Set 1 — Three-Phase Circuits and Power
Twenty worked problems on the system that carries essentially all the electrical energy in the world. A balanced three-phase circuit is not three circuits; it is one circuit written three times, and the whole art of the subject is refusing to solve it three times. Everything in this set follows from that: the per-phase equivalent, the two appearances of \(\sqrt{3}\), and the reason a balanced load needs no neutral wire at all. The last third of the set removes the balance and shows exactly what breaks.
Solve one phase, not three. In a balanced system every quantity in phase \(b\) is the phase-\(a\) quantity rotated by \(-120^\circ\), and phase \(c\) by \(+120^\circ\). Compute phase \(a\), then rotate. Doing the arithmetic three times is not thoroughness; it is three chances to make a mistake.
Star: \(V_L = \sqrt{3}\,V_{ph}\angle 30^\circ\), \(I_L = I_{ph}\). The \(\sqrt{3}\) comes from subtracting two phasors \(120^\circ\) apart, and the \(30^\circ\) lead comes with it. Dropping the angle is harmless in a magnitude-only problem and fatal in a phasor one.
Delta: \(V_L = V_{ph}\), \(I_L = \sqrt{3}\,I_{ph}\angle -30^\circ\). The same subtraction, now applied to currents at a corner of the delta. Star and delta each carry exactly one \(\sqrt{3}\) — never both, never neither.
Total power is \(P = \sqrt{3}\,V_L I_L\cos\phi\) in either connection. The \(\cos\phi\) is the angle between phase voltage and phase current — the load's own angle — never the angle between line quantities. This single point accounts for more lost marks than any other in the subject.
Convert a delta load to star before drawing a per-phase circuit. A balanced delta of \(Z_\Delta\) is a star of \(Z_\Delta/3\). Once converted, the per-phase equivalent is an ordinary single-phase circuit and every technique you already own applies.
A balanced star load carries no neutral current. Three equal phasors \(120^\circ\) apart sum to zero, so the neutral wire can be removed without changing anything. The moment the load is unbalanced this stops being true, and whether the neutral is present or absent changes the answer completely.
Two wattmeters measure any three-wire load, balanced or not. \(P = W_1 + W_2\) always; \(\tan\phi = \sqrt{3}\,(W_1-W_2)/(W_1+W_2)\) only if the load is balanced. A negative reading is information, not a fault.
A three-phase source produces the following set of voltages. Determine the phase sequence and write the three phasors in polar form.
The amplitudes given are peak values, so each phasor magnitude is \(400/\sqrt{2} = 282.8\) V. Writing the three phasors and reducing every angle to the range \(-180^\circ\) to \(+180^\circ\):
Note \(-250^\circ + 360^\circ = +110^\circ\). Leaving \(B\) at \(-250^\circ\) makes the comparison in the next step almost impossible to read, which is the entire reason for the reduction.
Now ask which phase lags \(A\) by \(120^\circ\). Subtracting \(120^\circ\) from the phase-\(A\) angle:
That is the angle of \(\mathbf{V}_{CN}\), not \(\mathbf{V}_{BN}\). So \(C\) follows \(A\), and \(B\) follows \(C\).
Confirming the second step: \(-130^\circ - 120^\circ = -250^\circ \equiv +110^\circ\), which is indeed \(\mathbf{V}_{BN}\). The cycle closes as \(A \to C \to B \to A\).
A star-connected alternator has a phase voltage \(\mathbf{V}_{an} = 230.94\angle 0^\circ\) V with \(abc\) sequence. Find the three line voltages, and verify that they sum to zero.
Line voltage is the difference of two phase voltages. Taking \(\mathbf{V}_{ab}\) first:
Expanding into rectangular form:
The factor \(|1.5 + j0.866| = \sqrt{2.25 + 0.75} = \sqrt{3}\). This is where the \(\sqrt{3}\) of the star connection comes from — it is not a convention or a definition, it is the magnitude of the sum of two unit phasors \(60^\circ\) apart.
The remaining two follow by rotation, since the system is balanced:
Checking the sum:
For the balanced Y–Y system shown, the source phase voltage is \(\mathbf{V}_{a'n} = 120\angle 0^\circ\) V with \(abc\) sequence. The generator winding impedance is \(Z_{ga} = (0.2 + j0.5)\,\Omega\), the line impedance is \(Z_{1a} = (0.8 + j1.5)\,\Omega\), and the load is \(Z_{A} = (39 + j28)\,\Omega\) per phase. Determine:
- the three line currents;
- the phase voltages at the load;
- the line voltages at the load;
- the phase voltages at the source terminals;
- the line voltages at the source terminals.
