Reactance and Impedance Diagrams

Set 4 — Reactance and Impedance Diagrams

Twenty worked problems on the drawing that every later calculation starts from. A single-line diagram is a picture of plant; a reactance diagram is a circuit. Getting from one to the other means choosing bases, propagating them through every transformer, converting each machine and line to per-unit, and then — because the ideal transformers have vanished — treating the whole thing as one ordinary series-parallel network. The last third of the set does exactly that and reduces it to a fault.

Textbook Chapter 4 · 20 solved · 12 practice · 3 challenge · 12 MCQs

i Method Recap

One base MVA, one base kV per zone. Transformers divide the system into zones; the base MVA is common to all of them and the base kV changes at each transformer in that transformer's own nameplate ratio. Set the bases first, before converting a single element.
Every element is converted by the same two rules. An impedance quoted in ohms is divided by \(Z_B = \text{kV}_B^2/\text{MVA}_B\) of its own zone. An impedance quoted in per-unit on its own rating is moved with \(X_{pu}^{\text{new}} = X_{pu}^{\text{old}}(\text{MVA}_{new}/\text{MVA}_{old})(\text{kV}_{old}/\text{kV}_{new})^2\).
The transformer disappears; the machine may not. A transformer whose nameplate ratio matches the ratio of the zone bases contributes only its impedance. A machine whose rated voltage differs from its zone's base voltage picks up a genuine \((\text{kV}_{old}/\text{kV}_{new})^2\) factor — that is not an error, it is the machine being described at other than its rated voltage.
Reactance diagram or impedance diagram? The impedance diagram keeps \(R\); the reactance diagram discards it, along with magnetising branches and static loads. For fault studies the discard is justified because \(X \gg R\) in transmission plant and the error is second-order — Problem 16 quantifies it. For load flow and loss calculation it is not justified at all.
Once converted, it is only a circuit. Series and parallel combination, delta–star conversion and Thévenin's theorem all apply unchanged. The whole point of the per-unit conversion was to earn the right to say that.
Reduce towards the fault, not away from it. Collapse the network to a single Thévenin reactance seen from the faulted bus with all emfs set to \(1.0\angle0^\circ\). Then \(I_f = 1/X_{th}\) per-unit and \(\text{MVA}_{sc} = \text{MVA}_B/X_{th}\).
Motors are sources during a fault. A synchronous or induction motor near the fault feeds current into it from its own stored magnetic energy and rotating inertia. Omitting motor contribution understates the fault duty, sometimes badly — Problem 19 finds it more than triples the current.

VideoWalkthrough

Problem 1Warm-upBase Impedance

A transmission line of reactance \(j40\ \Omega\) operates at 138 kV. Taking a base of 50 MVA, find the base impedance of that zone and the per-unit reactance of the line.

Solution

Base impedance of the 138 kV zone:

\[ Z_B = \frac{(\text{kV}_B)^{2}}{\text{MVA}_B} = \frac{(138)^{2}}{50} = 380.88\ \Omega \]

The line reactance is quoted in ohms, so it is simply divided by the base of its own zone:

\[ X_{pu} = \frac{\text{actual reactance}}{\text{base impedance}} = \frac{40}{380.88} = 0.105\ \text{p.u.} \]

A sanity check against the bands of Set 3: a transmission line on a 50 MVA base should land somewhere between roughly 0.02 and 0.5 p.u. At 0.105 this one is comfortably inside, so no base error is indicated.

Every element on a reactance diagram enters by one of exactly two doors. Ohms go through \(Z_B\) of their own zone, as here. Per-unit values on a nameplate go through the base-change formula. Nothing else is ever needed, and being able to say which door an element came through is the fastest way to check a diagram someone else has drawn.

Answer\(Z_B = 380.88\,\Omega\), \(X_{pu} = 0.105\) p.u.

Problem 2Exam levelWhole-System Conversion

A 50 Hz system has the following element reactances, each measured in ohms in its own zone. Draw the reactance diagram on a base of 30 MVA and 33 kV in the transmission zone, the generator zones being at 11 kV, 6.2 kV and 6.2 kV respectively.

Transmission line: \(20.5\ \Omega\) at 33 kV
Transformer \(T_1\): \(15.2\ \Omega\) referred to 33 kV
Transformer \(T_2\): \(16\ \Omega\) referred to 33 kV
Generator \(G_1\): \(1.6\ \Omega\) at 11 kV
Generator \(G_2\): \(1.2\ \Omega\) at 6.2 kV
Generator \(G_3\): \(0.56\ \Omega\) at 6.2 kV

Solution

Everything here is in ohms, so every element uses the same relation with the base impedance of its own zone. Writing it once in the convenient rearranged form:

\[ X_{pu} = \frac{X_\Omega}{\text{kV}_B^{2}/\text{MVA}_B} = \frac{X_\Omega \times \text{MVA}_B}{\text{kV}_B^{2}} \]

In this arrangement the base MVA multiplies and the base kV squared divides — worth writing out once and reusing, since six conversions follow.

Transmission zone, \(\text{kV}_B = 33\), so \(\text{kV}_B^2 = 1089\):

\[ X_{\text{line}} = \frac{20.5 \times 30}{1089} = 0.564\ \text{p.u.} \]

\[ X_{T1} = \frac{15.2 \times 30}{1089} = 0.418\ \text{p.u.}, \qquad X_{T2} = \frac{16 \times 30}{1089} = 0.440\ \text{p.u.} \]

Generator \(G_1\), whose zone is at 11 kV so \(\text{kV}_B^2 = 121\):

\[ X_{G1} = \frac{1.6 \times 30}{121} = 0.396\ \text{p.u.} \]

Generators \(G_2\) and \(G_3\), whose zone is at 6.2 kV so \(\text{kV}_B^2 = 38.44\):

\[ X_{G2} = \frac{1.2 \times 30}{38.44} = 0.936\ \text{p.u.}, \qquad X_{G3} = \frac{0.56 \times 30}{38.44} = 0.437\ \text{p.u.} \]

The reactance diagram is then six reactances of these values, connected in the topology of the single-line diagram, with each generator behind its own reactance driving \(1.0\angle0^\circ\) and all transformer symbols erased.

The base voltage was different in three of the six conversions, and that is the whole difficulty of this problem. Every element was divided by \(\text{kV}_B^2\) — but by the \(\text{kV}_B\) of its own zone, not the 33 kV of the base declaration. Using 33 kV throughout would have made \(G_2\) come out at 0.033 instead of 0.936, a factor of 28, and nothing in the arithmetic would have looked wrong. Marking the zone boundaries on the single-line diagram before converting anything is not fussiness; it is the only reliable defence.

AnswerLine 0.564, \(T_1\) 0.418, \(T_2\) 0.440, \(G_1\) 0.396, \(G_2\) 0.936, \(G_3\) 0.437 p.u.

Problem 3Warm-upBase Propagation

A base of 50 MVA, 138 kV is declared in the transmission zone of the system below. Find the base voltage in every other zone.

