Fundamentals Of Electrical Engineering Unit 3 Notes- AKTU B.Tech 1st Year

University: Dr. A.P.J. Abdul Kalam Technical University, AKTU

Programme: B.Tech First Year, Semester I and Semester II

Course Codes: AEE101A / AEE201A

Branches: Common to All

Unit: III

Unit Name: Three-Phase Systems

Introduction to Three-Phase Systems

A three-phase electrical system uses three alternating voltages having the same frequency and a defined phase relationship. In a balanced system, their RMS magnitudes are equal and successive phase voltages are separated by 120 electrical degrees.

Three-phase analysis builds on the sinusoidal waveform, phasor and power concepts developed in AC circuit analysis. The additional requirement is to distinguish the electrical quantities associated with one phase branch from the quantities measured in the external supply lines.

This distinction is essential because the voltage across a star-connected phase differs from the line voltage, while the current through a delta-connected phase differs from the line current. Applying the correct connection relationships is therefore the first step in a reliable numerical solution.

The unit covers balanced voltage generation, phase sequence, star and delta connections, total three-phase power, neutral-current behaviour and the two-wattmeter method. It also examines inverse calculations, reconnection problems, equivalent loads, unequal resistive branches and instrument range checks.

A source connection and a load connection must be identified independently. A star-connected generator can supply a delta-connected load. The voltage across a load branch is determined by the load’s terminals rather than by the appearance of the source symbol.

Throughout these notes, calculations use RMS values unless a quantity is explicitly marked instantaneous or peak. The phase sequence is ABC. An inductive phase impedance has a positive angle, while a capacitive phase impedance has a negative angle.

For the two-wattmeter arrangement used here, one instrument measures voltage AB with current A, while the other measures voltage CB with current C. Line B is the common pressure-coil reference. This convention fixes the reading-difference sign used in the reactive-power formula.

Unit 3 Topic Map

Syllabus Coverage

• Advantages of a three-phase supply.
• Balanced three-phase voltage generation.
• Phase sequence and electrical angle.
• Balanced phase-voltage phasors.
• Star and delta connections.
• Line and phase voltage relationships.
• Line and phase current relationships.
• Balanced three-phase active, reactive and apparent power.
• Constant total instantaneous power.
• Neutral-wire concept.
• Simple unequal resistive loads with an ideal neutral.
• Two-wattmeter power measurement.
• Power factor from balanced wattmeter readings.
• Zero and negative wattmeter indications.
• Instrument scales, range checks and integrated numerical problems.

Conventions and Calculation Habits

Use VL for line-to-line RMS voltage, Vph for the RMS voltage across one phase branch, IL for RMS current in a supply line and Iph for RMS current through one phase branch.

A quoted impedance per phase belongs to each individual phase branch. It is not the combined impedance of all three phases.

Choose the connection before selecting a formula. Convert line quantities into branch quantities, analyse one phase and then reconstruct the line quantities and total power.

Retain algebraic signs in phasor and wattmeter calculations. A negative reading or current reference must be interpreted rather than automatically replaced with a positive magnitude.

1. Meaning of a Three-Phase System

A three-phase supply contains three alternating voltages of the same frequency. For a balanced set, all three RMS magnitudes are equal and their phase angles are separated by 120 electrical degrees.

The three voltages do not reach their positive peaks at the same instant. Their peaks occur successively in the order specified by the phase sequence.

For positive ABC phase sequence, a convenient sine representation is:

vAN = Vm sin ωt

vBN = Vm sin(ωt − 120°)

vCN = Vm sin(ωt + 120°)

Here Vm is the peak phase-to-neutral voltage. The angular frequency is the same for all three waveforms.

When evaluating a numerical sample, keep angular units consistent. If ωt is calculated in radians, convert the 120° phase displacements into radians before combining the angles.

The instantaneous sum of these three balanced phase voltages is zero:

vAN + vBN + vCN = 0

This does not mean that each voltage is zero. It means that the positive and negative instantaneous values cancel algebraically.

The letters A, B and C identify conductors. They do not represent different frequencies. Alternative phase names are used in practice, but these notes retain A, B and C consistently.

The source neutral is denoted N. A load star point is denoted n. These points are connected only when the diagram explicitly includes a neutral conductor or when an ideal balanced result establishes equal potential.

Balanced Supply and Balanced Load

A balanced supply describes the three source voltages. A balanced load describes three equal phase impedances.

Equal supply voltage magnitudes do not guarantee equal currents when the connected branch impedances differ. Likewise, equal resistance magnitudes are not sufficient if the phase reactances or their signs differ.

The standard balanced formulas assume equal complex phase impedances and a balanced sinusoidal supply. Unequal-load examples explicitly identify an ideal connected neutral.

Three-Phase Generator Construction

Spatially arranged windings produce generated voltages with different time positions. Their physical arrangement supports the phase displacement represented in waveform and phasor diagrams.

The winding arrangement, electrical reference and phase sequence must agree. A physical cutaway gives context, while the labelled electrical diagram defines the calculation.

Worked Problem 3.1: Sampling the Three Phase Voltages

Problem: A balanced ABC supply has phase-to-neutral voltage 120 V RMS at 60 Hz. With vAN = Vm sin ωt, calculate all three instantaneous voltages at 2 ms.

Step 1: Convert RMS to peak.

Vm = √2 × 120.

Vm = 169.706 V.

Angular frequency is:

ω = 2π × 60.

ω = 376.991 rad/s.

Step 2: Find the A-phase angle.

t = 0.002 s.

ωt corresponds to:

43.2°.

The B-phase angle is 43.2° − 120° = −76.8°. The C-phase angle is 43.2° + 120° = 163.2°.

Step 3: Evaluate the samples.

vAN = 116.171 V.

vBN = −165.222 V.

vCN = 49.0503 V.

Step 4: Check the balanced sum.

116.171 − 165.222 + 49.0503 is approximately zero, allowing for displayed rounding.

The RMS magnitude of every phase remains 120 V. An instantaneous sample and an RMS value describe different properties.

Worked Problem 3.2: Sampling at an A-Phase Peak

Problem: A balanced ABC supply has phase-to-neutral voltage 240 V RMS at 50 Hz. Find the instantaneous phase voltages at 5 ms.

Peak phase voltage:

Vm = √2 × 240.

Vm = 339.411 V.

Angular frequency:

ω = 314.159 rad/s.

At 5 ms, the A-phase angle is 90°. The B-phase angle is −30° and the C-phase angle is 210°.

Therefore:

vAN = 339.411 V.

vBN = −169.706 V.

vCN = −169.706 V.

Their sum is zero. At the positive peak of A, the other two balanced phase voltages are each negative one half of that peak.

The three waveforms remain equal in amplitude even though their values at this particular instant differ.

2. Advantages over a Single-Phase Supply

Three-phase systems transfer power through three coordinated phase voltages. Their usefulness is connected to the way phase-displaced currents and voltages combine.

For a balanced sinusoidal load, total instantaneous power is constant. Each individual phase has a twice-frequency power variation, but these variations cancel in the three-phase sum.

A suitable three-phase winding also produces a rotating magnetic field. This is important in electrical machines, including induction motors.

A single isolated sinusoidal winding does not produce the same rotating-field pattern by itself. The coordinated spatial and time displacements of three phases provide the required field progression.

Practical Advantages

• Three-phase systems are widely used for power distribution and motor loads.
• Balanced operation provides steady total electrical power.
• A star system can offer line-to-neutral and line-to-line voltage levels.
• Balanced three-wire loads do not require a load-current neutral conductor.
• Suitable system design can use conductor material efficiently for a specified power transfer.
• Three-phase machine windings naturally support rotating-field operation.

A conductor-material comparison requires stated voltage limits, losses, conductor arrangements and power factor. It is not accurate to claim one universal percentage saving without defining the comparison.

Constant total power does not mean that individual phase currents become DC. Every line can still carry a sinusoidal current.

Equal current magnitudes also do not mean equal instantaneous direction. The currents reach corresponding points at different times.

Protected Three-Phase Trainer

A training panel provides a physical setting for line and branch measurements. Its appearance does not replace a connection diagram.

Before applying square-root-of-three relations, identify the load connection, supply balance, sequence and quantities being measured.

Worked Problem 3.3: Single-Phase and Three-Phase Current Comparison

Problem: Compare current for 9 kW supplied at 415 V RMS and power factor 0.9 by single phase and balanced three phase. Use the same numerical line voltage.

Single-phase current is:

I1 = P/(VPF).

I1 = 9000/(415 × 0.9).

I1 = 24.0964 A.

Balanced three-phase line current is:

I3 = P/(√3VLPF).

I3 = 9000/(√3 × 415 × 0.9).

I3 = 13.9121 A.

The ratio is:

I1/I3 = √3 = 1.73205.

This comparison holds active power, numerical voltage and power factor fixed. It does not by itself establish a complete conductor-material comparison because conductor count and voltage-to-earth constraints also matter.

Worked Problem 3.4: Unity-Power-Factor Comparison

Problem: Compare currents for 6 kW at 230 V RMS and unity power factor.

Single-phase current:

I1 = 6000/230.

I1 = 26.087 A.

Balanced three-phase current:

I3 = 6000/(√3 × 230).

I3 = 15.0613 A.

Again:

I1/I3 = √3.

The three-phase current is a current in one line. Do not add the three equal line-current magnitudes to obtain another supply-current value.

The phase displacement is already accounted for in the three-phase power expression.

3. Generation and Phase Sequence

An ideal three-phase alternator has three equivalent windings displaced by 120 electrical degrees.

A rotating magnetic field links the windings in the same pattern at different electrical positions. The resulting voltages have equal frequency and magnitude but different initial phases.

Phase sequence is the order in which the voltages pass through corresponding points, such as positive maxima.

ABC sequence means the A maximum is followed by the B maximum and then the C maximum.

With AN as zero-phase reference:

VAN = Vph at 0°.

VBN = Vph at −120°.

VCN = Vph at +120°.

The C angle +120° is equivalent to −240°. This is consistent with C occurring after B in the time progression.

Interchanging any two phase identities reverses sequence. Cyclic relabelling from ABC to BCA or CAB preserves the same sequence.

Electrical and Mechanical Angles

The 120° separation is electrical.

Mechanical displacement depends on the number of pole pairs in a machine. The simplified one-pole-pair sketch uses matching electrical and mechanical angles only as an introductory representation.

At frequency f, corresponding events in successive phases are separated by:

Δt = T/3 = 1/(3f)

Phase sequence describes order rather than a difference in voltage magnitude. Three equal RMS readings alone do not establish the sequence.

Worked Problem 3.5: Successive Peaks at 25 Hz

Problem: An ABC source operates at 25 Hz. If A reaches a positive maximum at zero time, find the next B and C maxima and the next A maximum.

Period:

T = 1/25.

T = 40 ms.

A 120° displacement is one third of a cycle:

Δt = 40/3.

Δt = 13.3333 ms.

Therefore:

• Next B maximum: 13.3333 ms.
• Next C maximum: 26.6667 ms.
• Next A maximum: 40 ms.

The sequence is about corresponding events. Comparing an A positive maximum with a B negative maximum would introduce another half-cycle and answer a different timing question.