The system is balanced, so the neutrals are at the same potential and a single phase can be solved in isolation. All three impedances of phase \(a\) are in series:
The line current in phase \(a\), and the other two by rotation:
In a star connection the line current is the phase current, so these are also the load phase currents. No \(\sqrt{3}\) appears anywhere in this step.
The load phase voltage is the current times the load impedance alone — not the total:
The load sees 115.22 V where the source produces 120 V — the missing 4.78 V is dropped across the winding and the line.
Load line voltages follow by the star relation \(\mathbf{V}_L = \sqrt{3}\angle30^\circ \cdot \mathbf{V}_{ph}\):
The source terminal voltage is the internal emf less the drop in the winding impedance only:
And the source line voltages, again by the star relation:
A balanced three-phase load connected in star consists of \((6 + j8)\,\Omega\) in each phase. It is supplied from a 400 V, 50 Hz three-phase system. Find the phase current, the line current, the power per phase and the total power.
A three-phase supply is always quoted by its line voltage, so 400 V is \(V_L\). In star the phase voltage is smaller by \(\sqrt{3}\):
The impedance in polar form:
Phase current, and hence line current, since star makes them equal:
Power per phase, using the load's own voltage, current and angle:
Total power, computed both ways as a check:
A third route confirms it: \(P_T = 3I_{ph}^2R = 3(23.094)^2(6) = 9600\) W. Any of the three is acceptable; agreeing on two of them is the point.
The same \((6 + j8)\,\Omega\) impedances of Problem 4 are now reconnected in delta across the same 400 V supply. Find the phase current, the line current and the total power, and compare with the star case.
In delta each impedance sits directly across a line pair, so the phase voltage is the full line voltage:
Phase current:
Line current, now larger by \(\sqrt{3}\) because two phase currents meet at each corner of the delta:
Power per phase and total:
Comparing the two connections of the same three impedances on the same supply:
Each phase of a three-phase alternator generates 3810.5 V and can carry a maximum current of 30 A. Find the line current, the line voltage and the total kVA capacity when the machine is connected (a) in star and (b) in delta.
The winding is the same physical object in both cases, so its two ratings are fixed: \(E_{ph} = 3810.5\) V is set by the number of turns and the flux, and \(I_{ph} = 30\) A by the cross-section of the conductor. Only the interconnection changes.
aStar connection. Line current equals phase current; line voltage is \(\sqrt{3}\) times phase voltage:
bDelta connection. Line voltage equals phase voltage; line current is \(\sqrt{3}\) times phase current:
The two capacities are identical:
In the equivalent Y–Y circuit shown, the load takes \((160 + j120)\) kVA per phase and the load line voltage is 600 V. The line impedance is \(Z_{1a} = (0.005 + j0.025)\,\Omega\) per phase. Determine the complex power supplied by the source and the complex power dissipated in the line, both for phase \(a\).
The load line voltage is 600 V, so the load phase voltage — taken as the reference — is
The line current comes from the definition of complex power, \(S_\phi = \mathbf{V}_\phi\mathbf{I}_\phi^{*}\). Note the conjugate — it is what makes a lagging load produce positive \(Q\):
The current lags the voltage by \(36.87^\circ\), consistent with an inductive load — a useful sanity check before going further.
The source terminal voltage is the load voltage plus the drop along the line:
Complex power delivered by the source in phase \(a\):
Complex power absorbed by the line impedance. Here \(S = |I|^2Z\) is the direct route, since the current through the line is already known:
Conservation check — source power must equal load plus line:
A balanced delta-connected load of \((18 + j24)\,\Omega\) per phase is supplied from a 400 V, 50 Hz three-phase source through a feeder of impedance \((0.5 + j1)\,\Omega\) per line. Find the line current, the voltage actually appearing across the load, the power consumed by the load and the power lost in the feeder.
A delta load cannot be put in series with a line impedance directly — the two are not in the same circuit topology. Convert the delta to its star equivalent first:
Now the per-phase circuit is a simple series loop driven by the source phase voltage:
Line current:
Voltage across the star-equivalent load, then converted back to a line voltage — which is what the actual delta-connected impedances see:
The load is starved of nearly 40 V by the feeder — a 10% drop, which in practice would be unacceptable and is exactly the sort of number a feeder-sizing calculation exists to prevent.
Returning to the real delta to confirm the conversion was consistent:
Powers, taken from the star equivalent since the line current is known there:
Verifying against the delta directly: \(3I_{\Delta,ph}^2R_\Delta = 3(12.01)^2(18) = 7.79\) kW. The star equivalent is genuinely equivalent — it reproduces the terminal behaviour exactly, including the power.