\(T_1\): 20 MVA, 138/20 kV, feeding generator \(G_1\)
\(T_2\): 20 MVA, 138/20 kV, feeding generator \(G_2\)
\(T_5\): 15 MVA, 138/13.8 kV, feeding a synchronous motor

Solution

The base MVA is 50 everywhere — it does not change at a transformer and it does not change anywhere else. Only the base voltage moves, and it moves in the transformer's nameplate ratio:

\[ \text{kV}_{B,\text{LT}} = \text{kV}_{B,\text{HT}} \times \frac{\text{LT voltage rating}}{\text{HT voltage rating}} \]

Through \(T_1\) and \(T_2\), both 138/20 kV:

\[ \text{kV}_B = 138 \times \frac{20}{138} = 20\ \text{kV} \]

Through \(T_5\), 138/13.8 kV:

\[ \text{kV}_B = 138 \times \frac{13.8}{138} = 13.8\ \text{kV} \]

So the four zones are:

\[ \begin{array}{lcc} \text{Zone} & \text{MVA}_B & \text{kV}_B \\ \hline \text{Generator } G_1 & 50 & 20 \\ \text{Generator } G_2 & 50 & 20 \\ \text{Transmission} & 50 & 138 \\ \text{Motor} & 50 & 13.8 \end{array} \]

Note that the generators are rated at 18 kV, not 20. Their zone base is nevertheless 20 kV, because the base is set by the transformer, never by the machine. That mismatch is genuine and is dealt with in Problem 5.

The base voltage follows the iron, not the copper. A transformer physically changes the voltage, so the base must change with it and by exactly its ratio; a generator does not, so it has no say in what its zone's base is. Once this is internalised the propagation becomes mechanical: walk outwards from the declared zone, multiply by each nameplate ratio as you cross it, and never touch the MVA.

Answer\(G_1, G_2\) zones: 20 kV · transmission: 138 kV · motor: 13.8 kV · all at 50 MVA

Problem 4Warm-upTransformer Conversion

Transformer \(T_1\) of Problem 3 is rated 20 MVA, 138/20 kV with \(X = 0.10\) p.u. on its own rating. Find its reactance on the 50 MVA, 138 kV base.

Solution

Apply the base-change formula. Working on the LT side, where the zone base is 20 kV and the transformer's own rating is also 20 kV:

\[ X_{pu}^{\text{new}} = X_{pu}^{\text{old}}\left(\frac{\text{MVA}_{B,new}}{\text{MVA}_{B,old}}\right)\left(\frac{\text{kV}_{B,old}}{\text{kV}_{B,new}}\right)^{2} \]

Substituting:

\[ X_{T1} = 0.10\left(\frac{50}{20}\right)\left(\frac{20}{20}\right)^{2} = 0.10(2.5)(1) = 0.25\ \text{p.u.} \]

Working on the HT side instead must give the same answer, and does — the base there is 138 kV and the rating is also 138 kV:

\[ X_{T1} = 0.10\left(\frac{50}{20}\right)\left(\frac{138}{138}\right)^{2} = 0.25\ \text{p.u.}\ \checkmark \]

The voltage ratio came out as unity on both sides, and that is the signature of a correctly chosen base. Because the zone bases were propagated through this transformer, its nameplate voltages necessarily match the bases on both sides, so only the MVA ratio can act. Whenever a transformer's voltage factor is not unity, the bases were not propagated through it — and the residual factor is the off-nominal ratio of Problem 18, which is a real feature of the network rather than a mistake.

Answer\(X_{T1} = 0.25\) p.u.

Problem 5Exam levelOff-Base Machine

Generator \(G_1\) is rated 20 MVA, 18 kV with \(X'' = 0.20\) p.u. on its own rating, and sits in a zone whose base is 50 MVA, 20 kV. Find its reactance on the system base, and explain why the voltage factor is not unity here when it was in Problem 4.

Solution

The machine's own base is 20 MVA and 18 kV; the zone base is 50 MVA and 20 kV. Both factors act:

\[ X_{G1} = 0.20\left(\frac{50}{20}\right)\left(\frac{18}{20}\right)^{2} \]

Evaluating each factor separately, since this is where sign and direction errors occur:

\[ \frac{50}{20} = 2.5, \qquad \left(\frac{18}{20}\right)^{2} = (0.9)^{2} = 0.81 \]

\[ X_{G1} = 0.20(2.5)(0.81) = 0.405\ \text{p.u.} \]

Why the factor is not unity. The zone base of 20 kV was fixed by transformer \(T_1\), whose LT winding is rated 20 kV. The generator connected to that winding happens to be an 18 kV machine. Nothing is wrong with either: an 18 kV generator feeding a transformer wound for 20 kV simply operates its transformer below nominal voltage, which is entirely ordinary.

Confirming through ohms, which must be invariant. On its own base \(Z_B = 18^2/20 = 16.2\ \Omega\), so \(X'' = 0.20(16.2) = 3.24\ \Omega\). On the zone base \(Z_B = 20^2/50 = 8.0\ \Omega\):

\[ X''_{pu} = \frac{3.24}{8.0} = 0.405\ \text{p.u.}\ \checkmark \]

A machine has no authority over its own base voltage. Transformers set the zone bases; whatever else sits in the zone must accept them. When the machine's rating and the zone base differ, the \((\text{kV}_{old}/\text{kV}_{new})^2\) factor is doing real physical work — it is expressing a 3.24 \(\Omega\) reactance in a system whose base impedance is 8 \(\Omega\) rather than 16.2. Suppressing the factor "because the numbers are close" understates this generator's reactance by 19% and its fault contribution correspondingly.

Answer\(X_{G1} = 0.405\) p.u.

Problem 6Warm-upLine Sections

The system of Problem 3 contains a \(j40\ \Omega\) line and a \(j20\ \Omega\) line, both in the 138 kV zone. Find the per-unit reactance of each on the 50 MVA base, and state what happens if two line sections of equal ohmic reactance sit in zones of the same base voltage.

Solution

Both lines occupy the same zone, so both use \(Z_B = 380.88\ \Omega\) from Problem 1:

\[ X_{40\Omega} = \frac{40}{380.88} = 0.105\ \text{p.u.}, \qquad X_{20\Omega} = \frac{20}{380.88} = 0.0525\ \text{p.u.} \]

The two are in the ratio 2:1, exactly as their ohmic values are — dividing both by the same base preserves ratios.

Hence the general statement: two sections of equal ohmic reactance lying in zones of the same base voltage have identical per-unit reactance, and need only be computed once.

Per-unit conversion is a scaling, and scalings preserve ratios within a zone. That makes a whole class of arithmetic unnecessary: a 60 \(\Omega\) section in the same zone is \(3 \times 0.0525 = 0.1575\) p.u. without touching \(Z_B\) again. Ratios are only broken when a zone boundary is crossed, which is another reason to mark the boundaries before starting.

Answer\(0.105\) and \(0.0525\) p.u.; equal ohms in the same zone give equal p.u.

Problem 7Exam levelMotor and its Transformer

A synchronous motor rated 30 MVA, 13.8 kV with \(X'' = 0.20\) p.u. is fed through transformer \(T_5\) rated 15 MVA, 138/13.8 kV with \(X = 0.10\) p.u. Find both reactances on the 50 MVA, 138 kV base.