Worked Problem 3.6: Successive Peaks at 50 Hz

Problem: Repeat the calculation for 50 Hz.

T = 1/50.

T = 20 ms.

Δt = T/3.

Δt = 6.66667 ms.

The next positive maxima occur at:

• B: 6.66667 ms.
• C: 13.3333 ms.
• A: 20 ms.

A cyclic change of labels preserves the order. Exchanging only two labels reverses it.

The time interval becomes smaller at higher frequency even though the electrical phase separation remains 120°.

Phase-Sequence Test Instrument

Sequence identifies phase progression, not merely the equality of three RMS voltages.

The direction convention used for the instrument, waveform diagram and rotating-field explanation must be consistent.

4. Balanced Phase-Voltage Phasors

Take VAN as the zero-degree reference.

The balanced ABC phase voltages are:

VAN = Vph at 0°.

VBN = Vph at −120°.

VCN = Vph at +120°.

All are RMS phasors.

In rectangular form:

VAN = Vph + j0.

VBN = −Vph/2 − j√3Vph/2.

VCN = −Vph/2 + j√3Vph/2.

Their real components cancel. Their imaginary components also cancel.

A line-to-line voltage is the difference of two conductor potentials:

VAB = VAN − VBN

It is not the sum of their magnitudes, and it is not their arithmetic magnitude difference.

Reversing terminal order gives:

VBA = −VAB

This reverses the vector by 180° while preserving magnitude.

Distinguish the Relevant Connection

A phase voltage is across one individual source winding or load impedance.

A line voltage is between two lines.

The relation between them depends on whether the particular winding or load is star or delta connected.

A star source feeding a delta load has one set of source phase quantities and another set of load phase quantities. The phrase phase voltage must therefore be attached to the intended component.

Worked Problem 3.7: Derive a Line Voltage Numerically

Problem: VAN is 100 V at zero degrees. Find VBN, VCN and VAB.

VBN = 100[cos(−120°) + j sin(−120°)].

VBN = −50 − j86.6025 V.

VCN = −50 + j86.6025 V.

Then:

VAB = 100 − (−50 − j86.6025).

VAB = 150 + j86.6025 V.

Magnitude:

173.205 V.

Angle:

+30°.

The line voltage is √3 times the phase magnitude and leads the corresponding A phase by 30° for these references.

Worked Problem 3.8: A 230 V Phase Supply

Problem: VAN = 230 V at zero degrees.

The other phase phasors are:

VBN = −115 − j199.186 V.

VCN = −115 + j199.186 V.

Their sum with VAN is zero.

Line voltage:

VAB = 230 − VBN.

VAB = 345 + j199.186 V.

Magnitude:

398.372 V.

Angle:

+30°.

The result shows why a phase-to-neutral value of 230 V corresponds to a line-to-line value close to 400 V in a balanced star set.

Worked Problem 3.9: A 240 V Phase Supply

Problem: VAN = 240 V at zero degrees.

VBN = −120 − j207.846 V.

VCN = −120 + j207.846 V.

VAB = 240 − VBN.

VAB = 360 + j207.846 V.

Magnitude:

415.692 V.

Angle:

+30°.

All three line voltages have equal magnitude under balance, but each has its own phase angle determined by the terminal order.

5. Line Quantities and Phase Quantities

Line voltage VL is an RMS voltage measured between two supply lines.

Line current IL is RMS current in one supply conductor.

Phase voltage Vph is the voltage across one source winding or one load phase impedance.

Phase current Iph is the current through that individual winding or impedance.

A stated 400 V three-phase supply normally means 400 V line-to-line RMS unless the question explicitly identifies another voltage.

An impedance quoted per phase belongs to each of the three branches.

Start Every Numerical with the Connection

For star, use Vph = VL/√3 and IL = Iph.

For delta, use Vph = VL and IL = √3Iph under balanced conditions.

Calculate the voltage across one phase before applying Ohm’s law to the per-phase impedance.

A neutral terminal is not necessary to define a phase branch. A delta phase is simply one branch connected between two lines.

The same total power must be obtained whether the calculation uses phase quantities or correctly converted line quantities.

Worked Problem 3.10: Star Line and Phase Values

Problem: A balanced star load has VL = 400 V and Iph = 10 A. Find Vph, IL and total apparent power.

Vph = 400/√3.

Vph = 230.94 V.

In star:

IL = Iph = 10 A.

Total apparent power:

S = 3VphIph.

S = 3 × 230.94 × 10.

S = 6928.2 VA.

The line expression gives the same result:

S = √3VLIL = 6928.2 VA.

Worked Problem 3.11: Delta Line and Phase Values

Problem: A balanced delta load has VL = 415 V and Iph = 8 A.

Each phase spans a line pair:

Vph = 415 V.

Line current:

IL = √3 × 8.

IL = 13.8564 A.

Total apparent power:

S = 3 × 415 × 8.

S = 9960 VA.

Checking with √3VLIL gives the same apparent power. The factor of three is already included in the line formula.

Worked Problem 3.12: Star Load at 230 V Line Voltage

Problem: A balanced star load has VL = 230 V and Iph = 6 A.

Vph = 230/√3.

Vph = 132.791 V.

Line current equals phase current:

IL = 6 A.

S = 3 × 132.791 × 6.

S = 2390.23 VA.

Do not use 230 V directly across each star branch. Doing so would overestimate phase current or branch power.

6. Star Connection: Circuit and Physical Meaning

In a star or Y connection, one terminal of each phase impedance is joined to a common star point n.

The other three terminals connect separately to lines A, B and C.

A neutral wire may connect n to the source neutral N, forming a four-wire system.

Each line conductor is in series with its associated phase impedance.

Therefore:

IL = Iph

This equality follows from the connection. There is no current-splitting junction between a line and its own star branch.

Under balanced conditions:

Vph = VL/√3

A 400 V balanced star load therefore has approximately 230.94 V across each phase.

Balanced Three-Wire Star

With equal phase impedances and a balanced supply, the load star point remains at the balanced neutral potential even when no neutral conductor is connected.

This result does not automatically apply to a floating unbalanced star.

A centre junction drawn in a star diagram does not necessarily mean that a fourth conductor returns to the source. The neutral link must actually be shown or specified.

The distinction matters because an ideal connected neutral fixes load phase voltages, while an unequal floating star can have shifted branch voltages.

Worked Problem 3.13: Balanced Resistive Star Load

Problem: Three 20 Ω resistors form a star across 400 V line-to-line RMS.

Phase voltage:

Vph = 400/√3.

Vph = 230.94 V.

Phase current:

Iph = 230.94/20.

Iph = 11.547 A.

Line current:

IL = 11.547 A.

One-phase power:

Vph²/R = 2666.67 W.

Total power:

P = 8000 W.

The line formula √3VLIL gives the same result at unity PF.

If a neutral is present, balanced neutral current is zero.

Balanced Load Training Module

Load balance and neutral connection are separate conditions. The balanced phase-voltage relation follows from the electrical model.

The labelled schematic determines the branch quantities even if a physical training module uses a different spatial arrangement.

Worked Problem 3.14: Resistive Star at 230 V Line Voltage

Problem: Three 10 Ω resistors form a balanced star across 230 V line voltage.

Vph = 230/√3.

Vph = 132.791 V.

Iph = 132.791/10.

Iph = 13.2791 A.

IL = 13.2791 A.

One-phase power:

1763.33 W.

Total active power:

P = 5290 W.

The complete balanced load has unity PF because every branch is resistive.

The current is not multiplied by √3 after calculating the branch current; that conversion belongs to balanced delta.

7. Derivation of the Star Voltage Relation

For ABC sequence:

VAB = VAN − VBN

Using the rectangular phase expressions:

VAB = Vph − Vph(−1/2 − j√3/2).

Therefore:

VAB = Vph(3/2 + j√3/2)

Its magnitude is:

Vph√[(3/2)² + (√3/2)²].

This simplifies to:

VL = √3Vph

The angle is:

tan⁻¹[(√3/2)/(3/2)].

Therefore:

VAB leads VAN by 30°.

With VAN at zero degrees, the line voltages are:

• VAB at +30°.
• VBC at −90°.
• VCA at +150°.

They remain equal in magnitude and separated by 120°.

Geometric Shift and Load Angle

The 30° displacement relates a specified line voltage to a corresponding phase-to-neutral voltage.

It is not automatically the load power-factor angle.

The load phase impedance contributes its own angle φ. Total three-phase power uses cos φ, where φ belongs to a phase voltage and its own phase current.

Reversing sequence or terminal order changes the phasor relationships. State both before reporting a lead or lag angle.

Worked Problem 3.15: Geometric Shift with an Inductive Load

Problem: VL = 400 V, VAN is at zero degrees and phase impedance angle is +30°. Find line-voltage angles and the angle between VAB and IA.

Vph = 400/√3.

Vph = 230.94 V.

The line angles are +30°, −90° and +150°.

IA lags VAN by 30°, so IA is at −30°.

The angle from IA to VAB is:

30° − (−30°).

60°.

However, load PF is:

cos 30° = 0.866025 lagging.

Using cos 60° in √3VLILcos φ would confuse a selected line-voltage/current angle with the load phase angle.

Worked Problem 3.16: Geometric Shift with a Leading Load

Problem: VL = 415 V and phase impedance angle is −20°.

Vph = 415/√3.

Vph = 239.6 V.

IA is at +20° relative to VAN.

VAB remains at +30°.

Their voltage-minus-current angle is:

30° − 20°.

10°.

The load PF is nevertheless:

cos 20° = 0.939693 leading.

The geometric relation does not redefine the impedance angle.

Worked Problem 3.17: A Unity-PF Star Load

Problem: VL = 230 V and phase impedance angle is zero.

Vph = 132.791 V.

IA is in phase with VAN and therefore has angle zero.

VAB is at +30°, so the selected line voltage and line current differ by 30°.

The load PF remains:

1, unity.

This example shows why the presence of a 30° line-angle difference does not imply a non-unity load power factor.

8. Balanced Star-Load Calculation

For a balanced star load:

Vph = VL/√3

Iph = Vph/Zph

IL = Iph

If:

Zph = R + jX

then:

|Zph| = √(R² + X²)

φ = atan2(X, R)

With VAN at zero degrees:

IA is at −φ.

The other current angles are −120° − φ and +120° − φ.

Total powers are:

P = 3Iph²R

Q = 3Iph²X

S = 3VphIph

Positive X gives positive Q and lagging current. Negative X gives negative Q and leading current.

Squaring X to find impedance magnitude removes its sign. Recover reactive direction from the original impedance rather than from the magnitude calculation.

Worked Problem 3.18: Complete Balanced Star Calculation

Problem: Find phase voltage, currents, powers, PF and current direction.

Impedance magnitude:

√(8² + 6²).

|Zph| = 10 Ω.

Phase angle:

φ = 36.8699°.

Phase voltage:

Vph = 230.94 V.

Current:

Iph = 230.94/10.

Iph = IL = 23.094 A.

Total powers:

P = 12,800 W.

Q = +9600 var.

S = 16,000 VA.

PF = 0.8 lagging.

With VAN at zero degrees, IA is at −36.8699°. The other line currents are displaced by 120° from it.