A 415 V, 50 Hz three-phase bus supplies two balanced loads in parallel:
- a star-connected load of \((12 + j9)\,\Omega\) per phase;
- a delta-connected load of \((45 + j60)\,\Omega\) per phase.
Find the total line current drawn from the bus and the overall power factor.
Two loads in parallel share a voltage, not a current, so both must be reduced to the same per-phase star reference before their currents can be added:
Per-phase bus voltage, taken as reference:
Each branch current, immediately resolved into rectangular form so they can be added:
Adding as phasors — never as magnitudes:
Adding the magnitudes would have given \(15.97 + 9.58 = 25.55\) A — close enough here to look plausible and still wrong, because the two loads have different angles.
The overall power factor is the cosine of the resultant current's angle:
Cross-checking through power. Summing \(P\) and \(Q\) separately:
Two wattmeters connected to measure the power taken by a balanced three-phase load read \(W_1 = 8\) kW and \(W_2 = 4\) kW. Determine the total power, the power factor and the reactive power of the load.
Total power is the algebraic sum of the two readings. This holds for any three-wire load, balanced or not:
The individual readings are \(W_1 = V_LI_L\cos(30^\circ - \phi)\) and \(W_2 = V_LI_L\cos(30^\circ + \phi)\). Expanding and taking the difference:
Dividing one by the other eliminates \(V_LI_L\) entirely, which is why the method needs no voltage or current reading at all:
Reactive power follows from the difference directly:
Check via the power triangle:
The two-wattmeter method is applied to a balanced three-phase load. One wattmeter reads 10 kW; the other reads \(-2.5\) kW, the pointer having gone backwards until the connections to its current coil were reversed. Determine the total power, the power factor and the reactive power.
The reversal is a measurement technique, not a correction: the true reading is \(-2.5\) kW and must enter the arithmetic with its sign.
The difference — now larger than the sum, which is the signature of a badly lagging load:
Hence
Reactive power:
Check:
The load draws nearly three times as much reactive power as real power — 22.9 kVA of plant capacity is tied up delivering 7.5 kW of useful work.
A 400 V, three-phase, four-wire system supplies an unbalanced star-connected load with
Find the three line currents and the current in the neutral conductor.
The neutral wire is present, so the load star point is tied to the source star point and each impedance sees its own full phase voltage regardless of what the others are doing. This is the whole reason four-wire distribution exists:
Each line current is now an independent single-phase calculation:
The neutral carries whatever the three lines fail to cancel:
Note the size of the result. The line currents are 11–23 A, but the neutral carries under 5 A, because three phasors spread around the circle cancel substantially even when their magnitudes differ by a factor of two.
The same three impedances of Problem 12 are supplied from the same 400 V source, but the neutral conductor is now removed. Find the displacement of the load star point, the voltage that now appears across each impedance, and the three line currents.
Without the neutral wire the load star point \(N\) is free to float away from the source star point \(O\). Its displacement follows from Millman's theorem — one nodal equation written at the floating node. Working in admittances:
The numerator is the sum of the three source-driven currents — which is exactly the neutral current computed in Problem 12:
Hence the neutral displacement voltage:
The load star point sits 25.4 V away from where the source thinks neutral is. Nothing in the circuit is faulty; this is simply what an unbalanced load does when denied a return path.
Each impedance now sees the source phase voltage less the displacement:
And the currents follow:
The essential check — with no neutral, the three line currents have nowhere else to go and must sum to zero:
A 415 V, 50 Hz three-phase induction motor draws 40 kW at 0.75 power factor lagging. Determine the kvar rating of a delta-connected capacitor bank needed to raise the power factor to 0.95 lagging, the capacitance per phase, and the reduction in line current.
Capacitors supply reactive power without consuming real power, so \(P\) is unchanged at 40 kW throughout. Only \(Q\) moves.
Reactive power before and after:
The capacitor bank supplies the difference:
In delta each capacitor carries one third of the total and stands across the full line voltage:
Line current before and after correction:
A three-phase, 415 V, 50 Hz induction motor delivers 30 kW of mechanical output at an efficiency of 90% and a power factor of 0.85 lagging. Find the input power, the line current, the apparent power and the reactive power drawn from the supply.
Efficiency relates mechanical output to electrical input. The 30 kW is what comes out of the shaft, not what goes into the terminals:
Line current from the three-phase power relation, using the input power:
Apparent power:
Reactive power, from the power triangle:
A balanced three-phase load of 15 kW at 0.8 power factor lagging is supplied at 400 V through a feeder of 0.4 \(\Omega\) resistance per line. Find the feeder loss when the load is connected (a) in star and (b) in delta. Then find the loss if the same 15 kW were delivered at unity power factor.