Solution

From Problem 3 the motor zone base is 13.8 kV, which matches both the motor's rating and the transformer's LT rating. Every voltage factor will therefore be unity and only the MVA ratios act.

Transformer \(T_5\), rated 15 MVA:

\[ X_{T5} = 0.10\left(\frac{50}{15}\right)\left(\frac{138}{138}\right)^{2} = 0.10(3.333) = 0.333\ \text{p.u.} \]

Motor, rated 30 MVA:

\[ X_M = 0.20\left(\frac{50}{30}\right)\left(\frac{13.8}{13.8}\right)^{2} = 0.20(1.667) = 0.333\ \text{p.u.} \]

The two happen to be equal, which is a coincidence of the ratings rather than anything structural: the transformer is half the motor's MVA but has half its per-unit reactance to start with.

Note that the transformer is rated 15 MVA while the motor it feeds is rated 30. That is not an error in the data — a motor's MVA rating is its own short-circuit base, set by its winding and its sub-transient reactance, not the load it actually draws. The transformer is sized for the running load; the motor's rating matters only when it turns into a source, as it does in Problem 19. Sizing the transformer from the motor's nameplate MVA would double its cost for no operational benefit.

Answer\(X_{T5} = X_M = 0.333\) p.u.

Problem 8Exam levelComplete Diagram

Assemble the complete reactance diagram of the system below on a base of 50 MVA, 138 kV, using the results of Problems 1 and 4 to 7. This system is used again in Problems 9, 13, 14, 15, 19 and 20.

\(G_1\): 20 MVA, 18 kV, \(X'' = 0.20\) p.u. — \(T_1\): 20 MVA, 138/20 kV, \(X = 0.10\) — \(j40\ \Omega\) line
\(G_2\): 20 MVA, 18 kV, \(X'' = 0.20\) p.u. — \(T_2\): 20 MVA, 138/20 kV, \(X = 0.10\) — \(j20\ \Omega\) line
Motor: 30 MVA, 13.8 kV, \(X'' = 0.20\) p.u. — \(T_5\): 15 MVA, 138/13.8 kV, \(X = 0.10\)

Two generators and a synchronous motor on a common 138 kV bus

Solution

Collecting every conversion already made, all on the 50 MVA base:

\[ \begin{array}{lccl} \text{Element} & \text{Own rating} & \text{Zone kV}_B & X_{pu} \\ \hline G_1,\ G_2 & 20\ \text{MVA},\ 18\ \text{kV} & 20 & 0.405 \\ T_1,\ T_2 & 20\ \text{MVA} & 138/20 & 0.250 \\ j40\ \Omega\ \text{line} & — & 138 & 0.105 \\ j20\ \Omega\ \text{line} & — & 138 & 0.0525 \\ T_5 & 15\ \text{MVA} & 138/13.8 & 0.333 \\ \text{Motor} & 30\ \text{MVA},\ 13.8\ \text{kV} & 13.8 & 0.333 \end{array} \]

The diagram is now a single connected network with no transformers in it. Three sources, each an emf of \(1.0\angle0^\circ\) behind its own reactance, all referred to one common node:

\[ \text{Branch 1: } G_1 \to \text{bus} = 0.405 + 0.250 + 0.105 = 0.760 \]

\[ \text{Branch 2: } G_2 \to \text{bus} = 0.405 + 0.250 + 0.0525 = 0.7075 \]

\[ \text{Branch 3: } \text{bus} \to \text{motor} = 0.333 \ \text{(then } 0.333 \text{ to the motor emf)} \]

Every element in each branch is simply in series, because the transformers that separated the voltage levels have been erased. That is the whole return on the per-unit conversion, and it is worth pausing on: a machine at 18 kV, a transformer, and a line at 138 kV have just been added together like three resistors.

Three things vanished from the single-line diagram and one thing was added. Gone are the transformer symbols, the voltage levels and the units. Added is the assumption that every emf is \(1.0\angle0^\circ\) — the flat-start approximation, exact only if the system was unloaded before the disturbance. For fault studies this is standard and its error is small, since pre-fault currents are an order of magnitude below fault currents. For anything else, the pre-fault load flow must be superposed.

AnswerBranch 1 = 0.760, Branch 2 = 0.7075, \(T_5\) = 0.333, motor = 0.333 p.u.

Problem 9Warm-upNetwork Reduction

Reduce the two generator branches of Problem 8 to a single reactance seen from the 138 kV bus.

Solution

Both generators drive the same bus through their own series chains, and both emfs are \(1.0\angle0^\circ\). Two equal sources in parallel behave as one source of the same emf behind the parallel combination of their reactances:

\[ X_{12} = \frac{X_1X_2}{X_1 + X_2} = \frac{(0.760)(0.7075)}{0.760 + 0.7075} \]

Evaluating:

\[ X_{12} = \frac{0.5377}{1.4675} = 0.366\ \text{p.u.} \]

A check worth making: the result must lie below the smaller of the two, and above half the smaller. Here \(0.354 < 0.366 < 0.7075\), so it does.

Equal emfs are what make this legitimate. Two sources of different emf in parallel cannot be collapsed this way — they would circulate current between themselves, and the correct reduction is Millman's theorem, giving an equivalent emf that is the admittance-weighted mean. The flat-start assumption of Problem 8, which set every emf to \(1.0\angle0^\circ\), is precisely what buys the simple parallel formula, and it is the main reason fault studies are so much easier than load flows.

Answer\(X_{12} = 0.366\) p.u. from the 138 kV bus

Problem 10Challenge-liteWhich Diagram?

Distinguish the impedance diagram from the reactance diagram. State precisely what is discarded in passing from one to the other, and for which studies each discard is admissible.

Solution

The impedance diagram retains, for every element, the full series impedance \(R + jX\), together with the shunt branches — line charging capacitance and transformer magnetising admittance — and represents static loads by their impedances.

The reactance diagram discards, in this order:

\[ \begin{array}{ll} \text{1. Resistance of all elements} & R \ll X \text{ in transmission plant} \\ \text{2. Transformer magnetising branch} & I_m \approx 1\text{–}2\%\ \text{of rated current} \\ \text{3. Line charging capacitance} & \text{negligible current against fault current} \\ \text{4. Static loads} & \text{impedance} \gg \text{source impedance} \end{array} \]

Why each discard is admissible for fault studies. A fault collapses the voltage at the fault point, so the current is set by the source impedance alone. Every branch that was carrying a small current in proportion to the system voltage — magnetising, charging, load — now carries a current negligible against the fault current, which is set by a much smaller impedance. Discarding them changes the answer by well under one per cent.

Why resistance in particular may go. For transmission plant \(X/R\) runs from about 10 for a line to 30 or more for a large machine. The magnitude of \(R + jX\) exceeds \(X\) by \(\sqrt{1 + (R/X)^2}\), which at \(X/R = 10\) is 1.005 — a half per cent, and always in the direction of overstating the fault current, which is the safe direction for rating switchgear.

Where the discards are not admissible. Load flow needs resistance, since it determines the real power loss and the whole of the \(P\)–\(\delta\) relationship at distribution voltages. Loss and efficiency calculations obviously need it. Distribution networks with \(X/R\) near unity need it. And the d.c. offset and time constant of a fault current depend on \(X/R\) explicitly, so breaker duty calculations need it even though the symmetrical current does not.