Industrial Induction Motor

A motor is a practical three-phase load, but its complete performance requires more than a fixed impedance model.

Electrical input and shaft output differ because of losses. If a numerical gives shaft output and efficiency, convert to electrical input before using an electrical power formula.

Worked Problem 3.19: A Capacitive Star Load

Problem: Zph = 10 − j7.5 Ω, VL = 230 V and f = 50 Hz.

|Zph| = √(10² + 7.5²).

|Zph| = 12.5 Ω.

φ = −36.8699°.

Phase voltage:

132.791 V.

Current:

Iph = IL = 10.6232 A.

Total powers:

P = 3385.6 W.

Q = −2539.2 var.

S = 4232 VA.

PF = 0.8 leading.

IA leads VAN by 36.8699°.

Its peak is 15.0235 A. With a sine reference:

iA = 15.0235 sin(314.159t + 0.643501) A

The positive current phase agrees with the negative capacitive impedance angle.

Worked Problem 3.20: Star Load at 440 V

Problem: Zph = 15 + j20 Ω and VL = 440 V.

Impedance magnitude:

25 Ω.

Impedance angle:

53.1301°.

Phase voltage:

440/√3.

254.034 V.

Phase and line current:

254.034/25.

10.1614 A.

Total active power:

4646.4 W.

Reactive power:

6195.2 var.

Apparent power:

7744 VA.

Power factor:

0.6 lagging.

With VAN as reference, IA is at −53.1301°.

9. Recovering a Star Phase Impedance

If line voltage and line current are known for a balanced star load:

|Zph| = VL/(√3IL)

If total active power is known:

R = P/(3IL²)

This follows because star phase current equals line current.

Reactance magnitude is:

|X| = √(|Zph|² − R²)

The load description supplies its sign.

For an inductive branch:

L = XL/(2πf)

For a capacitive branch:

C = 1/(2πfXC)

Measurements of voltage, current and active power determine power-factor magnitude, but do not alone distinguish inductive from capacitive behaviour.

Consistency Tests

For the assumed passive series-resistance/reactance model:

R ≤ |Zph|

Also:

P ≤ √3VLIL

A violated bound may indicate inconsistent data, an incorrect connection assumption or mixed RMS and peak values.

Remember that P is total power. A per-phase resistance must be calculated using one third of the balanced total power.

Worked Problem 3.21: Star Phase Values from Line Data

Problem: VL = 400 V, IL = 10 A, PF = 0.8 lagging and f = 50 Hz.

Vph = 230.94 V.

|Zph| = 230.94/10.

|Zph| = 23.094 Ω.

R = |Zph|PF.

R = 18.4752 Ω.

XL = |Zph|√(1 − PF²).

XL = 13.8564 Ω.

L = XL/(2π × 50).

L = 0.0441063 H per phase.

Total power:

P = 5542.56 W.

Checking 3IL²R gives the same total rather than the power of only one branch.

Worked Problem 3.22: A Lower-PF Star Load

Problem: VL = 415 V, IL = 8 A, PF = 0.6 lagging and f = 50 Hz.

Phase voltage:

239.6 V.

Impedance magnitude:

29.95 Ω.

Resistance:

R = 17.97 Ω.

Inductive reactance:

XL = 23.96 Ω.

Inductance:

L = 0.0762672 H per phase.

Total active power:

P = 3450.25 W.

The value of R is a phase-branch resistance. Treating total power as one-phase power would produce a factor-of-three error.

Worked Problem 3.23: A 0.9 PF Star Load

Problem: VL = 230 V, IL = 12 A, PF = 0.9 lagging and f = 50 Hz.

Vph = 132.791 V.

|Zph| = 132.791/12.

|Zph| = 11.0659 Ω.

R = 11.0659 × 0.9.

R = 9.95929 Ω.

XL = |Zph|√(1 − 0.9²).

XL = 4.82351 Ω.

L = 0.0153537 H per phase.

Total active power:

4302.41 W.

A high lagging PF means the resistive part is a larger fraction of impedance magnitude, not that inductance is necessarily zero.

10. Delta Connection and Branch Quantities

In delta connection, the three phase impedances form a closed loop.

The branches connect between A and B, B and C, and C and A. The supply lines connect to the three corners.

There is no common load star point.

Therefore:

Vph = VL

For a balanced load:

Iph = VL/|Zph|

At each corner, line current is formed by the phasor combination of two branch currents.

Choose branch directions:

• IAB from A to B.
• IBC from B to C.
• ICA from C to A.

KCL gives:

IA = IAB − ICA

IB = IBC − IAB

IC = ICA − IBC

The minus sign is determined by the arrows at the junction.

Current Magnitudes Must Not Be Added

Balanced branch currents are phase displaced. Their difference has magnitude √3Iph, not 2Iph.

The line current is also not zero merely because two branch-current magnitudes are equal. Their directions in the complex plane differ.

A negative instantaneous branch current means actual instantaneous flow opposes its reference arrow. KCL remains valid throughout the cycle.

Worked Problem 3.24: Balanced Resistive Delta

Phase voltage:

Vph = 400 V.

Phase current:

Iph = 400/60.

Iph = 6.66667 A.

Line current:

IL = √3Iph.

IL = 11.547 A.

One-phase power:

400²/60.

2666.67 W.

Total active power:

8000 W.

The load PF is unity because each resistor current is in phase with its own branch voltage.

Worked Problem 3.25: Delta at 415 V

Problem: Three 100 Ω resistors form a balanced delta across 415 V.

Vph = VL.

Vph = 415 V.

Iph = 415/100.

Iph = 4.15 A.

IL = √3 × 4.15.

IL = 7.18801 A.

One-phase power:

1722.25 W.

Total active power:

5166.75 W.

The line-power expression gives the same result. Do not use line current in one resistor’s I²R calculation because that resistor carries phase current.

Three-Branch Resistive Load Bank

The problem’s electrical values establish balance. Similar-looking hardware does not independently prove equal impedances.

For a delta analysis, apply line voltage to each phase branch and then obtain line current through the balanced current relation.

11. Derivation of the Delta Current Relation

Take VAB as reference and let phase impedance angle be φ.

For ABC sequence:

IAB = Iph at −φ

ICA = Iph at 120° − φ

At corner A:

IA = IAB − ICA

Factor out the common magnitude and load angle:

IA = Iph exp(−jφ)[1 − exp(j120°)].

The bracket becomes:

3/2 − j√3/2

Its magnitude is √3 and its angle is −30°.

Therefore:

IL = √3Iph

and:

IA lags IAB by 30°

for the stated branch and sequence references.

The angle between VAB and IA is 30° + φ.

That selected line-voltage/current angle is not the load PF angle. Load PF belongs to the voltage and current of the same phase branch.

The √3 magnitude relation requires balanced voltages and equal phase impedances. For an arbitrary unbalanced delta, use KCL with the actual branch phasors.

Worked Problem 3.26: Resistive Delta Branch Phasors

Problem: Branch current magnitude is 10 A and phase impedance angle is zero. Take VAB at zero degrees.

IAB = 10 at 0°.

IAB = 10 + j0 A.

ICA = 10 at 120°.

ICA = −5 + j8.66025 A.

IA = IAB − ICA.

IA = 15 − j8.66025 A.

Magnitude:

17.3205 A.

Angle:

−30°.

The line current lags the corresponding branch current by 30°, although load PF is unity.

Worked Problem 3.27: Inductive Delta Branch Phasors

Problem: Iph = 8 A and φ = 36.8699°.

IAB is at −36.8699°:

IAB = 6.4 − j4.8 A.

ICA is at 83.1301°:

ICA = 0.956922 + j7.94256 A.

IA = IAB − ICA.

IA = 5.44308 − j12.7426 A.

Magnitude:

13.8564 A.

Angle relative to VAB:

−66.8699°.

The line current has the additional 30° geometric lag relative to the AB branch current.

Worked Problem 3.28: Leading Delta Branch Phasors

Problem: Iph = 6 A and phase impedance angle is −30°.

IAB is at +30°:

IAB = 5.19615 + j3 A.

ICA is at +150°:

ICA = −5.19615 + j3 A.

Subtract:

IA = 10.3923 + j0 A.

Therefore:

IL = 10.3923 A.

The A-line current is in phase with VAB for this particular leading phase angle. The phase-branch PF is cos 30° leading, not unity.

12. Balanced Delta-Load Calculation

For balanced delta:

Vph = VL

Iph = Vph/|Zph|

IL = √3Iph

The total active power is:

P = 3Iph²R

Reactive power is:

Q = 3Iph²X

Apparent power is:

S = 3VphIph

Using line quantities:

P = √3VLIL cos φ

Q = √3VLIL sin φ

S = √3VLIL

For delta, IL² = 3Iph². Therefore total active power can also be written IL²R when R is the resistance in each balanced delta phase.

This is a derived total-power shortcut. The direct phase formula remains a reliable starting point.

If each phase impedance doubles at fixed line voltage and angle, both current magnitudes halve and active power halves.

If voltage doubles at fixed phase impedance, currents double and active and reactive powers become four times as large.

Worked Problem 3.29: Complete Delta Calculation

Impedance magnitude:

30 Ω.

Impedance angle:

36.8699°.

Phase current:

400/30.

Iph = 13.3333 A.

Line current:

IL = 23.094 A.

Total powers:

P = 12,800 W.

Q = 9600 var.

S = 16,000 VA.

PF = 0.8 lagging.

With VAN at zero degrees, IA is at −36.8699°. The AB branch voltage is at +30°, so IAB is at −6.8699°.

Worked Problem 3.30: Delta at 415 V

Problem: Zph = 40 + j30 Ω and VL = 415 V.

|Zph| = 50 Ω.

φ = 36.8699°.

Vph = 415 V.

Iph = 415/50.

Iph = 8.3 A.

IL = 14.376 A.

Total powers:

P = 8266.8 W.

Q = 6200.1 var.

S = 10,333.5 VA.

PF = 0.8 lagging.

The A-line current angle relative to VAN is −36.8699°. The AB phase current is 30° ahead of that line-current direction.

Worked Problem 3.31: Capacitive Delta

Problem: Zph = 20 − j15 Ω, VL = 230 V and f = 50 Hz.

|Zph| = 25 Ω.

φ = −36.8699°.

Phase current:

Iph = 9.2 A.

Line current:

IL = 15.9349 A.

Total powers:

P = 5078.4 W.

Q = −3808.8 var.

S = 6348 VA.

PF = 0.8 leading.

With VAN as reference, IA is at +36.8699°. IAB is at +66.8699°.

The line-current peak is 22.5353 A:

iA = 22.5353 sin(314.159t + 0.643501) A

Worked Problem 3.32: Delta at 440 V

Problem: Zph = 30 + j40 Ω and VL = 440 V.

Impedance magnitude:

50 Ω.

Angle:

53.1301°.

Phase current:

8.8 A.

Line current:

15.242 A.

Total active power:

6969.6 W.

Reactive power:

9292.8 var.

Apparent power:

11,616 VA.

PF = 0.6 lagging.

With VAN at zero degrees, IA is at −53.1301° and IAB is at −23.1301°.