The line current is fixed entirely by the power, the line voltage and the power factor. Nothing in that expression refers to how the load is connected internally:
abSince the feeder carries the line current in both cases, the loss is the same for star and for delta:
This is not an approximation or a coincidence. The feeder is outside the load; it can only see terminal quantities, and the connection is invisible from the terminals once \(P\), \(V_L\) and \(\cos\phi\) are specified.
At unity power factor the same real power needs less current:
The penalty for operating at 0.8 instead of unity:
A 400 V, 50 Hz three-phase bus supplies three loads simultaneously:
- a motor taking 25 kW at 0.8 power factor lagging;
- a lighting load of 12 kW at unity power factor;
- a capacitor bank rated 10 kvar.
Find the total real, reactive and apparent power, the line current and the overall power factor.
Loads specified in kW and power factor are best combined through complex power, where \(P\) and \(Q\) each add as ordinary numbers. Taking each in turn:
The capacitor's reactive power is negative by convention — it supplies vars rather than absorbing them. Getting this sign wrong turns a correction bank into an additional inductive load and roughly doubles the error.
Summing:
Apparent power and power factor:
Line current:
Worth noting what the capacitors bought. Without them, \(Q_T = 18.75\) kvar, giving \(S_T = 41.48\) kVA, \(\cos\phi = 0.892\) and \(I_L = 59.87\) A — so 10 kvar of capacitance removed 5 A of line current.
A phase-sequence indicator is built from a capacitor of reactance \(X_C = 100\ \Omega\) in phase \(A\) and two identical lamps of resistance \(R = 100\ \Omega\) in phases \(B\) and \(C\), all connected in star with the star point floating, across a 400 V three-phase supply. For \(ABC\) sequence, find the voltage across each lamp and state which burns brighter.
The star point floats, so this is the unbalanced three-wire problem of Problem 13 again. Working in admittances, with \(V_{ph} = 230.94\) V:
The numerator, taking \(ABC\) sequence. The two lamp terms are equal in magnitude and \(240^\circ\) apart, so they sum to \(-1\) times a single one:
Hence the star-point displacement:
Voltage across each lamp is the source phase voltage minus this displacement:
The verdict:
Lamp \(B\) stands at 345.6 V and lamp \(C\) at only 92.6 V. Since lamp brightness goes roughly as \(V^2\), one is blazing and the other barely glowing — an unmistakable indication rather than a marginal one.
Check that the three currents sum to zero, as they must with no neutral:
A single wattmeter has its current coil in line \(A\) and its pressure coil connected across lines \(B\) and \(C\). Show that for a balanced load the reading is \(V_LI_L\sin\phi\), so that \(Q = \sqrt{3}\,W\). If the instrument reads 5 kW on a balanced 400 V load drawing 20 A, find the reactive power, the power factor and the real power.
A wattmeter reads the product of its coil quantities and the cosine of the angle between them. Taking \(\mathbf{V}_{AN} = V_{ph}\angle0^\circ\) with \(ABC\) sequence, the current coil carries
The pressure coil sees the line voltage \(\mathbf{V}_{BC}\), which lags \(\mathbf{V}_{AN}\) by \(90^\circ\) — a fact that makes this connection work at all:
The angle between pressure and current quantities is therefore \((-90^\circ) - (-\phi) = \phi - 90^\circ\), and
Since the total reactive power of a balanced load is \(Q = \sqrt{3}V_LI_L\sin\phi\), the relation follows immediately:
Applying it to the given readings:
Real power, for completeness:
A balanced three-phase load draws a line current of 30 A from a 415 V supply and absorbs 18 kW. Find the power factor, the reactive power, and the per-phase impedance if the load is (a) star-connected and (b) delta-connected.
Apparent power from the line quantities, which is the one relation that needs no knowledge of the connection:
Power factor and angle:
Reactive power:
aStar. Phase current equals line current; phase voltage is \(V_L/\sqrt{3}\):
bDelta. Phase voltage is the full \(V_L\); phase current is \(I_L/\sqrt{3}\):
And the expected relation between the two:
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. A star-connected load of \((8 + j6)\,\Omega\) per phase is supplied at 400 V. Find the line current and the total power.
Show answer
\(V_{ph} = 230.94\) V, \(|Z| = 10\) Ω, so \(I_L = 23.09\) A and \(P = 3(23.09)^2(8) = \mathbf{12.8}\) kW.P2. The same impedances are connected in delta on the same supply. Find the new line current and total power without repeating the whole calculation.