The reactance diagram is not a simplification of the impedance diagram — it is a different tool. Each discard is justified by a specific inequality that holds in a specific study. Carrying a reactance diagram into a load flow gives lossless lines and nonsense; carrying a full impedance diagram into a hand fault calculation gives complex arithmetic for a half-per-cent improvement. Knowing which inequalities are being invoked is what makes the choice deliberate rather than habitual.

AnswerReactance diagram drops \(R\), magnetising and charging branches and static loads — valid for fault studies, invalid for load flow and loss calculation

Problem 11Exam levelLoad as a Branch

A load of 30 MVA at 0.9 power factor lagging is connected at a 13.8 kV bus whose base is 50 MVA, 13.8 kV. Represent it as a shunt impedance on the impedance diagram, assuming the bus voltage is 1.0 p.u.

Solution

Express the load as a per-unit complex power:

\[ S_{pu} = \frac{30\angle 25.84^\circ}{50} = 0.6\angle 25.84^\circ\ \text{p.u.} \]

The angle is \(\cos^{-1}(0.9) = 25.84^\circ\), positive because the load is lagging and therefore absorbing vars.

The shunt impedance that draws this power at 1.0 p.u. follows from \(S = V^2/Z^{*}\):

\[ Z = \frac{|V|^{2}}{S^{*}} = \frac{1.0^{2}}{0.6\angle -25.84^\circ} = 1.667\angle 25.84^\circ\ \text{p.u.} \]

In rectangular form, which is what goes on the diagram:

\[ Z = 1.667(\cos 25.84^\circ + j\sin 25.84^\circ) = 1.500 + j0.727\ \text{p.u.} \]

Compare this with the source reactances of Problem 8, which run from 0.05 to 0.41 p.u. The load impedance is four to thirty times larger.

That size comparison is why loads are dropped from a reactance diagram. A branch of 1.67 p.u. sitting in parallel with a source path of 0.23 p.u. diverts about 12% of the current — but during a fault the bus voltage collapses towards zero, so this constant-impedance branch draws almost nothing while the fault path draws everything. Retaining it would change the fault current by well under one per cent and would cost a complex parallel combination at every bus. For a load flow, where the voltage stays near 1.0 p.u., the same branch is the entire point of the calculation.

Answer\(Z_{\text{load}} = 1.500 + j0.727 = 1.667\angle25.84^\circ\) p.u.

Problem 12Warm-upDelta Load

A delta-connected load of \((45 + j30)\ \Omega\) per phase is connected at a 13.8 kV bus. On a base of 50 MVA, 13.8 kV, find its per-unit impedance.

Solution

Per-unit impedance is a per-phase, star-referred quantity, so the delta must be converted first. For a balanced load:

\[ Z_Y = \frac{Z_\Delta}{3} = \frac{45 + j30}{3} = (15 + j10)\ \Omega \]

Base impedance of the zone:

\[ Z_B = \frac{(13.8)^{2}}{50} = \frac{190.44}{50} = 3.809\ \Omega \]

Hence:

\[ Z_{pu} = \frac{15 + j10}{3.809} = 3.938 + j2.626\ \text{p.u.} \]

\[ |Z_{pu}| = \frac{|15 + j10|}{3.809} = \frac{18.03}{3.809} = 4.733\ \text{p.u.} \]

Converting after dividing by \(Z_B\) would give \((45+j30)/3.809 = 11.81 + j7.88\) and then a division by 3 — the same answer, since both operations are scalings. Converting not at all gives three times too much and is the error to guard against.

Everything in per-unit is per phase and star-referred, always. The base impedance \(\text{kV}_{LL}^2/\text{MVA}_{3\phi}\) was constructed in Set 3 Problem 16 to be exactly the line-to-neutral impedance base, so a delta impedance simply is not the quantity it divides. This applies equally to delta-connected capacitor banks, delta motor windings and delta transformer windings — divide by three before dividing by \(Z_B\).

Answer\(Z_{pu} = 3.938 + j2.626 = 4.733\angle33.7^\circ\) p.u.

Problem 13Warm-upParallel Lines

The \(j40\ \Omega\) and \(j20\ \Omega\) lines are now connected between the same pair of 138 kV buses, in parallel. Find the combined per-unit reactance, and verify the result in ohms.

Solution

From Problem 6 the two per-unit values are 0.105 and 0.0525. In parallel:

\[ X = \frac{(0.105)(0.0525)}{0.105 + 0.0525} = \frac{0.005513}{0.1575} = 0.035\ \text{p.u.} \]

Verifying in ohms — combine first, convert second:

\[ X_\Omega = \frac{(40)(20)}{60} = 13.33\ \Omega, \qquad \frac{13.33}{380.88} = 0.035\ \text{p.u.}\ \checkmark \]

The two routes agree because both lines lie in the same zone and therefore share a base impedance. Had they been at different voltage levels the ohmic combination would have been meaningless, while the per-unit one would still have been correct.

Per-unit combination works across zone boundaries; ohmic combination does not. That is the practical superiority of the diagram, and it is why a second parallel circuit is such a favourable investment: adding a line of the same reactance halves the transfer reactance, which raises the stability limit \(P = EV\sin\delta/X\) of Set 24 by a factor of two and cuts the voltage drop in half, for the cost of one more circuit on the same towers.

Answer\(X = 0.035\) p.u., equivalently \(13.33\ \Omega\)

Problem 14Exam levelThévenin Reactance

For the system of Problem 8, find the Thévenin reactance seen from the motor's 13.8 kV terminals, (a) with the motor omitted and (b) with the motor included as a source.

Solution

aMotor omitted. Looking back from the motor terminals, the path is \(T_5\) in series with the two generator branches in parallel. From Problem 9 the latter is 0.366:

\[ X_{\text{sys}} = 0.366 + 0.333 = 0.700\ \text{p.u.} \]

bMotor included. The motor is itself an emf of \(1.0\angle0^\circ\) behind 0.333 p.u., connected directly at the terminals in question. Setting all emfs to zero to find the Thévenin impedance, its reactance appears in parallel with the system:

\[ X_{th} = \frac{(0.700)(0.333)}{0.700 + 0.333} = \frac{0.2331}{1.033} = 0.226\ \text{p.u.} \]

The motor has cut the Thévenin reactance by more than two thirds — from 0.700 to 0.226.

The motor is electrically closer to its own terminals than the whole rest of the system is. Its 0.333 p.u. sits directly across the bus, while the generators reach it only through two transformers and a line totalling 0.700. During a fault the machine nearest the fault dominates, whatever its rating — a principle that recurs throughout fault analysis and is the reason a large motor bus is often the highest fault-duty point in an industrial plant, higher than the incoming supply itself.

Answer(a) \(0.700\) p.u. · (b) \(X_{th} = 0.226\) p.u.

Problem 15Exam levelFault Level

For the system of Problem 8, find the Thévenin reactance, fault MVA and symmetrical fault current for a three-phase fault on the 138 kV bus.