13. Finding Delta Impedance and Component Values

For balanced delta:

Iph = IL/√3

Vph = VL

Therefore:

|Zph| = √3VL/IL

If total active power is known:

R = P/(3Iph²)

Using the balanced current relation:

R = P/IL²

If PF is known:

R = |Zph|PF

|X| = |Zph|√(1 − PF²)

The sign of X comes from leading or lagging information.

For identical VL, IL and PF, the inferred delta phase impedance is three times the equivalent star phase impedance.

A line measurement set does not identify the physical branch impedance uniquely unless the connection is also known.

Worked Problem 3.33: Recover Delta R and L

Problem: VL = 400 V, IL = 20 A, PF = 0.8 lagging and f = 50 Hz.

Iph = 20/√3.

Iph = 11.547 A.

|Zph| = 400/11.547.

|Zph| = 34.641 Ω.

R = 34.641 × 0.8.

R = 27.7128 Ω.

XL = 34.641 × 0.6.

XL = 20.7846 Ω.

L = XL/(2π × 50).

L = 0.0661595 H per phase.

Total active power is 11,085.1 W. Checking P/IL² gives the same phase resistance for this balanced delta.

Worked Problem 3.34: Delta at 0.6 Lagging PF

Problem: VL = 415 V, IL = 12 A, PF = 0.6 and f = 50 Hz.

Phase current:

6.9282 A.

Phase impedance magnitude:

59.9001 Ω.

Resistance:

35.9401 Ω.

Inductive reactance:

47.9201 Ω.

Inductance:

0.152534 H per phase.

Total active power:

5175.37 W.

The phase current rather than line current belongs in the direct three-branch loss expression 3Iph²R.

Worked Problem 3.35: Delta R and C

Problem: VL = 230 V, IL = 15 A, PF = 0.9 leading and f = 50 Hz.

Iph = 15/√3.

Iph = 8.66025 A.

|Zph| = 230/8.66025.

|Zph| = 26.5581 Ω.

R = |Zph| × 0.9.

R = 23.9023 Ω.

Capacitive reactance magnitude:

XC = 11.5764 Ω.

C = 1/(2πfXC).

C = 274.964 μF per phase.

Total active power is 5378.02 W. The impedance imaginary part is negative, while its resistance and average active absorption remain positive.

14. Star and Delta with the Same Supply and Impedances

Suppose the same three equal phase impedances are first connected in delta and then reconnected in star across the same line voltage.

In delta, each phase receives VL.

In star, each receives VL/√3.

Therefore star phase current is 1/√3 of the original delta phase current.

Since delta line current is √3 times delta phase current, while star line current equals star phase current:

IL,Y = IL,Δ/3

For unchanged impedance angle:

PY = PΔ/3

The one-third result requires unchanged phase impedances and unchanged line voltage.

It is not a universal comparison between arbitrary star and delta loads. If impedances or supply conditions differ, recalculate the actual branch conditions.

This is a fixed-impedance circuit result rather than a complete machine-starting analysis. Machine impedance and torque can vary with operating condition.

Worked Problem 3.36: Reconnect 8 + j6 Ω Phases

Problem: Three impedances 8 + j6 Ω are connected first in delta and then star across 400 V.

Magnitude:

|Zph| = 10 Ω.

Power factor remains 0.8 lagging.

In delta:

Iph = 400/10.

Iph,Δ = 40 A.

IL,Δ = 69.282 A.

In star:

Iph = 400/(√3 × 10).

IL,Y = Iph,Y = 23.094 A.

Total powers:

PΔ = 38,400 W.

PY = 12,800 W.

Both line current and active power have the required one-third ratio.

Six-Terminal Motor Interface

Six accessible winding terminals allow alternative links when the device is designed for those connections.

The permitted connection and actual terminal identification come from device information. A photograph does not establish the required electrical links by itself.

Worked Problem 3.37: Reconnect Resistive Phases

Problem: Three 20 Ω resistors are connected first in delta and then star across 230 V.

In delta:

Vph = 230 V.

Iph = 11.5 A.

IL = 19.9186 A.

In star:

Vph = 132.791 V.

IL = Iph = 6.63953 A.

Total powers are:

PΔ = 7935 W.

PY = 2645 W.

Power factor is unity in both connections. The current reduction comes from the reduced phase voltage and changed line-current relationship, not from a change in resistance.

Worked Problem 3.38: Reconnect Capacitive Impedances

Problem: Three impedances 15 − j20 Ω are connected first in delta and then star across 440 V.

Magnitude:

25 Ω.

Power factor:

0.6 leading.

Delta phase current:

17.6 A.

Delta line current:

30.4841 A.

Star phase voltage:

254.034 V.

Star line current:

10.1614 A.

Total powers:

PΔ = 13,939.2 W.

PY = 4646.4 W.

The one-third active-power ratio also holds for this leading load because its complete phase impedances remain unchanged.

15. Balanced Star–Delta Equivalent Loads

A balanced delta load has the same external line behaviour as a balanced star load when:

ZY = ZΔ/3

or:

ZΔ = 3ZY

This is an equivalent-load conversion.

It differs from reconnecting the same impedances. In an equivalent conversion, phase impedances are deliberately changed to preserve line currents and total powers.

The factor of three applies to the full complex impedance.

If:

ZY = 4 + j3 Ω

then:

ZΔ = 12 + j9 Ω

Both resistance and reactance scale. Scaling only resistance changes phase angle and does not preserve the load.

For inductive equivalents, inductance triples at the same frequency.

For series RC equivalents, capacitive reactance triples, which means capacitance becomes one third.

The simple rule applies to balanced equal branches. General unequal star–delta transformations require other expressions.

Worked Problem 3.39: Equivalent Delta for 4 + j3 Ω Star

Problem: ZY = 4 + j3 Ω at VL = 400 V.

Equivalent delta:

ZΔ = 12 + j9 Ω.

Star magnitude is 5 Ω. Delta magnitude is 15 Ω.

Star line current:

400/(√3 × 5).

IL,Y = 46.188 A.

Delta phase current:

400/15.

Iph,Δ = 26.6667 A.

Delta line current:

√3 × 26.6667.

IL,Δ = 46.188 A.

Both loads absorb:

P = 25,600 W.

Their external line currents and PF agree despite different branch currents and impedances.

Worked Problem 3.40: Equivalent Delta at 415 V

Problem: ZY = 10 + j7.5 Ω and VL = 415 V.

Equivalent delta:

ZΔ = 30 + j22.5 Ω.

The star magnitude is 12.5 Ω and the delta magnitude is 37.5 Ω.

Star line current:

19.168 A.

Delta phase current:

11.0667 A.

Delta line current:

19.168 A.

Total active power in either case:

11,022.4 W.

The phase impedance angle remains unchanged, so the PF remains 0.8 lagging.

Worked Problem 3.41: Capacitive Equivalent

Problem: ZY = 8 − j6 Ω and VL = 230 V.

Equivalent delta:

ZΔ = 24 − j18 Ω.

Star line current:

230/(√3 × 10).

13.2791 A.

Delta phase current:

230/30.

7.66667 A.

Delta line current:

13.2791 A.

Both total active powers are:

4232 W.

Both loads have 0.8 leading PF.

This is equivalence with different phase impedances, not the one-third-current reconnection of unchanged impedances.

16. Power in Balanced Three-Phase Circuits

Each balanced phase has active power:

Pph = VphIph cos φ

Total active power is:

P = 3VphIph cos φ

Substituting either the star or delta relations gives:

P = √3VLIL cos φ

Reactive power is:

Q = 3VphIph sin φ

or:

Q = √3VLIL sin φ

Apparent power is:

S = 3VphIph = √3VLIL

Therefore:

S² = P² + Q²

PF = P/S

Use watts or kilowatts for active power, var or kvar for reactive power and VA or kVA for apparent power.

The Angle in the Formula

φ is the phase-impedance angle.

It is the angle between a phase-branch voltage and that branch’s own current.

A selected line-voltage/current pair can include a 30° geometric shift. That extra shift must not be inserted as the load PF angle.

Inductive Q is positive. Capacitive Q is negative. Apparent power is a positive magnitude in both cases.

Worked Problem 3.42: Power from Line Measurements

Problem: VL = 400 V, IL = 20 A and PF = 0.8 lagging.

S = √3 × 400 × 20.

S = 13,856.4 VA.

P = S × 0.8.

P = 11,085.1 W.

Phase magnitude:

36.8699°.

Q = S × 0.6.

Q = +8313.84 var.

Checking √(P² + Q²) reproduces S. No additional factor of three is applied after using the line-power formula.

Worked Problem 3.43: Power at 0.6 Lagging PF

Problem: VL = 415 V, IL = 12 A and PF = 0.6.

S = √3 × 415 × 12.

S = 8625.61 VA.

P = 0.6S.

P = 5175.37 W.

Q = 0.8S.

Q = 6900.49 var.

Phase angle:

53.1301°.

The reactive component is larger than active power in numerical magnitude, but the total active absorption remains positive.

Worked Problem 3.44: Power at Leading PF

Problem: VL = 230 V, IL = 10 A and PF = 0.9 leading.

S = 3983.72 VA.

P = 3585.35 W.

Phase angle:

φ = −25.8419°.

Reactive power:

Q = −1736.46 var.

The negative Q gives capacitive behaviour. It does not make the apparent-power magnitude negative or imply that the load has negative active resistance.

Worked Problem 3.45: Unity-PF Total Power

Problem: VL = 440 V, IL = 15 A and PF = 1.

S = √3 × 440 × 15.

S = 11,431.5 VA.

At unity PF:

P = 11,431.5 W.

Q = 0 var.

φ = 0°.

Active and apparent powers are numerically equal but retain their distinct units and meanings.

The three line currents remain phase displaced even at unity power factor.

17. Constant Total Balanced Instantaneous Power

For one phase with RMS voltage Vph, RMS current Iph and current lag φ:

pA = VphIph[cos φ − cos(2ωt − φ)]

The B and C powers have the same mean term.

Their twice-frequency terms are displaced by 240° and 480°. Those three cosine terms form a balanced set and sum to zero.

Therefore:

pA + pB + pC = 3VphIph cos φ

at every instant.

For a pure reactive balanced load, cos φ is zero and total instantaneous active input is zero under the ideal model.

Individual phases can still exchange energy. Nonzero currents and stored energies are compatible with zero total active absorption.

Conditions Behind Constant Power

The cancellation requires balanced sinusoidal operation.

Unequal phase impedances, unequal waveform magnitudes or distortion can leave a time-varying total power.

Inspecting only one phase is insufficient. When A-phase voltage is zero, the other phases can still account for the complete total power.

Worked Problem 3.46: A-Phase Voltage at Zero

Problem: Each phase has 100 V RMS and 5 A RMS at 0.8 lagging PF. Find phase powers when the A voltage angle is zero.

φ = cos⁻¹0.8.

φ = 36.8699°.

Use:

p = 2VphIph sin α sin(α − φ)

with phase voltage angles 0°, −120° and +120°.

The phase powers are:

pA = 0 W.

pB = 340.192 W.

pC = 859.808 W.

Their sum is:

1200 W.

The balanced expression gives 3 × 100 × 5 × 0.8 = 1200 W, confirming the result.

Worked Problem 3.47: Unity-PF Phase Powers

Problem: Each phase has 230 V RMS and 2 A RMS at unity PF. The A voltage angle is 30°.