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Both are three times the star values: \(I_L = \mathbf{69.28}\) A and \(P = \mathbf{38.4}\) kW. No fresh arithmetic is needed.P3. Two wattmeters read 6 kW and 6 kW on a balanced load. What is the power factor?
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Equal readings give \(\tan\phi = 0\), so \(\phi = 0\) and pf \(= \mathbf{1}\). The load is purely resistive.P4. Two wattmeters read 9 kW and 0 kW. What is the power factor, and why is the zero reading not a fault?
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\(\tan\phi = \sqrt{3}(9)/9 = \sqrt{3}\), so \(\phi = 60^\circ\) and pf \(= \mathbf{0.5}\) lagging. Zero is exactly the boundary at which \(\cos(30^\circ+\phi)\) vanishes.P5. A balanced delta load of \(30\angle 40^\circ\,\Omega\) per phase is supplied at 415 V. Find the phase current, the line current and the total power.
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\(I_{ph} = 415/30 = 13.83\) A, \(I_L = 23.96\) A, \(P = \sqrt3(415)(23.96)\cos40^\circ = \mathbf{13.19}\) kW.P6. A 400 V, four-wire system feeds resistive loads of 5 kW, 5 kW and 5 kW on the three phases. What is the neutral current? What if the third load is switched off?
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Balanced: \(I_N = \mathbf{0}\). With one phase off, two equal currents \(120^\circ\) apart remain; their sum has the same magnitude as one of them, so \(I_N = 5000/230.94 = \mathbf{21.65}\) A.P7. A 415 V load takes 50 kW at 0.7 pf lagging. Find the kvar needed to correct to 0.9 lagging.
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\(\tan\phi_1 = 1.0202\), \(\tan\phi_2 = 0.4843\), so \(Q_C = 50(1.0202 - 0.4843) = \mathbf{26.8}\) kvar.P8. Three identical impedances take 12 kW in star. What do they take in delta on the same supply, and what is the ratio of line currents?
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\(\mathbf{36}\) kW; the line-current ratio is also \(\mathbf{3}\), not \(\sqrt3\). The \(\sqrt3\) relates line to phase within one connection, not one connection to another.P9. A phase sequence is given by \(\mathbf{V}_{AN} = 100\angle 20^\circ\), \(\mathbf{V}_{BN} = 100\angle 140^\circ\), \(\mathbf{V}_{CN} = 100\angle -100^\circ\). Identify it.
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\(20^\circ - 120^\circ = -100^\circ\), which is \(C\). Sequence is \(\mathbf{ACB}\).P10. A balanced load draws 40 A at 0.6 pf lagging from a 400 V supply. Find \(P\), \(Q\) and \(S\).
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\(S = \sqrt3(400)(40) = 27.71\) kVA, \(P = \mathbf{16.63}\) kW, \(Q = \mathbf{22.17}\) kvar.P11. A feeder of 0.3 Ω per line supplies 20 kW at 400 V and 0.85 pf. Find the feeder loss and the loss as a percentage of the load.
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\(I_L = 20\,000/(\sqrt3\cdot400\cdot0.85) = 33.96\) A, loss \(= 3(33.96)^2(0.3) = \mathbf{1038}\) W, i.e. \(\mathbf{5.2\%}\).P12. A 400 V three-wire supply feeds an unbalanced star load of \(20\,\Omega\), \(20\,\Omega\) and \(20\,\Omega\). What is the star-point displacement? Now change one impedance to \(40\,\Omega\) — is the displacement still zero?
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Balanced: \(\sum\mathbf{V}Y = 0\) so \(\mathbf{V}_{NO} = \mathbf{0}\). With one branch at 40 Ω the numerator no longer vanishes and \(\mathbf{V}_{NO} = \mathbf{46.19}\) V — the star point moves the moment symmetry is broken.
Challenge Problems
Each of these needs an idea rather than a formula. Decide what the governing principle is before opening the answer.
C1. Prove that the instantaneous power delivered by a balanced three-phase system to a balanced load is constant in time, whereas the instantaneous power in a single-phase system pulsates at twice supply frequency. What mechanical consequence does this have, and why does it fail the moment the load becomes unbalanced?