Solution

Seen from the 138 kV bus there are two paths: the two generators in parallel, and the motor through \(T_5\).

\[ X_{\text{gens}} = 0.366\ \text{p.u.}, \qquad X_{\text{motor path}} = 0.333 + 0.333 = 0.667\ \text{p.u.} \]

These two are in parallel at the faulted bus:

\[ X_{th} = \frac{(0.366)(0.667)}{0.366 + 0.667} = \frac{0.2442}{1.033} = 0.236\ \text{p.u.} \]

Fault MVA:

\[ \text{MVA}_{sc} = \frac{\text{MVA}_B}{X_{th}} = \frac{50}{0.236} = 211.5\ \text{MVA} \]

Base current at 138 kV, and hence the fault current:

\[ I_B = \frac{50\times10^{6}}{\sqrt3(138\,000)} = 209.2\ \text{A} \]

\[ I_f = \frac{I_B}{X_{th}} = \frac{209.2}{0.236} = 885\ \text{A} \]

Cross-check through the fault MVA:

\[ I_f = \frac{211.5\times10^{6}}{\sqrt3(138\,000)} = 885\ \text{A}\ \checkmark \]

Compare this with the 0.226 p.u. of Problem 14 and the geography becomes clear. A fault at the 138 kV bus sees 0.236 p.u.; a fault at the motor terminals sees 0.226. The motor bus is the more severely stressed of the two despite lying behind a transformer, because the motor itself sits directly on it. Fault level is a property of a location, not of a system, and the only way to find the worst location is to reduce the network afresh from each candidate bus.

Answer\(X_{th} = 0.236\) p.u., \(\text{MVA}_{sc} = 211.5\) MVA, \(I_f = 885\) A

Problem 16Challenge-liteNeglecting Resistance

A line has an impedance of \((8 + j40)\ \Omega\). Find the percentage error in fault current introduced by using the reactance alone, and repeat for elements of \(X/R = 3\) and \(X/R = 10\). State whether the error is on the safe side.

Solution

The true impedance magnitude, against the reactance alone:

\[ |Z| = \sqrt{8^{2} + 40^{2}} = \sqrt{1664} = 40.79\ \Omega, \qquad X = 40\ \Omega \]

Fault current is inversely proportional to impedance, so using \(X\) alone overstates it by

\[ \frac{I_X}{I_Z} = \frac{|Z|}{X} = \frac{40.79}{40} = 1.0198 \quad\Rightarrow\quad 1.98\% \]

The general expression is worth extracting, since it makes the dependence explicit:

\[ \frac{|Z|}{X} = \sqrt{1 + \left(\frac{R}{X}\right)^{2}} \]

Evaluating at the two other ratios:

\[ \begin{array}{lcc} X/R & |Z|/X & \text{Error} \\ \hline 3 & \sqrt{1 + 1/9} = 1.0541 & 5.41\% \\ 5 & \sqrt{1 + 1/25} = 1.0198 & 1.98\% \\ 10 & \sqrt{1 + 1/100} = 1.0050 & 0.50\% \end{array} \]

Direction of the error. Discarding \(R\) reduces the impedance, so the computed fault current is always larger than the true one. Switchgear selected on that figure is therefore over-rated rather than under-rated — the error is on the safe side.

The safe direction is what makes the approximation respectable, and the square root is what makes it small. Because the error enters as \(\sqrt{1 + (R/X)^2}\) rather than \(R/X\) itself, a 20% resistance costs only 2% in the answer. This breaks down in distribution networks, where \(X/R\) can approach or fall below unity — at \(X/R = 1\) the error is 41%, and a reactance-only calculation there is not an approximation but a mistake. It also breaks down for breaker duty even at high \(X/R\), since the d.c. offset and its decay time constant \(L/R\) depend on the resistance explicitly.

Answer1.98% at \(X/R=5\); 5.41% at 3; 0.50% at 10 — always overstating, hence safe

Problem 17Exam levelThree-Winding Transformer

A three-winding transformer has star-equivalent impedances \(Z_p = 0.02\), \(Z_s = 0.05\) and \(Z_t = 0.07\) p.u. on a base of 15 MVA, 66 kV. Convert these to a 50 MVA base and describe how the unit appears on the reactance diagram.

Solution

All three arms share the same base and the voltage bases are propagated through the transformer itself, so only the MVA ratio acts:

\[ \text{multiplier} = \frac{50}{15} = 3.333 \]

Applying it:

\[ Z_p = 0.02(3.333) = 0.0667, \quad Z_s = 0.05(3.333) = 0.1667, \quad Z_t = 0.07(3.333) = 0.2333\ \text{p.u.} \]

On the diagram the unit is a three-terminal star: a common internal node with \(Z_p\) running to the primary bus, \(Z_s\) to the secondary bus and \(Z_t\) to the tertiary bus. It is not a single series element and it cannot be drawn as one.

Checking that the pairwise impedances survive the conversion, as they must:

\[ Z_{ps} = 0.0667 + 0.1667 = 0.2333 = 0.07(3.333)\ \checkmark \]

The internal star node is fictitious and carries no physical meaning whatever. No terminal exists there, no voltage can be measured there, and — as noted in Set 3 Problem 12 — one of the three arms is frequently negative in real data. What is physical is the set of three pairwise impedances, and the star is simply the most convenient three-parameter network that reproduces them. It matters on the diagram because a tertiary winding, usually delta-connected, provides a path for zero-sequence and third-harmonic currents that the sequence networks of Set 22 depend on.

Answer\(Z_p = 0.0667\), \(Z_s = 0.1667\), \(Z_t = 0.2333\) p.u. on 50 MVA, as a three-terminal star

Problem 18Challenge-liteOff-Nominal Ratio

A transformer nominally 138/13.8 kV with \(X = 0.10\) p.u. is operated on a tap that makes its actual ratio 138/14.5 kV, while the zone bases remain 138 kV and 13.8 kV. Find the off-nominal ratio and show how the transformer must now be represented.

Solution

The bases stand in the ratio

\[ \frac{\text{kV}_{B,HT}}{\text{kV}_{B,LT}} = \frac{138}{13.8} = 10 \]

while the transformer's actual ratio is \(138/14.5 = 9.517\). The condition of Set 3 Challenge C1 — bases in the turns ratio — is therefore violated, and the ideal transformer will not vanish.

Expressing the actual ratio in per-unit. With 1.0 p.u. (138 kV) applied to the primary, the secondary produces 14.5 kV, which on a 13.8 kV base is

\[ t = \frac{14.5}{13.8} = 1.0507 \]

The representation. The transformer becomes its ordinary series reactance of 0.10 p.u. in series with an ideal transformer of ratio \(1 : t = 1 : 1.0507\). The reactance still converts by the base-change formula as usual; the ideal transformer is what remains and it must be kept.

What it does: for a primary at 1.0 p.u. and no load, the secondary sits at 1.0507 p.u. rather than 1.0 — a 5.07% boost, which is exactly what the tap was moved to achieve.

The alternative is to redeclare the zone base as 14.5 kV so that the bases are in the turns ratio. That removes the ideal transformer but forces every other element in that zone onto an awkward base, and prevents comparison with any other 13.8 kV zone. It is almost never worth doing.

