a^{2} = 0.950625\):

Both shunts have reversed sign, as they must.

Reading the physics. The interpretation runs through the series element, not the shunts:

Raising the tap on the bus 2 side raises the voltage at bus 4 relative to bus 2 — which is the whole purpose of the device and the reason it is the network's principal voltage-control tool.

The effect on the complete matrix. With line 2–4 removed and this transformer inserted, the row sums become \(+j0.29176\) at bus 2 and \(-j0.2031\) at bus 4, against \(+j0.085\) and \(+j0.055\) before. Neither is a shunt element that exists — they are the tap's fictitious shunts plus the remaining line charging, and a negative row sum in a network with no reactors is the signature of an off-nominal tap.

And why the tap is a load-flow variable, not a parameter. \(a\) appears inside \(\mathbf{Y}_{\text{bus}}\), so an on-load tap changer regulating a bus voltage changes the matrix at every iteration. Load-flow programs treat tap-controlled buses by adjusting \(a\) between iterations and rebuilding the affected four entries — cheap, because Problem 8 showed the modification is local.

The four entries, from the general result of Problem 14 with \(y = -j5.0\) and \(|a| = 1\):

\(Y_{ij} \ne Y_{ji}\). They have the same magnitude and conjugate angles — the real parts differ in sign. The admittance matrix of a network containing a phase shifter is not symmetric, and every algorithm that assumes symmetry to halve its storage or its arithmetic must be told about it.

No \(\pi\) can reproduce this. A \(\pi\) circuit of passive elements has \(Y_{ij} = Y_{ji} = -y_s\) by construction — the series element is the same element seen from either side. Two unequal off-diagonals cannot be matched by any choice of three passive admittances, so the phase shifter has no lumped equivalent and must be carried in the matrix as four independent numbers.

The physical statement. The device is non-reciprocal: a volt applied at bus \(i\) produces a different current at bus \(j\) than the same volt applied at \(j\) produces at \(i\). Reciprocity — the \(AD-BC = 1\) of Set 11 and the symmetry of every matrix so far — holds for all passive bilateral elements and fails here because the ideal transformer's complex ratio treats the two directions differently.

What it is for. A tap changer moves reactive power by changing a voltage magnitude; a phase shifter moves real power by changing an angle:

Inserting \(\alpha = 3^\circ\) into a branch whose natural angle difference is 5° raises its flow by 60%, without touching a single generator. It is the only device that controls the division of power between parallel paths, which is otherwise fixed entirely by the impedances.

The scale of the effect here. Between buses 2 and 4 the natural angle difference in the load flow of Set 19 turns out to be under 2°. A 3° phase shifter would therefore more than double that branch's flow — which is why phase shifters are rated in degrees of a few units and why their control is slow and stepped.

The general case combines both: \(a = |a|\angle\alpha\) with \(|a| \ne 1\) gives a device that controls magnitude and angle together. Then \(Y_{ii} = y/|a|^{2}\) differs from \(Y_{jj} = y\) and the off-diagonals differ from each other — all four entries independent, which is the most general two-terminal element a power network contains.

The count here. Five diagonals, always non-zero, plus two entries for each of the seven branches:

76% occupied — hardly sparse. A five-bus network is too small to show the effect.

The general expression. Real transmission networks have a branch-to-bus ratio of roughly 1.5, so \(e \approx 1.5n\) and

The fill falls as \(1/n\). This is the crucial fact: a bus has a fixed number of neighbours no matter how large the system is, because it is a physical substation with a physical number of circuits leaving it.

The numbers:

The storage consequence. A 10 000-bus matrix stored as a full array of double-precision complex numbers needs

A factor of 2500. Storing the zeros is not merely wasteful; it is the difference between a study that runs and one that does not.

The solution consequence is larger still. Gaussian elimination on a full matrix costs \(O(n^{3})\); on a sparse one with good ordering it is close to \(O(n)\):

Seven orders of magnitude. Real-time contingency analysis, which solves thousands of such systems a minute, exists only because of this.