The phase angles are 30°, −90° and 150°.

Using p = 2VphIph sin²α:

pA = 230 W.

pB = 920 W.

pC = 230 W.

Total:

1380 W.

The constant balanced value is:

3 × 230 × 2.

1380 W.

Equal mean phase powers do not imply equal instantaneous phase powers at every moment.

Worked Problem 3.48: A Different Instant and PF

Problem: Each phase has 120 V RMS and 4 A RMS at 0.6 lagging PF. The A voltage angle is 60°.

φ = 53.1301°.

The three voltage angles are 60°, −60° and 180°.

Instantaneous phase powers are:

pA = 99.4462 W.

pB = 764.554 W.

pC = 0 W.

Total:

864 W.

The balanced constant is:

3 × 120 × 4 × 0.6.

864 W.

The calculation confirms that the zero power of one phase does not mean zero total system power.

18. Current, Impedance and PF from Power Data

If total P, line voltage and PF are given:

IL = P/(√3VLPF)

This expression applies to either balanced star or balanced delta because it uses line quantities.

Recovering phase impedance afterward requires the actual connection.

If total P and Q are known:

S = √(P² + Q²)

IL = S/(√3VL)

PF = P/S

The sign of Q gives leading or lagging direction.

For balanced phases:

Pph = P/3

Qph = Q/3

Units and Interpretation

Use watts with volts and amperes, or use another deliberately consistent unit system.

A value of 12 kW must become 12,000 W before substitution into a formula expecting watts.

At fixed active power and line voltage, unity PF gives the minimum line current associated with that active-power requirement. Lower PF increases apparent power and current.

Worked Problem 3.49: Current from 12 kW and 9 kvar

Problem: P = 12 kW, Q = +9 kvar and VL = 400 V.

S = √(12² + 9²).

S = 15 kVA.

IL = 15,000/(√3 × 400).

IL = 21.6506 A.

PF = 0.8 lagging.

φ = 36.8699°.

Per phase:

Pph = 4000 W.

Qph = 3000 var.

Sph = 5000 VA.

Connection is not required for this line-current calculation, but is required for physical branch voltage and impedance.

Worked Problem 3.50: A Leading Load from P and Q

Problem: P = 6 kW, Q = −8 kvar and VL = 415 V.

S = 10 kVA.

IL = 10,000/(√3 × 415).

IL = 13.9121 A.

PF = 0.6 leading.

φ = −53.1301°.

Per-phase complex power is:

2000 − j2666.67 VA.

Per-phase apparent-power magnitude is:

3333.33 VA.

The negative reactive component must remain signed when combining this load with another load.

Worked Problem 3.51: An 18 kW Load

Problem: P = 18 kW, Q = 13.5 kvar and VL = 440 V.

S = √(18² + 13.5²).

S = 22.5 kVA.

IL = 22,500/(√3 × 440).

IL = 29.5236 A.

PF = 0.8 lagging.

Per phase:

Pph = 6000 W.

Qph = 4500 var.

Sph = 7500 VA.

The three-phase total has already been used; multiplying by three again would be incorrect.

Worked Problem 3.52: Unity-PF Load from Power

Problem: P = 8 kW, Q = 0 and VL = 230 V.

S = 8 kVA.

IL = 8000/(√3 × 230).

IL = 20.0817 A.

PF = 1.

Per phase:

Pph = 2666.67 W.

Qph = 0.

The phase voltage and current separately still depend on whether the load is physically star or delta.

19. Neutral Wire: Purpose and Balanced Cancellation

A neutral wire connects the load star point to source neutral in a four-wire star system.

It provides a return path for unequal phase currents and, in the ideal model, keeps the load star point at the source neutral potential.

Choose line currents positive from source to load. Choose neutral current positive from load star point back to source.

KCL gives:

IN = IA + IB + IC

For balanced sinusoidal currents, the equal magnitudes are separated by 120° and sum to zero.

Zero neutral current does not mean zero line current. Each line can carry substantial RMS current while the vector resultant is zero.

Neutral and Protective Earth

Neutral is an intended circuit conductor in the stated system.

Protective earth has a separate protective function.

The terms should not be used as interchangeable names for a load-current return path.

The basic cancellation result assumes balanced sinusoidal currents. Some harmonic components in nonlinear loads can add in the neutral. Detailed harmonic analysis is outside these introductory calculations.

Worked Problem 3.53: Balanced Unity-PF Neutral Cancellation

Problem: Each star phase carries 10 A at unity PF. Take VAN as zero degrees.

Current phasors:

IA = 10 + j0 A.

IB = −5 − j8.66025 A.

IC = −5 + j8.66025 A.

Add real components:

10 − 5 − 5 = 0.

Add imaginary components:

0 − 8.66025 + 8.66025 = 0.

Therefore:

IN = 0 A.

The neutral cancellation is a vector result. Adding the three RMS magnitudes to obtain 30 A would ignore their phase relationship.

Three-Phase Conductors and Neutral

The actual wiring arrangement must match the system model.

A neutral-current calculation requires defined current directions and phases. Balanced cancellation is different from an arithmetic sum of magnitudes.

Worked Problem 3.54: Balanced Non-Unity-PF Cancellation

Problem: Each phase carries 12 A at 0.6 lagging PF.

φ = 53.1301°.

Current angles are −φ, −120° − φ and +120° − φ.

Rectangular values are:

IA = 7.2 − j9.6 A.

IB = −11.9138 − j1.43538 A.

IC = 4.71384 + j11.0354 A.

The real and imaginary sums are zero apart from displayed rounding.

IN = 0 A.

Balanced neutral cancellation does not require unity PF. Equal current magnitudes and equal phase displacement are the relevant conditions.

20. Simple Unequal Loads with an Ideal Neutral

With an ideal connected neutral, the source fixes the load star point.

Each branch can therefore be calculated independently using its own phase-to-neutral voltage and impedance.

The neutral returns the phasor imbalance.

For unequal resistive branches under a balanced supply:

IA = Vph/RA at 0°.

IB = Vph/RB at −120°.

IC = Vph/RC at +120°.

Add rectangular current components to find IN.

For this specific resistive case:

IN² = IA² + IB² + IC² − IAIB − IBIC − ICIA

The shortcut assumes 120° current separation. It does not apply unchanged when unequal branches also have different impedance angles.

Floating-Neutral Boundary

If an unequal star has no neutral connection, its star point can shift.

Do not continue assigning VL/√3 to every branch without analysis.

The simple calculations here explicitly use an ideal connected neutral. They illustrate the neutral’s role rather than the more general floating-star problem.

Worked Problem 3.55: Unequal Resistive Branches

Problem: Vph = 230 V and resistances A, B and C are 23 Ω, 46 Ω and 46 Ω.

Current magnitudes:

IA = 10 A.

IB = 5 A.

IC = 5 A.

Their angles are 0°, −120° and +120°.

Neutral sum:

IN = 10 + 5 at −120° + 5 at +120°.

IN = 5 + j0 A.

|IN| = 5 A.

Total active power:

230(10 + 5 + 5).

P = 4600 W.

This scalar power sum is valid because every resistive branch has unity phase PF.

Worked Problem 3.56: Three Different Resistances

Problem: Vph = 120 V and RA = 12 Ω, RB = 24 Ω and RC = 40 Ω.

Current magnitudes:

10 A, 5 A and 3 A.

Rectangular neutral current:

Real part = 10 − 5/2 − 3/2.

Real part = 6 A.

Imaginary part = −5√3/2 + 3√3/2.

Imaginary part = −1.73205 A.

Therefore:

IN = 6 − j1.73205 A.

Magnitude:

6.245 A.

Total active power:

120(10 + 5 + 3).

2160 W.

Do not insert one arbitrary unequal line current into the balanced √3VLIL formula.

Worked Problem 3.57: Two Equal Higher Currents

Problem: Vph = 240 V and resistances are 24 Ω, 24 Ω and 48 Ω.

Current magnitudes:

IA = 10 A.

IB = 10 A.

IC = 5 A.

Neutral current:

IN = 2.5 − j4.33013 A.

Magnitude:

|IN| = 5 A.

Total active power:

240(10 + 10 + 5).

P = 6000 W.

Equal A and B magnitudes do not cancel each other because their phase separation is 120°, not 180°.

Worked Problem 3.58: One Open Phase Branch

Problem: Vph = 100 V, RA = 10 Ω, RB = 20 Ω and the C branch is open.

Current magnitudes:

IA = 10 A.

IB = 5 A.

IC = 0 A.

Neutral current:

IN = 10 + 5 at −120°.

IN = 7.5 − j4.33013 A.

Magnitude:

8.66025 A.

Total active power:

100(10 + 5).

P = 1500 W.

The ideal connected neutral keeps the remaining branch voltages at their specified values despite the open C branch.

21. Two-Wattmeter Method: Connection

For a three-phase three-wire circuit, place one current coil in line A and the second current coil in line C.

Connect their pressure coils across AB and CB respectively.

Line B is common to the pressure-coil connections.

Define:

WA = Average of vABiA

WC = Average of vCBiC

Total active power is:

P = WA + WC

Use the original algebraic readings.

A negative reading is retained in the sum. The method does not require opening a delta or reconnecting it into star.

Scope of the Method

The active-power sum applies to an ideal three-wire measurement boundary, including an unbalanced three-wire load.

The simple difference formulas for reactive power and PF require balanced sinusoidal operation.

An arbitrary four-wire load may have a nonzero neutral current. The three line currents then do not necessarily sum to zero, so the same two-element proof cannot generally be applied.

A balanced four-wire load with zero neutral current is a special case, not a reason to extend the result to every four-wire system.

Worked Problem 3.59: Unity-PF Wattmeter Readings

Problem: VL = 400 V, IL = 10 A and PF = 1.

φ = 0°.

Instrument angles are both 30°.

WA = 400 × 10 × cos 30°.

WA = 3464.1 W.

WC = 3464.1 W.

Total power:

P = 6928.2 W.

Reactive power:

Q = 0 var.

The readings are equal and positive. Apparent power is also 6928.2 VA.

Worked Problem 3.60: Readings at 0.8 Lagging PF

Problem: VL = 415 V, IL = 12 A and PF = 0.8 lagging.

φ = 36.8699°.

Angles in the instruments:

30° + φ = 66.8699°.

30° − φ = −6.8699°.

Readings:

WA = 1956.25 W.

WC = 4944.25 W.

Total active power:

P = 6900.49 W.

Reactive power:

Q = 5175.37 var.

Apparent power:

S = 8625.61 VA.

The larger WC reading agrees with the positive lagging angle under this particular labelling convention.

Worked Problem 3.61: Readings at Leading PF

Problem: VL = 230 V, IL = 8 A and PF = 0.6 leading.

Signed phase angle:

φ = −53.1301°.

Instrument angles:

−23.1301° and 83.1301°.

Readings:

WA = 1692.09 W.

WC = 220.092 W.

Total active power:

P = 1912.18 W.

Reactive power:

Q = −2549.58 var.

Apparent power:

S = 3186.97 VA.

WA is larger for this leading load, while both readings remain positive because the PF magnitude is above 0.5.