The off-nominal ratio is not an error to be eliminated — it is the control variable of the power system. Every on-load tap changer in a network is a deliberately off-nominal transformer, and \(t\) is precisely the quantity a voltage-control scheme adjusts. Load-flow programs carry it as a per-branch parameter for exactly this reason, and the voltage control of Set 34 is the study of how to choose it. The per-unit system removes the nominal ratio so that what remains in the model is only the part an operator can change.

Answer\(t = 1.0507\); represented as \(j0.10\) p.u. in series with an ideal transformer \(1{:}1.0507\)

Problem 19Exam levelMotor Contribution

A three-phase fault occurs at the motor terminals of the system of Problem 8. Using superposition, find the fault current contributed separately by the motor and by the rest of the system, and hence the total. Comment on the consequence of omitting the motor.

Solution

Every source drives \(1.0\angle0^\circ\) into the same short circuit, so each contributes independently and the contributions add.

System contribution, through \(T_5\) and the two generator branches — the 0.700 p.u. of Problem 14(a):

\[ I_{\text{sys}} = \frac{1.0}{0.700} = 1.429\ \text{p.u.} \]

Motor contribution, straight through its own sub-transient reactance onto the faulted bus:

\[ I_{M} = \frac{1.0}{0.333} = 3.000\ \text{p.u.} \]

Total:

\[ I_f = 1.429 + 3.000 = 4.429\ \text{p.u.} \]

Confirming against the Thévenin route of Problem 14(b):

\[ I_f = \frac{1.0}{X_{th}} = \frac{1.0}{0.226} = 4.425\ \text{p.u.}\ \checkmark \]

The small discrepancy is rounding in \(X_{th}\) only.

Consequence of omission. Ignoring the motor gives 1.429 p.u. against a true 4.429 — an understatement by a factor of 3.1, or 68%.

A motor does not stop being a machine when it is called a load. On losing its terminal voltage, a synchronous motor is driven by its own rotor flux and inertia and behaves for the first few cycles exactly as a generator would; an induction motor does the same, though its contribution decays within two or three cycles as the rotor flux dies. Switchgear on an industrial motor bus that was rated from the incoming supply alone would be under-rated by two thirds here — which is why standards require motor contribution to be included in interrupting duty, and why the industrial bus of Problem 15 turned out to be more severely stressed than the transmission bus above it.

AnswerSystem 1.429, motor 3.000, total \(4.429\) p.u. — omitting the motor understates by 68%

Problem 20Exam levelDiagram to Amperes

Complete the study of the system of Problem 8: for the three-phase fault at the motor terminals, find the fault MVA and the fault current in amperes, broken down by contribution.

Solution

From Problem 14, the Thévenin reactance at the motor bus:

\[ X_{th} = 0.226\ \text{p.u.} \]

Fault MVA:

\[ \text{MVA}_{sc} = \frac{50}{0.226} = 221.4\ \text{MVA} \]

Base current in the 13.8 kV zone — note this is the zone of the fault, not of the base declaration:

\[ I_B = \frac{50\times10^{6}}{\sqrt3(13\,800)} = 2092\ \text{A} \]

Total fault current, from the 4.429 p.u. of Problem 19:

\[ I_f = 4.429 \times 2092 = 9265\ \text{A} \approx 9.27\ \text{kA} \]

Broken down by source:

\[ \begin{array}{lcc} \text{Source} & \text{p.u.} & \text{Amperes} \\ \hline \text{Motor} & 3.000 & 6276 \\ \text{System through } T_5 & 1.429 & 2989 \\ \hline \text{Total} & 4.429 & 9265 \end{array} \]

Cross-checking through the fault MVA:

\[ I_f = \frac{221.4\times10^{6}}{\sqrt3(13\,800)} = 9264\ \text{A}\ \checkmark \]

Look back at what this calculation crossed. Two 18 kV generators, two transformers, two transmission lines at 138 kV, a third transformer and a 13.8 kV motor — three voltage levels, seven elements — reduced to a single division and a single multiplication. Not one impedance was referred through a turns ratio, and not one \(\sqrt3\) appeared until the final conversion to amperes. That is the entire case for building the reactance diagram before attempting anything else, and everything from Set 16 onwards assumes it has been built.

Answer\(\text{MVA}_{sc} = 221.4\) MVA, \(I_f = 9.27\) kA — motor 6.28 kA, system 2.99 kA

Practice

Practice Problems

Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.

P1. Find the base impedance of a 220 kV zone on a 100 MVA base, and the per-unit reactance of a \(j75\ \Omega\) line in it.

Show answer
\(Z_B = 220^2/100 = 484\ \Omega\); \(X_{pu} = 75/484 = \mathbf{0.155}\) p.u.
P2. A base of 100 MVA, 11 kV is set at a generator. A 11/220 kV transformer then feeds a 220/66 kV transformer. Give the base voltage in all three zones.

Show answer
\(\mathbf{11}\), \(\mathbf{220}\) and \(\mathbf{66}\) kV — base MVA stays at 100 throughout.
P3. A 40 MVA, 11/132 kV transformer has \(X = 0.12\) p.u. Find its reactance on a 100 MVA base with bases of 11 and 132 kV.

Show answer
\(0.12(100/40)(1)^2 = \mathbf{0.30}\) p.u. The voltage factor is unity because the bases were propagated through this transformer.
P4. A 25 MVA, 13.2 kV generator with \(X'' = 0.15\) p.u. sits in a zone whose base is 100 MVA, 13.8 kV. Find its reactance on the system base.

Show answer
\(0.15(100/25)(13.2/13.8)^2 = 0.15(4)(0.9149) = \mathbf{0.549}\) p.u.
P5. Two generator branches of 0.5 and 0.3 p.u. feed a common bus, both emfs at \(1.0\angle0^\circ\). Find the Thévenin reactance at that bus and the fault MVA on a 100 MVA base.

Show answer
\((0.5)(0.3)/0.8 = \mathbf{0.1875}\) p.u.; \(100/0.1875 = \mathbf{533}\) MVA.
P6. A delta load of \((30 + j40)\ \Omega\)/phase sits at 11 kV on a 50 MVA base. Find its per-unit impedance.

Show answer
\(Z_Y = 10 + j13.33\ \Omega\); \(Z_B = 121/50 = 2.42\ \Omega\); \(Z_{pu} = \mathbf{4.13 + j5.51}\) p.u.
P7. A load of 20 MVA at 0.8 p.f. lagging sits on a 100 MVA base at 1.0 p.u. voltage. Find its shunt impedance in per-unit.

Show answer
\(S_{pu} = 0.2\angle36.87^\circ\); \(Z = 1/0.2\angle-36.87^\circ = 5\angle36.87^\circ = \mathbf{4 + j3}\) p.u.
P8. An element has \(Z = 5 + j20\ \Omega\). What percentage error in fault current results from using \(X\) alone, and in which direction?

Show answer
\(\sqrt{1 + (1/4)^2} = 1.0308\), so \(\mathbf{3.08\%}\) overstated — the safe direction.
P9. A motor of \(X'' = 0.25\) p.u. is connected at a bus whose system Thévenin reactance is 0.40 p.u. Find the fault current at that bus with and without the motor.