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Take \(v_k = V_m\cos(\omega t - \theta_k)\) and \(i_k = I_m\cos(\omega t - \theta_k - \phi)\) with \(\theta_k = 0, 120^\circ, 240^\circ\). Each phase contributesThe constant term is the same in all three. The double-frequency terms have arguments differing by \(2(120^\circ) = 240^\circ\) and \(2(240^\circ) = 480^\circ \equiv 120^\circ\) — three phasors equally spaced round the circle — so they sum to zero exactly.\[ p_k = \tfrac{V_mI_m}{2}\big[\cos\phi + \cos(2\omega t - 2\theta_k - \phi)\big] \]
Hence \(p(t) = \tfrac{3}{2}V_mI_m\cos\phi = 3V_{ph}I_{ph}\cos\phi\), a constant. In single phase only one term survives and the power swings between zero and twice its average every half cycle.
Mechanical consequence: a three-phase machine develops constant torque, so it does not vibrate at \(2f\) and needs no flywheel to smooth it; a single-phase machine of the same rating shakes its mounting at 100 Hz. This is a large part of why three phase won.
Why unbalance breaks it: the cancellation depended on the three double-frequency terms being equal in magnitude. Unequal magnitudes leave a residual pulsation, which appears as a torque ripple and, in a motor, as audible noise and additional rotor heating — the practical reason unbalanced supply is a machine-rating issue, not merely an accounting one.C2. A balanced three-phase load is fed through a three-wire line. Explain, using Blondel's theorem, why two wattmeters are sufficient, why three are needed if a neutral is present, and what physically goes wrong if you attempt the two-wattmeter method on a four-wire unbalanced load.
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Blondel's theorem: the power in an \(n\)-wire system can be measured by \(n-1\) wattmeters, with all pressure coils returned to the \(n\)th wire chosen as common.
Why it works: instantaneous power is \(p = \sum_k v_k i_k\) with all \(v_k\) measured to an arbitrary datum. Choose the datum on wire \(n\); then \(v_n = 0\) identically and that term disappears, whatever \(i_n\) is doing. KCL supplies the missing information: \(i_n = -\sum_{k\ne n} i_k\), so no current measurement is lost either.
Three wires → two meters; four wires → three meters. The count is structural and has nothing to do with balance. It is a common error to suppose that two meters suffice for three wires because the load is balanced; balance is needed only for the \(\tan\phi\) formula, never for \(P = W_1+W_2\).
What goes wrong on four wires: the neutral now carries an independent current that KCL no longer determines from the two measured lines. The two meters see \(v_A i_A + v_B i_B\) referred to line \(C\), but the true power contains a neutral term they cannot observe. On a balanced four-wire load the neutral current is zero and the answer comes out right by accident; the moment it is unbalanced the reading is simply wrong, with no warning that it is.C3. A three-phase induction motor runs from a 400 V supply. One of the three supply fuses blows while the motor is running under load. Explain what the motor becomes electrically, why it usually keeps running, why it may nevertheless burn out, and why it will not restart if stopped.
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What it becomes: with one line open, the two remaining windings are in series across a single line voltage. The machine is now a single-phase load, and its stator produces a pulsating field rather than a rotating one.
Why it keeps running: a pulsating field decomposes into two counter-rotating fields of half amplitude each. The rotor is already turning with one of them, so it continues to develop torque from the forward component. This is precisely the principle of a single-phase induction motor.
Why it may burn out: to deliver the same shaft power from two windings instead of three, the current in those two must rise by roughly \(\sqrt3\), so their \(I^2R\) loss rises by about three. Worse, the backward-rotating field induces rotor currents at nearly \(2f\), adding rotor loss and a \(2f\) torque ripple. The machine draws more current, runs hotter and hums — the classic "single-phasing" failure, and the reason motor protection includes a phase-failure relay rather than relying on overload alone.
Why it will not restart: at standstill the forward and backward fields are exactly equal, so their torques cancel and the net starting torque is zero. The motor sits stalled, drawing locked-rotor current, and burns out in seconds unless protected. The asymmetry that let it keep running was supplied entirely by the rotor's own motion.