22. Proof That the Wattmeter Sum Gives Total Power

For a three-wire circuit:

iA + iB + iC = 0

The instantaneous instrument products are:

pWA = (vA − vB)iA

pWC = (vC − vB)iC

Add them:

pWA + pWC = vAiA + vCiC − vB(iA + iC).

KCL gives:

iA + iC = −iB.

Therefore:

pWA + pWC = vAiA + vBiB + vCiC

The right-hand side is total instantaneous power entering the three-wire load.

Averaging gives:

P = WA + WC

Equal impedances are not required for this proof. The essential condition is the three-wire current boundary.

A separate neutral carrying current prevents the same KCL reduction unless its contribution is included through an appropriate measurement arrangement.

Average Products

A wattmeter measures the average of an instantaneous product.

It does not multiply the separately averaged voltage and current.

Both AC voltage and current can have zero signed average while their product has positive average active power.

Worked Problem 3.62: Algebraic Sum with Two Positive Readings

Problem: WA = 2000 W and WC = 3500 W in a three-wire circuit.

Total active input:

P = 5500 W.

This sum is valid without requiring equal phase impedances.

If balanced sinusoidal operation is also established:

Q = √3(3500 − 2000).

Q = 2598.08 var.

Without the balance information, report the active-power total but do not claim the calculated difference is the actual total reactive power.

Worked Problem 3.63: One Negative Reading

Problem: WA = −500 W and WC = 2500 W.

P = −500 + 2500.

P = 2000 W.

For an additionally established balanced sinusoidal load:

Q = √3[2500 − (−500)].

Q = 5196.15 var.

Adding absolute readings would give 3000 W, which is not the original algebraic total.

The negative instrument value is part of the measurement result and must be preserved.

Worked Problem 3.64: Larger A Instrument Reading

Problem: WA = 3000 W and WC = 1500 W.

Total active power:

P = 4500 W.

If balanced sinusoidal operation applies:

Q = √3(1500 − 3000).

Q = −2598.08 var.

The negative value indicates leading behaviour under the stated connection.

If the load is unbalanced, the active sum remains valid but the simple reactive difference is not generally justified.

23. Balanced-Load Wattmeter Formulas

Use ABC sequence and VAN at zero degrees.

VAB is at +30°, while IA is at −φ.

The angle between them is 30° + φ. Therefore:

WA = VLIL cos(30° + φ)

VCB is at +90°, while IC is at +120° − φ.

Their angle difference has cosine equal to cos(30° − φ). Therefore:

WC = VLIL cos(30° − φ)

Adding:

WA + WC = 2VLIL cos 30° cos φ.

Thus:

WA + WC = √3VLIL cos φ

Subtracting in the defined order:

WC − WA = VLIL sin φ

Therefore:

Q = √3(WC − WA)

Label Convention

Some textbooks interchange wattmeter numbers or use another common line. Their difference formula can have the opposite order.

Here WA is AB/A and WC is CB/C.

For this convention:

• Lagging load: WC is larger.
• Leading load: WA is larger.
• Unity PF: equal readings.

Match the formula to the labelled coils rather than relying on a remembered W1 or W2 name without checking the circuit.

Worked Problem 3.65: Balanced Readings at 0.8 Lagging

Problem: VL = 400 V, IL = 15 A and PF = 0.8 lagging.

φ = 36.8699°.

WA = VLIL cos(66.8699°).

WA = 2356.92 W.

WC = VLIL cos(−6.8699°).

WC = 5956.92 W.

Total:

P = 8313.84 W.

Reactive power:

Q = 6235.38 var.

Apparent power:

S = 10,392.3 VA.

The ratio P/S reproduces 0.8.

Worked Problem 3.66: Balanced Readings at 0.9 Leading

Problem: VL = 415 V, IL = 10 A and PF = 0.9 leading.

φ = −25.8419°.

Instrument angles are:

4.15807° and 55.8419°.

Readings:

WA = 4139.08 W.

WC = 2330.13 W.

Total active power:

6469.21 W.

Reactive power:

−3133.18 var.

Apparent power:

7188.01 VA.

The signed difference gives the leading direction. Both instruments are positive because neither voltage-current angle exceeds 90°.

Worked Problem 3.67: One Zero Reading

Problem: VL = 230 V, IL = 20 A and PF = 0.5 lagging.

φ = 60°.

WA angle is 90°, so:

WA = 0 W.

WC angle is −30°, so:

WC = 3983.72 W.

Total active power:

P = 3983.72 W.

Reactive power:

Q = 6900 var.

Apparent power:

S = 7967.43 VA.

The zero reading does not mean zero current in that instrument. Its voltage and current are in quadrature.

Worked Problem 3.68: One Negative Reading at Low PF

Problem: VL = 440 V, IL = 8 A and PF = 0.3 lagging.

φ = 72.5424°.

The WA angle is 102.542°, giving:

WA = −764.41 W.

The WC angle is −42.5424°, giving:

WC = 2593.46 W.

Total active power:

P = 1829.05 W.

Reactive power:

Q = 5815.99 var.

Apparent power:

S = 6096.82 VA.

Keeping the negative sign is necessary to recover the specified 0.3 PF.

24. Finding Power Factor from Two Signed Readings

For a balanced sinusoidal load with positive total active power:

tan φ = √3(WC − WA)/(WA + WC)

Use the original signed readings.

An alternative procedure is:

P = WA + WC

Q = √3(WC − WA)

S = √(P² + Q²)

PF = P/S

This method keeps the reactive-power direction visible.

If P is zero, do not divide Q by P.

Equal and opposite nonzero readings correspond to a pure reactive balanced limit. Use the physical interpretation or atan2(Q, P).

If VL is known:

IL = S/(√3VL)

Recovering physical phase impedance afterward requires knowing star or delta connection.

Balance Condition

The active sum remains valid more generally for three-wire loads.

The difference-based PF calculation is not a general formula for an arbitrary unbalanced load.

Balance must be established before interpreting the reading difference as total reactive power.

Worked Problem 3.69: PF and Current from 2 kW and 4 kW

Problem: WA = 2000 W, WC = 4000 W and VL = 400 V for a balanced load.

P = 2000 + 4000.

P = 6000 W.

Q = √3(4000 − 2000).

Q = 3464.1 var.

S = √(6000² + 3464.1²).

S = 6928.2 VA.

PF = 0.866025 lagging.

φ = 30°.

IL = 6928.2/(√3 × 400).

IL = 10 A.

Power and Apparent-Power Measurement

A measurement needs defined voltage and current conventions, connections and ranges.

Active, reactive and apparent powers remain distinct even when an instrument displays several quantities on the same panel.

Worked Problem 3.70: Leading PF from Unequal Readings

Problem: WA = 4500 W, WC = 1500 W and VL = 440 V.

P = 6000 W.

Q = √3(1500 − 4500).

Q = −5196.15 var.

S = 7937.25 VA.

PF = 0.755929 leading.

φ = −40.8934°.

Line current:

IL = 10.4149 A.

Taking the absolute difference before assigning direction would lose the leading information even if the PF magnitude were calculated correctly.

Worked Problem 3.71: Equal Positive Readings

Problem: WA = WC = 3000 W and VL = 400 V.

P = 6000 W.

Q = 0 var.

S = 6000 VA.

PF = 1.

φ = 0°.

IL = 6000/(√3 × 400).

IL = 8.66025 A.

For this nonzero balanced sinusoidal case, equal readings establish unity PF under the defined references.

Worked Problem 3.72: Equal and Opposite Readings

Problem: WA = 500 W, WC = −500 W and VL = 230 V.

P = 0 W.

Q = √3(−500 − 500).

Q = −1732.05 var.

S = 1732.05 VA.

PF = 0, leading.

φ = −90°.

IL = 1732.05/(√3 × 230).

IL = 4.34783 A.

Use the pure-reactive interpretation. The tangent expression would require division by zero and is not the appropriate direct step.

25. Zero and Negative Wattmeter Readings

For the defined balanced ABC connection:

WA = VLIL cos(30° + φ)

WC = VLIL cos(30° − φ)

Reading signs depend on the instrument voltage-current angles.

At low lagging PF, 30° + φ exceeds 90°. Its cosine becomes negative.

An individual wattmeter reading is not the active power of one isolated physical load phase. A negative reading therefore does not mean one passive resistor is generating energy.

Documented Coil Reversal

An analogue instrument may be described as having one coil reference reversed to obtain an upscale indication.

Restore the original negative sign before applying the original connection formulas.

A positive display after reversal is not automatically a positive original reading.

Always distinguish displayed magnitude from the algebraic reading corresponding to the declared measurement references.

Two-Wattmeter Hardware

The two readings must be interpreted with their labelled coil connections and signs.

Adding absolute values changes the measured total whenever an original reading is negative.

Worked Problem 3.73: A 0.25 Lagging Load

Problem: VL = 400 V, IL = 10 A and PF = 0.25 lagging.

φ = 75.5225°.

WA angle:

105.522°.

WC angle:

−45.5225°.

Readings:

WA = −1070.47 W.

WC = 2802.52 W.

Total active power:

P = 1732.05 W.

Reactive power:

Q = 6708.2 var.

Apparent power:

S = 6928.2 VA.

The total remains positive even though WA is negative.

Worked Problem 3.74: A 0.5 Leading Load

Problem: VL = 400 V, IL = 10 A and PF = 0.5 leading.

φ = −60°.

WA angle is −30°:

WA = 3464.1 W.

WC angle is 90°:

WC = 0 W.

Total:

P = 3464.1 W.

Reactive power:

Q = −6000 var.

Apparent power:

S = 6928.2 VA.

Which meter becomes zero distinguishes the leading case from the corresponding 0.5 lagging case under this fixed convention.

Worked Problem 3.75: Pure Inductive Limit

Problem: VL = 400 V, IL = 10 A and PF = 0 lagging.

φ = 90°.

WA angle is 120°:

WA = −2000 W.

WC angle is −60°:

WC = +2000 W.

Total active power:

P = 0 W.

Reactive power:

Q = 6928.2 var.

Apparent power:

S = 6928.2 VA.

Equal and opposite readings give zero total active power while line current remains nonzero.

26. Wattmeter Ranges, Consistency and Load Identification

Each current coil carries its associated line current.

Each pressure coil is connected across two lines.

Therefore its voltage-range check uses:

VL

rather than the star phase voltage VL/√3.

If full-scale watts equal the selected range product over D divisions:

K = VrangeIrange/D

in watts per division.

Use a supplied scale constant or multiplying factor directly when the problem gives one.

For balanced data, compare S obtained from P and Q with √3VLIL.

Disagreement can indicate a wrong reading sign, missed scale factor, incorrect balance assumption or incompatible numerical data.

At low PF, small active watts do not imply small line current.

For a motor, the method measures electrical input power. If mechanical output and efficiency are supplied, obtain electrical input before calculating line current.

Worked Problem 3.76: Scale Readings with Original References

Problem: Each instrument has 50 W per division. WA indicates 40 divisions and WC indicates 80 divisions, both with original references.

WA = 40 × 50.

WA = 2000 W.

WC = 80 × 50.

WC = 4000 W.

Total:

P = 6000 W.

Reactive power:

Q = 3464.1 var.

Apparent power:

S = 6928.2 VA.