Show answer
Without: \(1/0.40 = 2.5\) p.u. With: \(2.5 + 1/0.25 = \mathbf{6.5}\) p.u. — the motor contributes more than the whole system.
P10. A 66/11 kV transformer is tapped to give an actual ratio of 66/11.55 kV, with bases of 66 and 11 kV. Find the off-nominal ratio.

Show answer
\(t = 11.55/11 = \mathbf{1.05}\) — a 5% boost, represented by an ideal transformer \(1{:}1.05\) in series with the impedance.
P11. Two identical lines of \(j0.18\) p.u. run in parallel between two buses. One is switched out. By what factor does the transfer reactance change, and what does that do to the steady-state stability limit \(P = EV\sin\delta/X\)?

Show answer
From 0.09 to 0.18 — a doubling, which halves the stability limit. This is why a double-circuit outage is a far more serious contingency than a single one.
P12. A three-winding transformer has \(Z_p = 0.04\), \(Z_s = -0.01\), \(Z_t = 0.09\) p.u. Is the negative arm an error?

Show answer
No. Only the pairwise sums are physical: \(Z_{ps} = 0.03\), \(Z_{pt} = 0.13\), \(Z_{st} = 0.08\) — all positive. The star node is a computational fiction.

Challenge

Challenge Problems

Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.

C1. The reactance diagram sets every generator emf to \(1.0\angle0^\circ\). Examine this assumption: when is it exact, what does it cost when it is not, and how would you correct a fault calculation for a heavily loaded pre-fault system?

Show answer
When exact: only if the system was unloaded before the fault, so no current flowed and every internal emf equalled the terminal voltage, which the flat profile puts at 1.0 p.u.

What it costs otherwise: two separate errors. First, the internal emf of a loaded generator is not 1.0 — for a machine delivering \(S\) at terminal voltage \(V\), \(E = V + jX''I\), which for a lagging load makes \(|E| > 1\) and understates the fault current. Second, the emfs are not all at the same angle: power transfer requires angular separation, so the sources are genuinely out of step with one another and the simple parallel combination of Problem 9 is not exactly right.

The correction — superposition. The rigorous method treats the fault as the superposition of two states: (i) the pre-fault load flow, with its actual voltages and currents; and (ii) a "pure fault" network in which all sources are dead and a single source of \(-V_f\), the pre-fault voltage at the fault point, is applied there. The true post-fault current in any branch is the sum of the two. This is exactly the Thévenin construction, and it is why the \(Z\)-bus of Set 18 is the natural tool — column \(k\) of \(Z_{\text{bus}}\) gives every branch's share of a fault at bus \(k\) in one step.

Why the approximation survives in practice: pre-fault currents are typically 1 p.u. or less while fault currents are 5 to 20 p.u., so the superposed load component shifts the answer by a few per cent — and standards apply a voltage factor (commonly 1.1) that deliberately covers it, in the conservative direction.
C2. Problem 15 found the 138 kV bus at 0.236 p.u. and Problem 14 found the motor bus at 0.226 p.u., so the lower-voltage bus is the more severely stressed. Explain why this is not paradoxical, and identify what network features generally make a bus a fault-level hot spot.

Show answer
Why not paradoxical: fault level in MVA and fault current in amperes are different quantities. Here \(\text{MVA}_{sc}\) is 211.5 at the 138 kV bus and 221.4 at the motor bus — genuinely higher at the motor. But the current differs far more dramatically, 885 A against 9265 A, simply because the base current at 13.8 kV is ten times that at 138 kV. Low-voltage buses always carry larger fault currents for the same fault MVA, and it is current that switchgear must interrupt and busbars must withstand mechanically.

What makes a hot spot:
— Rotating plant connected directly, with no intervening impedance. The motor's 0.333 p.u. sat straight across its bus while the generators reached it only through 0.700.
— Many parallel infeeds. Each added source lowers the Thévenin reactance; a strongly meshed bus is a strong bus in the stability sense and a difficult one in the switchgear sense.
— Low transformer impedance upstream. As Set 3 Problem 6 showed, halving the transformer impedance doubles the fault level below it.
— Low voltage. Purely arithmetic, but it dominates: the same MVA at a tenth the voltage is ten times the current.

The practical consequence is that industrial and generator-terminal buses, not transmission buses, usually set the switchgear specification — and that fault level must be recomputed from every bus, since no single reduction answers the question for all of them.
C3. A colleague argues that since the per-unit system eliminates transformers, one could equally analyse the system in ohms referred to a single voltage level, and that per-unit therefore offers convenience but no new capability. Assess this claim carefully.

Show answer
The claim is technically correct and practically wrong, and it is worth being precise about which is which.

Correct: referring every impedance to one chosen voltage level by \(Z' = a^2Z\) does produce a single connected network in ohms, and it gives identical answers. Nothing in per-unit is mathematically unavailable in ohms. This was in fact how calculations were done before the 1920s.

Where the convenience becomes capability:
— Error detection. Per-unit values fall in narrow, memorable bands — machines near 0.15, transformers 0.05–0.15, lines 0.05–0.5. An impedance in ohms carries no such signal; 40 \(\Omega\) is unremarkable at 138 kV and absurd at 13.8. Set 3 Problem 14 showed a base error of a factor of 16 passing silently, and the only thing that catches it is the band.
— Manufacturer data. Nameplates quote per-unit or percent. Working in ohms means converting every one of them in and back out again, adding conversions rather than removing them.
— Scale independence. A 10 MVA and a 1000 MVA machine have nearly the same per-unit reactance, so intuition transfers between systems. In ohms they differ by a factor of a hundred and no intuition survives.
— Numerical conditioning. A load-flow Jacobian assembled in ohms across three voltage levels spans several orders of magnitude and conditions badly. In per-unit every entry is order unity — which for an iterative solver on a large network is the difference between convergence and failure, and is a genuine capability rather than a convenience.
— The \(\sqrt3\) cancels by construction (Set 3 Problem 16), so a stray one is a reliable error flag. In ohms there is no such flag.

Verdict: for a two-bus hand calculation the colleague is right and per-unit is merely tidy. For a 2000-bus network solved iteratively, the conditioning argument alone makes it indispensable — and the error-detection argument makes it indispensable even by hand.