Multiple-Choice Questions
MCQ 1. In a balanced star-connected system, the line voltage:
(a) equals the phase voltage (b) is \(\sqrt3\) times the phase voltage and leads it by \(30^\circ\) (c) is \(\sqrt3\) times the phase voltage and lags it by \(30^\circ\) (d) is three times the phase voltageShow answer
(b). Option (c) describes the current relation in delta, and swapping the two is the single commonest error in this topic.MCQ 2. Three identical impedances take power \(P\) when connected in star across a supply. Reconnected in delta across the same supply they take:
(a) \(P/3\) (b) \(\sqrt3 P\) (c) \(3P\) (d) \(P\)Show answer
(c) \(3P\). Each impedance sees \(\sqrt3\) times the voltage, so it dissipates \((\sqrt3)^2 = 3\) times the power. This is the basis of the star–delta starter.MCQ 3. In the two-wattmeter method, one wattmeter reads zero. The power factor is:
(a) 0 (b) 0.5 (c) 0.866 (d) 1Show answer
(b) 0.5. \(W_2 = V_LI_L\cos(30^\circ+\phi) = 0\) requires \(\phi = 60^\circ\).MCQ 4. A wattmeter in the two-wattmeter method reads negative. This indicates:
(a) a wiring fault (b) a leading power factor (c) a power factor below 0.5 (d) an unbalanced loadShow answer
(c). A negative reading occurs whenever \(\phi > 60^\circ\), i.e. pf below 0.5, whether the load leads or lags. It is a normal indication, not a fault.MCQ 5. The neutral current in a balanced four-wire star system is:
(a) equal to the line current (b) \(\sqrt3\) times the phase current (c) three times the phase current (d) zeroShow answer
(d) zero. Three equal phasors \(120^\circ\) apart sum to zero, which is why the neutral can be omitted entirely on a balanced load.MCQ 6. A balanced delta load of \(Z_\Delta\) per phase is equivalent to a star load of:
(a) \(3Z_\Delta\) (b) \(Z_\Delta/3\) (c) \(\sqrt3 Z_\Delta\) (d) \(Z_\Delta/\sqrt3\)Show answer
(b) \(Z_\Delta/3\). The factor is 3, not \(\sqrt3\) — impedance transformation goes as the square of the ratio of voltages.MCQ 7. Total power in a balanced three-phase load is \(\sqrt3 V_LI_L\cos\phi\), where \(\phi\) is the angle between:
(a) \(V_L\) and \(I_L\) (b) phase voltage and phase current (c) \(V_L\) and phase current (d) any two line currentsShow answer
(b). The angle is the load's own impedance angle. Using the angle between line quantities introduces a spurious \(30^\circ\) and is the reason this formula is so often misapplied.MCQ 8. Removing the neutral from an unbalanced four-wire star load causes:
(a) no change (b) the phase voltages to become unequal (c) the line currents to become equal (d) the load to draw more powerShow answer
(b). The star point is displaced, and the lightly loaded phase suffers an overvoltage — the mechanism behind broken-neutral damage in distribution networks.MCQ 9. A three-phase alternator is reconnected from star to delta. Its kVA rating:
(a) increases by \(\sqrt3\) (b) decreases by \(\sqrt3\) (c) is unchanged (d) increases threefoldShow answer
(c) unchanged. The rating is \(3E_{ph}I_{ph}\) and both winding limits are fixed. Only the terminal voltage and current change, in compensating directions.MCQ 10. Feeder \(I^2R\) loss for a fixed real power and fixed line voltage varies as:
(a) \(\cos\phi\) (b) \(\cos^2\phi\) (c) \(1/\cos\phi\) (d) \(1/\cos^2\phi\)Show answer
(d). \(I \propto 1/\cos\phi\), and loss goes as \(I^2\). Correcting from 0.8 to unity cuts the loss by 36%.MCQ 11. Instantaneous power in a balanced three-phase system is:
(a) constant (b) pulsating at \(f\) (c) pulsating at \(2f\) (d) pulsating at \(3f\)Show answer
(a) constant. The three double-frequency terms are \(120^\circ\) apart and cancel exactly — see Challenge C1. Single-phase power pulsates at \(2f\).MCQ 12. Two loads of 0.8 pf and 0.6 pf are connected to the same bus. The resultant power factor is:
(a) 0.7 (b) 0.693 (c) between 0.6 and 0.8, depending on their sizes (d) 1.4Show answer
(c). Power factors have no arithmetic of their own. Resolve each load into \(P\) and \(Q\), add those, and recompute — the result depends entirely on the relative magnitudes.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| Star — voltage | \(V_L = \sqrt3\,V_{ph}\angle 30^\circ\) | Line leads phase by \(30^\circ\) |
| Star — current | \(I_L = I_{ph}\) | No \(\sqrt3\) anywhere in the current |
| Delta — voltage | \(V_L = V_{ph}\) | Each element across a line pair |
| Delta — current | \(I_L = \sqrt3\,I_{ph}\angle -30^\circ\) | Line lags phase by \(30^\circ\) |
| Real power | \(P = \sqrt3\,V_LI_L\cos\phi = 3V_{ph}I_{ph}\cos\phi\) | \(\phi\) is the load angle, always |