PF = 0.866025 lagging.

No sign reversal is needed because the problem explicitly retains both original references.

Worked Problem 3.77: Restore a Reversed Reading

Problem: Each instrument has 40 W per division. WA shows 20 divisions after one coil reference is reversed. WC shows 90 divisions with its original reference.

The original WA reading is:

WA = −20 × 40 = −800 W.

WC is:

WC = 90 × 40 = 3600 W.

Total:

P = 2800 W.

Q = √3[3600 − (−800)].

Q = 7621.02 var.

S = 8119.11 VA.

PF = 0.344865 lagging.

Adding displayed positive magnitudes would give 4400 W and an incorrect result.

Worked Problem 3.78: Scale Readings Giving Leading PF

Problem: Scale constant is 20 W per division. WA shows 75 divisions and WC shows 25 divisions with original references.

WA = 1500 W.

WC = 500 W.

Total active power:

P = 2000 W.

Reactive power:

Q = −1732.05 var.

Apparent power:

S = 2645.75 VA.

PF = 0.755929 leading.

The larger A reading gives negative Q under the specified B-common connection.

Worked Problem 3.79: Coil Ranges at Low PF

Problem: A balanced load has VL = 400 V, IL = 8 A and PF = 0.2 lagging. Each meter uses a 450 V pressure range and a 10 A current range.

φ = 78.463°.

Predicted readings:

WA = −1013.42 W.

WC = 2121.93 W.

Total:

P = 1108.51 W.

Voltage check:

400 V ≤ 450 V.

Current check:

8 A ≤ 10 A.

Both range selections pass. Opposite reading signs are compatible with the low PF and do not indicate that the range is incorrect.

Worked Problem 3.80: Current Range Exceeded

Problem: VL = 415 V, IL = 12 A and PF = 0.2 lagging. Ranges remain 450 V and 10 A.

Predicted readings:

WA = −1577.13 W.

WC = 3302.25 W.

Total:

P = 1725.12 W.

Voltage range passes:

415 V ≤ 450 V.

Current range fails:

12 A > 10 A.

The small net watt total does not make the selected current range acceptable. Each current coil carries the actual line current.

Worked Problem 3.81: A Lower-Voltage Range Check

Problem: VL = 230 V, IL = 4 A and PF = 0.2 lagging. Selected ranges are 300 V and 5 A.

Readings:

WA = −291.357 W.

WC = 610.055 W.

Total:

P = 318.697 W.

Voltage check:

230 V ≤ 300 V.

Current check:

4 A ≤ 5 A.

Both ranges pass.

The pressure-coil check still uses full line voltage even if the connected load is star.

27. Integrated Three-Phase Numerical Method

A complete solution distinguishes the source, external lines and individual load phases.

Identify whether a stated voltage is line-to-line or phase-to-neutral.

Identify whether power means combined electrical input, one-phase power or mechanical output.

Identify whether impedance is per phase and whether the load is balanced.

Reliable Examination Sequence

• Identify star or delta.
• Check balanced-load assumptions.
• Determine voltage across one phase.
• Find phase impedance magnitude and angle.
• Calculate phase current.
• Convert to line current.
• Calculate P, Q, S and PF.
• State current lead or lag.
• Predict signed wattmeter readings if requested.
• Verify using independent phase and line calculations.

For neutral problems, declare the return-current reference and add phasors.

For reconnection problems, state what remains unchanged.

For equivalent-load problems, scale the complete complex impedance.

Three useful power checks are the sum of phase powers, the line-power formula and the algebraic wattmeter sum. Their agreement checks connection conversion, phase angle and meter signs.

Worked Problem 3.82: Integrated Star Load

Problem: Analyse the load at 50 Hz.

Impedance magnitude:

15 Ω.

Angle:

36.8699°.

Phase voltage:

230.94 V.

Phase and line current:

15.396 A.

Total powers:

P = 8533.33 W.

Q = 6400 var.

S = 10,666.7 VA.

PF = 0.8 lagging.

With VAN as reference, IA is at −36.8699°.

Peak line current is 21.7732 A:

iA = 21.7732 sin(314.159t − 0.643501) A

Predicted wattmeter readings:

WA = 2419.15 W.

WC = 6114.19 W.

Their sum agrees with the total active power, allowing for displayed rounding.

Worked Problem 3.83: Integrated Delta Load

Impedance magnitude:

50 Ω.

Angle:

53.1301°.

Phase voltage:

415 V.

Phase current:

8.3 A.

Line current:

14.376 A.

Total powers:

P = 6200.1 W.

Q = 8266.8 var.

S = 10,333.5 VA.

PF = 0.6 lagging.

With VAN at zero degrees, IA is at −53.1301° and IAB is at −23.1301°.

Peak line current is 20.3308 A:

iA = 20.3308 sin(314.159t − 0.927295) A

Wattmeter readings:

WA = 713.63 W.

WC = 5486.47 W.

Their sum is 6200.1 W.

Worked Problem 3.84: Integrated Leading Star Load

Problem: Zph = 8 − j6 Ω, VL = 230 V and f = 50 Hz.

Magnitude:

10 Ω.

Angle:

−36.8699°.

Phase voltage:

132.791 V.

Phase and line current:

13.2791 A.

Total powers:

P = 4232 W.

Q = −3174 var.

S = 5290 VA.

PF = 0.8 leading.

IA leads VAN by 36.8699°.

Peak line current is 18.7794 A:

iA = 18.7794 sin(314.159t + 0.643501) A

Wattmeter readings:

WA = 3032.25 W.

WC = 1199.75 W.

Their sum is 4232 W, and the negative difference-based Q agrees with the capacitive phase impedance.

Worked Problem 3.85: Integrated Leading Delta Load

Problem: Zph = 40 − j30 Ω, VL = 400 V and f = 50 Hz.

Magnitude:

50 Ω.

Angle:

−36.8699°.

Phase voltage:

400 V.

Phase current:

8 A.

Line current:

13.8564 A.

Total powers:

P = 7680 W.

Q = −5760 var.

S = 9600 VA.

PF = 0.8 leading.

With VAN at zero degrees, IA is at +36.8699° and IAB is at +66.8699°.

Peak line current is 19.5959 A:

iA = 19.5959 sin(314.159t + 0.643501) A

Wattmeter readings:

WA = 5502.77 W.

WC = 2177.23 W.

Their sum and signed difference agree with the independently calculated powers.

28. Formula Sheet and Examination Shortcuts

For either balanced connection:

P = √3VLIL cos φ

Q = √3VLIL sin φ

S = √3VLIL

S² = P² + Q²

PF = P/S

Balanced Equivalent Loads

ZΔ = 3ZY

ZY = ZΔ/3

Scale the complete complex impedance.

For unchanged impedances reconnected from delta to star at fixed VL:

IL,Y = IL,Δ/3

PY = PΔ/3

Do not confuse these two different comparisons.

Neutral Current

With line currents toward the load and neutral return toward the source:

IN = IA + IB + IC

For balanced sinusoidal currents:

IN = 0

For unequal resistive currents separated by 120°:

IN² = IA² + IB² + IC² − IAIB − IBIC − ICIA

Use direct phasor addition for more general unequal current angles.

Two Wattmeters with Line B Common

WA = VLIL cos(30° + φ)

WC = VLIL cos(30° − φ)

P = WA + WC

For balanced sinusoidal operation:

Q = √3(WC − WA)

For positive nonzero active power:

tan φ = √3(WC − WA)/(WA + WC)

Fast Checks

• In star, phase voltage is below line voltage.
• In star, phase current equals line current.
• In delta, phase voltage equals line voltage.
• In delta, phase current is below line current under balance.
• The balanced power formula already includes all three phases.
• The load PF angle belongs to matching phase voltage and current.
• Balanced neutral current is zero even with nonzero phase currents.
• Equal positive balanced wattmeter readings give unity PF.
• One zero reading gives PF magnitude 0.5 for a nonzero balanced load.
• A negative original reading remains negative in the sum.

Conditions Behind the Shortcuts

The √3 magnitude conversions require the appropriate balanced conditions.

The wattmeter active-power sum has broader three-wire validity than the balanced difference formula.

The reconnection ratio requires unchanged impedance and line voltage.

A dimensional check alone does not prove that the correct branch or connection has been used.

29. Short Questions with Answers

1. What Makes a Three-Phase Supply Balanced?

Answer: The three sinusoidal phase voltages have equal RMS magnitudes, equal frequency and 120° phase separation.

A balanced load additionally requires equal complex phase impedances.

Supply balance and load balance are separate statements. Equal source voltages can supply unequal currents when the connected impedances differ.

2. Why Is Line Voltage Larger Than Phase Voltage in Star?

Answer: A line voltage is the phasor difference of two phase voltages separated by 120°.

The resulting magnitude is √3 times one phase voltage.

It is not twice the phase magnitude because the two phasors are not opposite. The result also includes a 30° phase shift for the specified sequence and terminal references.

3. Why Is Line Current Equal to Phase Current in Star?

Answer: Each line conductor is directly in series with its associated phase branch.

There is no current-splitting junction between them.

Therefore IL = Iph follows from the actual connection. The √3 factor used for balanced delta current should not be introduced into a star branch-current calculation.

4. Why Does Delta Have a √3 Current Factor?

Answer: At each delta corner, line current is the vector difference of two phase-branch currents.

For balanced branch currents separated by 120°, that difference has magnitude √3Iph.

The currents must be subtracted as phasors using defined branch directions. Their equal magnitudes cannot be simply added or cancelled arithmetically.

5. Must Source and Load Connections Be the Same?

Answer: No.

A star-connected source can supply a delta-connected load.

The load phase voltage is determined by the terminals of its phase impedance. Keep source winding quantities separate from load branch quantities when the source and load use different connections.

6. Which Angle Belongs in the Total-Power Formula?

Answer: Use the phase-impedance angle, equal to the angle between a phase voltage and its own phase current.

A chosen line-voltage/current pair may include an additional geometric 30° shift.

Using that shifted angle instead of φ gives an incorrect power factor even when the current and voltage magnitudes are correct.

7. Can Phase Currents Be Nonzero While Neutral Current Is Zero?

Answer: Yes.

Balanced sinusoidal currents have equal magnitudes and 120° separation, so their vector sum is zero.

Each line can carry a substantial RMS current. Zero resultant neutral current does not mean that individual branch currents or phase powers are zero.

8. Does Every Unbalanced Star Have VL/√3 Across Each Branch?

Answer: No.

An ideal connected neutral can maintain the specified phase-to-neutral branch voltages. A balanced floating star also has the corresponding balanced result.

An unequal star without a neutral can develop a displaced star-point potential. Its branch voltages must then be determined from the actual network.

9. Why Are Two Wattmeters Sufficient in a Three-Wire Circuit?

Answer: The three line currents satisfy iA + iB + iC = 0.

This constraint allows two appropriately selected voltage-current products to reproduce total instantaneous input power.

Averaging those products gives the total active power. A separate current-carrying neutral removes the simple three-wire constraint.

10. Is the Two-Wattmeter Sum Valid Only for Balanced Loads?

Answer: No.

The ideal active-power sum applies to three-wire loads without requiring equal impedances.