The catch is fill-in. Elimination creates non-zeros where there were none: eliminating bus \(k\) connects every pair of its neighbours. Eliminating a bus with \(d\) neighbours creates up to \(d(d-1)/2\) new entries, so the order matters enormously — eliminating the hub bus 2 first, with four neighbours, would create six new terms, while eliminating bus 1 or 5 first, with two each, creates one. Eliminating low-degree buses first is the standard heuristic, and it is what makes the count above achievable in practice rather than merely in principle.

The definition. A matrix is diagonally dominant if

Strictly dominant if the inequality is strict for at least one row and the matrix is irreducible.

Without charging. Then \(Y_{ii} = \sum_{j\ne i}y_{ij}\) exactly, so the test compares \(\left|\sum y_{ij}\right|\) with \(\sum|y_{ij}|\):

Equality in every row — dominant, but only weakly.

Why exactly equal. The triangle inequality gives \(\left|\sum y\right| \le \sum|y|\) with equality only when all the terms share an angle. In this test system every line has \(x/r = 3\), so every admittance lies at \(-71.57^\circ\) and the sum's magnitude is the sum of the magnitudes. In a real network with mixed \(x/r\) ratios the diagonal would be strictly smaller than the off-diagonal sum, and the equality here is an artefact of the data.

With charging the diagonal moves in the wrong direction, because the shunt is capacitive and subtracts from an inductive diagonal:

Every row now fails the test, by about a quarter of one per cent.

The implication, stated carefully. Diagonal dominance is a sufficient condition for the convergence of Gauss–Seidel, not a necessary one:

The load flow of Set 19 converges on this matrix in about 20 iterations. The failed test predicts nothing, and it would be wrong to conclude from it that the method is unsafe here.

What actually governs convergence. The spectral radius of the iteration matrix, which for Gauss–Seidel on a load flow depends on the nonlinearity as much as on \(\mathbf{Y}_{\text{bus}}\) — the load-flow equations are not linear, and the linear-algebra theorems apply only to the linear system solved at each step. In practice Gauss–Seidel fails on power networks for a different reason entirely: heavily loaded systems near the nose of Problem 17 in Set 15, where the Jacobian is near-singular and no first-order method converges usefully.

The calculation is one matrix–vector product:

Bus 1 in full:

All five:

Buses 4 and 5 give the check. Both connect only to buses at 1.0, so no current flows in any of their series branches and the injection is purely their own shunt:

Exactly the row sums of Problem 6 — the row-sum test, appearing again as a special case of this calculation.

Reading the real powers. Bus 1, held 6% high, exports 0.3975 pu — 39.75 MW — into the network. Buses 2 and 3 absorb 30 MW and 7.5 MW, and buses 4 and 5 absorb nothing at all, being too far from the disturbance to feel it at a flat start. The sum \(0.3975 - 0.30 - 0.075 = 0.0225\) pu is the 2.25 MW of loss.

What this is not. These are not the load flow's answer. The load flow specifies the injections and solves for the voltages; here the voltages were assumed and the injections computed — the inverse problem, and the easy one. The flat start's injections bear no resemblance to the actual generation and load, and the whole business of Sets 19 and 20 is adjusting the voltages until they do.

Where this calculation is used for real. Three places, all of them important: computing the mismatch vector at each load-flow iteration; computing bus injections from a state estimator's voltage solution; and computing fault currents once \(\mathbf{Z}_{\text{bus}}\) has given the faulted voltage profile. Every one of them is this same product.

Step 1 — remove line 2–4 (\(y = 1.6667-j5.0\), \(B/2 = 0.020\)) from the original:

Step 2 — insert the transformer, using the contributions computed in Problem 15:

Step 3 — add the capacitor at bus 5:

The matrix:

Note that \(Y_{24}\) is now purely imaginary — the transformer was specified as a pure reactance, so its contribution has no real part.

The verification, and it is the interesting part:

Accounting for rows 2 and 4. Row 2 should equal its remaining line charging plus the transformer's fictitious shunt:

Both reconcile exactly. The negative row sum at bus 4 is entirely the transformer's inductive fictitious shunt overwhelming the line charging, and it is correct — not a sign that anything is wrong.

The matrix is still symmetric, because the tap is real. Had it been the phase shifter of Problem 16, \(Y_{24}\) and \(Y_{42}\) would differ and the verification would have to be done on both rows and columns.