| Reactive power | \(Q = \sqrt3\,V_LI_L\sin\phi\) | Positive for lagging loads |
| Apparent power | \(S = \sqrt3\,V_LI_L = \sqrt{P^2+Q^2}\) | Connection-independent |
| Complex power | \(S = \mathbf{V}\mathbf{I}^{*} = |I|^2Z\) | Conjugate the current, never the voltage |
| Delta → star | \(Z_Y = Z_\Delta/3\) | Balanced loads only; factor 3, not \(\sqrt3\) |
| Star vs delta power | \(P_\Delta = 3P_Y\) | Same impedances, same supply voltage |
| Two wattmeters — power | \(P = W_1 + W_2\) | Valid balanced or unbalanced, 3-wire |
| Two wattmeters — angle | \(\tan\phi = \sqrt3\,\dfrac{W_1-W_2}{W_1+W_2}\) | Balanced loads only |
| Two wattmeters — vars | \(Q = \sqrt3\,(W_1-W_2)\) | Balanced loads only |
| Blondel's theorem | \(n\) wires need \(n-1\) wattmeters | 3-wire → 2 meters; 4-wire → 3 meters |
| Neutral current | \(\mathbf{I}_N = \mathbf{I}_A+\mathbf{I}_B+\mathbf{I}_C\) | Zero if and only if balanced |
| Star-point displacement | \(\mathbf{V}_{NO} = \dfrac{\sum \mathbf{V}_kY_k}{\sum Y_k}\) | Millman; 3-wire unbalanced star |
| PF correction | \(Q_C = P(\tan\phi_1 - \tan\phi_2)\) | \(P\) unchanged by capacitors |
| Delta capacitor bank | \(C = \dfrac{Q_C}{3V_L^2\omega}\) | Star bank needs \(3\times\) the capacitance |
| Feeder loss | \(P_{\text{loss}} = 3I_L^2R \propto 1/\cos^2\phi\) | At fixed \(P\) and \(V_L\) |
| Instantaneous power | \(p(t) = 3V_{ph}I_{ph}\cos\phi\) | Constant — the reason for constant torque |
Common Mistakes
Putting the \(\sqrt3\) on the wrong quantity. Star has it on voltage, delta on current — one each, never both and never neither. Write down which connection you are in before writing anything else.
Taking the quoted supply voltage as a phase voltage. A three-phase supply is always quoted by its line voltage. "400 V three-phase" means \(V_L = 400\) V and, in star, \(V_{ph} = 231\) V — as in Problems 4 and 8.
Using the angle between line voltage and line current as \(\phi\). That angle is \(\phi \pm 30^\circ\). The \(\phi\) in every power formula is the load's own impedance angle.
Multiplying the line current by the total series impedance to get a load voltage. In Problem 3 that simply reproduces the source emf. Multiply by the impedance whose voltage you actually want.
Adding apparent powers or power factors arithmetically. Only \(P\) and \(Q\) add. Problem 17 gives 53.25 kVA the wrong way and 38.02 kVA the right way.
Giving a capacitor bank positive reactive power. Capacitors supply vars, so \(Q_C\) enters the sum negative. Getting the sign wrong roughly doubles the error rather than merely omitting the term.
Assuming \(I_N = 0\) without checking balance. True only for balanced loads. Problem 12 has 4.91 A flowing in a wire many students assume is dead.
Treating a three-wire unbalanced star as if the neutral were still there. Without the neutral the star point moves and every phase voltage changes — Problem 13. Applying the four-wire method gives three plausible and entirely wrong currents.
Reversing a negative wattmeter reading and then entering it as positive. The reversal is how you read the magnitude; the sign stays negative in the arithmetic. Problem 11 gives 7.5 kW correctly and 12.5 kW incorrectly.
Forgetting to convert delta to star before adding a line impedance. The two are not in series until they are in the same topology — Problem 8.
Using the mechanical output power to find the line current of a motor. The supply delivers the input power. Divide by efficiency first, as in Problem 15.
Applying \(\tan\phi = \sqrt3(W_1-W_2)/(W_1+W_2)\) to an unbalanced load. The sum is still the true power, but the deduced angle means nothing at all.
Every problem in this set has been solved in volts, amperes and ohms, and every one of them has required a decision about \(\sqrt3\) — whether the quantity in hand was a line or a phase value, and whether the connection was star or delta. On a single load that decision is a minor irritation. On a network with four transformers, three voltage levels and a dozen machines it becomes the dominant source of error.
Set 2 examines the economics that decide those voltage levels in the first place, and Set 3 removes the irritation altogether. The per-unit system normalises every quantity to a base chosen per voltage level, so that transformer ratios vanish from the impedance diagram and the \(\sqrt3\) factors cancel identically if the bases are chosen consistently. It is not a computational trick; it is the language in which every load-flow and fault study from Set 16 onwards is written.