Self-Test

Multiple-Choice Questions

MCQ 1. In a reactance diagram the base MVA:
(a) changes at every transformer (b) is common to the whole network (c) equals each machine's rating (d) changes at every voltage level

Show answer
(b). Only the base kV changes, and only at transformers, in the transformer's own ratio.
MCQ 2. An impedance given in ohms is converted to per-unit by dividing by:
(a) the base MVA (b) \(Z_B\) of the declared zone (c) \(Z_B\) of its own zone (d) the base current

Show answer
(c). Using the declared zone's \(Z_B\) everywhere is the error that Problem 2 is built to expose.
MCQ 3. A generator rated 18 kV sits in a zone whose base is 20 kV. Its per-unit reactance on the system base includes a factor of:
(a) \(18/20\) (b) \(20/18\) (c) \((18/20)^2\) (d) \((20/18)^2\)

Show answer
(c). Old base voltage over new, squared — Problem 5.
MCQ 4. A transformer whose nameplate ratio equals the ratio of its two zone bases contributes to the diagram:
(a) its impedance only (b) its impedance and an ideal transformer (c) an ideal transformer only (d) nothing

Show answer
(a). The ideal transformer vanishes precisely when the bases stand in the turns ratio.
MCQ 5. A delta-connected impedance of \(Z_\Delta\) per phase enters the diagram as:
(a) \(Z_\Delta/Z_B\) (b) \(Z_\Delta/(3Z_B)\) (c) \(3Z_\Delta/Z_B\) (d) \(Z_\Delta/(\sqrt3 Z_B)\)

Show answer
(b). Convert delta to star first — per-unit quantities are always per phase and star-referred.
MCQ 6. Passing from an impedance diagram to a reactance diagram discards:
(a) resistance only (b) resistance and shunt branches (c) resistance, shunt branches and static loads (d) nothing of consequence

Show answer
(c). All three, each justified by a specific inequality that holds during a fault — Problem 10.
MCQ 7. Neglecting resistance in a fault calculation makes the computed current:
(a) too small (b) too large (c) unchanged (d) either, depending on \(X/R\)

Show answer
(b) too large, since \(X < |Z|\) always. This is the safe direction for rating switchgear.
MCQ 8. Two sources of equal emf may be replaced by one source behind the parallel combination of their reactances because:
(a) their reactances are equal (b) no current circulates between them (c) the network is linear (d) the fault is symmetrical

Show answer
(b). Unequal emfs would circulate current and require Millman's theorem instead — Problem 9.
MCQ 9. A synchronous motor during a nearby three-phase fault:
(a) draws no current (b) draws its normal load current (c) feeds current into the fault (d) may be ignored

Show answer
(c). Its rotor flux and inertia make it a source. Problem 19 finds it contributing more than the entire rest of the system.
MCQ 10. The star arms of a three-winding transformer equivalent:
(a) are always positive (b) correspond to physical windings (c) may be negative, which is normal (d) are equal to one another

Show answer
(c). Only the pairwise sums are measurable and constrained positive.
MCQ 11. An off-nominal tap ratio appears on the diagram as:
(a) a modified impedance (b) an ideal transformer \(1{:}t\) in series with the impedance (c) a shunt branch (d) nothing, since per-unit removes it

Show answer
(b). Per-unit removes the nominal ratio; the deviation from nominal is real and is the control variable of voltage regulation.
MCQ 12. Fault level is a property of:
(a) the system as a whole (b) the generators only (c) a particular bus (d) the base MVA chosen

Show answer
(c) a particular bus. It must be recomputed by reducing the network afresh from each candidate location — Problems 14 and 15 differ for exactly this reason.

Reference

Key Formulas

Quantity	Relation	Notes
Base impedance	\(Z_B = \text{kV}_B^2/\text{MVA}_B\)	Of the element's own zone
Ohms to p.u.	\(X_{pu} = X_\Omega\,\text{MVA}_B/\text{kV}_B^2\)	Convenient rearrangement
Base change	\(X_{pu}^{new} = X_{pu}^{old}\dfrac{\text{MVA}_{new}}{\text{MVA}_{old}}\left(\dfrac{\text{kV}_{old}}{\text{kV}_{new}}\right)^2\)	For nameplate p.u. values
Base kV propagation	\(\text{kV}_{B,LT} = \text{kV}_{B,HT}\times\dfrac{\text{LT rating}}{\text{HT rating}}\)	MVA base never changes
Delta to star	\(Z_Y = Z_\Delta/3\)	Before dividing by \(Z_B\)
Load as impedance	\(Z = \|V\|^2/S^{*}\)	Valid at the stated voltage only
Series branch	reactances add	Legitimate because transformers vanished
Parallel sources	\(X = X_1X_2/(X_1+X_2)\)	Only if the emfs are equal
Thévenin at a bus	all emfs shorted, look in	Recompute for every candidate bus
Fault current	\(I_f = 1/X_{th}\) p.u.	Flat pre-fault profile assumed
Fault MVA	\(\text{MVA}_{sc} = \text{MVA}_B/X_{th}\)
Fault amperes	\(I_f = I_B/X_{th}\)	\(I_B\) of the faulted zone
Superposition	\(I_f = \sum_k 1/X_k\)	Each source contributes independently
Error from dropping \(R\)	\(\|Z\|/X = \sqrt{1 + (R/X)^2}\)	Overstates current — safe direction
Three-winding star	\(Z_p = \tfrac12(Z_{ps}+Z_{pt}-Z_{st})\)	Cyclic; arms may be negative
Off-nominal ratio	\(t = \dfrac{V_{2,\text{actual}}/V_{2,B}}{V_{1,\text{actual}}/V_{1,B}}\)	Ideal transformer \(1{:}t\) retained

Diagnostics

Common Mistakes

Dividing every ohmic value by the declared zone's \(Z_B\). Each element uses the base impedance of the zone it actually occupies. In Problem 2 this would misstate \(G_2\) by a factor of 28, silently.
Changing the base MVA somewhere in the network. It is one number for the whole system; only the base kV moves.
Suppressing the voltage factor for a machine whose rating is "close" to the base. An 18 kV machine on a 20 kV base carries a genuine factor of 0.81 — Problem 5.
Inverting the base-change formula. MVA the right way up, kV inverted and squared.
Dividing a delta impedance by \(Z_B\) without converting to star first. Gives three times too much — Problem 12.
Combining impedances in ohms across a zone boundary. Legitimate only within one zone; the whole point of per-unit is that the combination works across boundaries.
Collapsing sources of unequal emf by the parallel formula. Valid only under the flat-start assumption, which makes every emf \(1.0\angle0^\circ\) — Problem 9.
Omitting motors from a fault study. Problem 19 understates the fault current by 68% when the motor is left out.
Using the base current of the declared zone rather than the faulted zone. Problem 20 needs \(I_B\) at 13.8 kV, not at 138 kV — a factor of ten.
Assuming the highest-voltage bus has the highest fault duty. It rarely does. Fault level belongs to a bus and must be recomputed from each one.
Treating a negative three-winding star arm as an error. Only the pairwise sums are physical — Problem 17.
Carrying a reactance diagram into a load-flow or loss calculation. Without \(R\) the lines are lossless and the answer is meaningless — Problem 10.

Looking Ahead

The diagram assembled in Problem 8 and reduced in Problem 20 is the last one in this book that will be drawn by hand. Its elements came from nameplates and from ohmic values supplied in the question; nothing so far has said where those ohms come from.

Sets 5 to 8 answer that. The series reactance of a line follows from the geometry of the conductors and their spacing, the shunt capacitance from their diameter and height above ground, and the corona loss from the surface gradient those same dimensions produce. Sets 9 to 15 then ask what the line does with those parameters — how much voltage it drops, how much power it can carry, and at what point the lumped model of Problem 8 stops being adequate. From Set 16 the diagram returns, but as a matrix rather than a picture, and the hand reduction of Problem 9 becomes a matrix inversion.