The simple reactive-power and PF expressions based on reading difference require balanced sinusoidal operation. The validity of the sum should not be used to justify every balanced-load formula for an arbitrary unbalanced load.

11. What Does One Negative Reading Indicate?

Answer: For the defined balanced passive-load connection, one instrument becomes negative below PF magnitude 0.5.

Its measured voltage-current angle exceeds 90°.

The sign is retained in total power. An individual reading is not the power of one isolated physical load phase, so negativity does not mean one passive resistor generates energy.

12. Do Equal Wattmeter Readings Always Prove Unity PF?

Answer: For a nonzero balanced sinusoidal passive load with the defined references, equal readings give zero Q and unity PF.

That conclusion should not be extended automatically to arbitrary unbalanced conditions or a zero-reading case.

The assumptions and original meter references remain part of the result.

13. Why Must Absolute Readings Not Be Added?

Answer: Replacing a negative original reading with a positive magnitude changes the measured active-power total.

If a coil was reversed to obtain an upscale indication, restore the original sign first.

The method uses an algebraic sum. Display direction and reference reversal must therefore be documented before calculation.

14. How Does an Equivalent Delta Differ from Reconnection?

Answer: An equivalent delta uses phase impedance three times the equivalent star impedance so that external line behaviour remains unchanged.

Reconnection keeps the same phase impedances and changes only the wiring.

At fixed line voltage, reconnecting unchanged delta phases into star reduces line current and power to one third.

15. Is Neutral the Same as Protective Earth?

Answer: No.

Neutral is an intended circuit conductor and reference or return path in the stated system.

Protective earth has a different protective role.

They should not be treated as interchangeable load-current conductors. The basic neutral-current equations refer to the electrical neutral connection defined by the circuit model.

30. Complete Examination Answers

1. Explain Balanced Three-Phase Generation

Answer:

An ideal three-phase generator has three equivalent windings displaced by 120 electrical degrees.

A rotating magnetic field links each winding in the same way at a different electrical position. The induced voltages therefore have equal amplitudes and frequency but successive phase displacement.

For ABC sequence, a convenient sine representation is:

vAN = Vm sin ωt

vBN = Vm sin(ωt − 120°)

vCN = Vm sin(ωt + 120°)

The instantaneous sum is zero.

Phase sequence identifies the order of corresponding events, usually positive maxima. After an A maximum, B reaches the next corresponding maximum one third of a period later and C reaches it another third-period later.

Interchanging two phase identities reverses sequence. A cyclic relabelling preserves the same progression.

The 120° separation is electrical. Its mechanical interpretation depends on machine pole arrangement.

Balanced three-phase operation supports a rotating magnetic field in a suitable machine winding. It also gives constant total instantaneous power to a balanced sinusoidal load.

These advantages do not mean that individual phase currents are constant. Each phase can still carry sinusoidal current and have pulsating phase power.

A complete explanation should distinguish balanced source voltages from balanced load impedances and should state the phase reference used in its equations.

2. Derive Star Line and Phase Relations

Answer:

In a star connection, one end of each phase joins a common star point, while the remaining ends connect separately to the three lines.

Each line is directly in series with one phase branch. Therefore:

IL = Iph

For ABC sequence with VAN at zero degrees:

VAN = Vph

VBN = Vph(−1/2 − j√3/2)

The AB line voltage is:

VAB = VAN − VBN

Substitution gives:

VAB = Vph(3/2 + j√3/2)

Its magnitude is:

Vph√[(3/2)² + (√3/2)²].

Therefore:

VL = √3Vph

The angle of the bracket is +30°, so VAB leads VAN by 30° for these references.

Thus each balanced star phase receives:

Vph = VL/√3

The voltage relationship depends on the balanced phasor set. For an unequal floating star, branch voltages cannot automatically be assigned using the balanced result.

The geometric 30° shift must also be kept separate from the phase-impedance angle. Total balanced power uses the angle between a phase branch voltage and its own current.

3. Derive Delta Line and Phase Relations

Answer:

A delta load has three phase impedances connected between AB, BC and CA.

Each branch spans two lines. Therefore:

Vph = VL

Choose branch-current references IAB from A to B, IBC from B to C and ICA from C to A.

KCL at A gives:

IA = IAB − ICA

For balanced ABC operation with VAB as reference:

IAB = Iph at −φ

ICA = Iph at 120° − φ

Factor out the common magnitude and impedance angle:

IA = Iph exp(−jφ)[1 − exp(j120°)].

The bracket becomes:

3/2 − j√3/2

Its magnitude is √3 and its angle is −30°.

Therefore:

IL = √3Iph

The A-line current lags the corresponding AB branch current by 30° for the specified directions.

This geometric shift does not change the phase-branch PF, which remains cos φ.

Use Iph in the direct total phase-power expression 3Iph²R. Use IL in √3VLILcos φ.

Mixing these currents produces a factor-of-three power error.

For an unbalanced delta, retain the actual branch phasors and apply KCL rather than assuming the balanced √3 magnitude conversion.

4. Derive Balanced Three-Phase Power Formulas

Answer:

For one phase:

Pph = VphIph cos φ

A balanced load has three identical phase power contributions, so:

P = 3VphIph cos φ

For star:

Vph = VL/√3

Iph = IL

Substituting gives:

P = √3VLIL cos φ

For delta:

Vph = VL

Iph = IL/√3

Substitution gives the same line expression.

Similarly:

Q = 3VphIph sin φ

Therefore:

Q = √3VLIL sin φ

Apparent-power magnitude is:

S = 3VphIph = √3VLIL

The power triangle gives:

S² = P² + Q²

and:

PF = P/S

Inductive Q is positive and PF is lagging. Capacitive Q is negative and PF is leading.

The angle φ is the phase-impedance angle. It is not an arbitrary angle between one line voltage and one line current.

All voltage and current magnitudes in these expressions are RMS values. Peak values require the corresponding factor from the sinusoidal power derivation.

5. Explain Constant Instantaneous Power

Answer:

Let each balanced phase have RMS voltage Vph and RMS current Iph, with a common current lag φ.

For the A phase:

vA = √2Vph sin ωt.

iA = √2Iph sin(ωt − φ).

Multiplying and applying the sine-product identity gives:

pA = VphIph[cos φ − cos(2ωt − φ)]

The B and C phase powers have the same average term, VphIph cos φ.

Their twice-frequency terms are displaced by twice the original 120° phase displacement.

The three cosine terms sum to zero because they form an equally displaced balanced set.

Therefore:

ptotal = 3VphIph cos φ

at every instant.

Each phase still has pulsating power, and each phase current remains sinusoidal. Constant total power is a property of their sum.

The result requires balanced sinusoidal operation. Unequal phase magnitudes, unequal phase angles or waveform distortion can leave a varying total component.

For an ideal pure-reactive balanced load, the constant total active power is zero. Individual phases can still exchange energy and carry current.

The derivation should therefore not be interpreted as proof that every phase has constant power or that reactive current disappears.

6. Prove the Two-Wattmeter Active-Power Sum

Answer:

Connect current coils in lines A and C. Connect the pressure coils across AB and CB so that line B is common.

The instantaneous products measured by the two ideal elements are:

pWA = (vA − vB)iA

pWC = (vC − vB)iC

Add them:

pWA + pWC = vAiA + vCiC − vB(iA + iC).

For a three-wire circuit:

iA + iB + iC = 0

Hence:

iA + iC = −iB

Substitution gives:

pWA + pWC = vAiA + vBiB + vCiC

This is total instantaneous power crossing the three-line measurement boundary.

Averaging over the appropriate interval gives:

P = WA + WC

The result uses algebraic readings. Negative original readings remain negative.

The proof does not require equal phase impedances. It applies to an ideal three-wire load more generally.

However, it does not establish the balanced reactive-power difference formula for an arbitrary unbalanced load.

If a separate neutral carries current across the measurement boundary, the three-line current sum is not generally zero and this particular reduction is not sufficient.

7. Derive PF from Balanced Wattmeter Readings

Answer:

Use balanced ABC operation and the B-common connection.

With VAN at zero degrees, VAB is at +30° and IA is at −φ.

Therefore:

WA = VLIL cos(30° + φ)

For the other instrument, VCB is at +90° while IC is at +120° − φ.

Thus:

WC = VLIL cos(30° − φ)

Adding the cosine expressions:

WA + WC = 2VLIL cos30° cosφ.

Therefore:

P = WA + WC = √3VLIL cosφ

Subtracting in the defined order:

WC − WA = 2VLIL sin30° sinφ.

Therefore:

WC − WA = VLIL sinφ

Multiplying by √3:

Q = √3(WC − WA)

For positive nonzero total active power:

tanφ = Q/P

Hence:

tanφ = √3(WC − WA)/(WA + WC)

The sign of Q determines lagging or leading under this convention.

Positive Q gives positive φ and lagging behaviour. Negative Q gives leading behaviour.

The derivation requires balanced sinusoidal operation. Preserve original reading signs and match the formula to the actual labelled meter coils.

If P is zero, use the pure-reactive interpretation or an appropriate two-argument angle function rather than dividing by zero.

8. Explain the Purpose and Current of a Neutral

Answer:

A neutral conductor links the load star point to the source neutral in a four-wire system.

With an ideal source and neutral, this keeps each branch supplied by its specified phase-to-neutral voltage.

For unequal loads, branch currents differ and their phasor sum is generally nonzero. The neutral provides the return path for that imbalance.

Choose line currents from source toward load and neutral current from load star point back toward source.

KCL gives:

IN = IA + IB + IC

For balanced sinusoidal currents, all magnitudes are equal and phase displacement is 120°.

Their phasor sum is zero, so:

IN = 0

This does not mean the individual currents are zero.

For unequal resistive branches with a fixed neutral, calculate each branch current independently and add its rectangular phasor components.

An unbalanced star without a connected neutral can have a shifted star point. Its phase voltages must then be calculated from the full network.

Neutral and protective earth have different functions and are not interchangeable circuit-current paths.

The simple zero-neutral result also assumes the stated sinusoidal balanced-current model. More advanced harmonic behaviour requires separate analysis.

Final Revision and Answer Checklist

Check the Connection

Identify whether the physical load is star or delta.

Do not infer load connection solely from the source winding symbol.

Confirm whether a centre point is merely a junction or is actually connected to source neutral.

For every phase impedance, identify the two terminals across which its voltage is measured.

Check the Quantities

Mark VL, Vph, IL and Iph before substitution.

In star:

Vph = VL/√3 and IL = Iph

In balanced delta:

Vph = VL and IL = √3Iph

A per-phase impedance belongs to one branch. A total power belongs to all three phases.

Check the Phase Angle

Use the phase-impedance angle for load PF.

Keep the additional geometric 30° line relationship separate.

State ABC sequence and voltage terminal order whenever reporting absolute phasor angles.

A leading load has negative impedance angle and negative Q under the convention used here.

Check Reconnection and Equivalence

Unchanged impedances reconnected at the same line voltage produce different line currents and powers.

Equivalent balanced loads use different impedances to preserve external line behaviour.

Do not use the one-third reconnection result as though it described an equivalent-load conversion.

Check Neutral Current

Add currents as phasors.

State the neutral-current direction.

Use the connected-neutral assumption for branch-by-branch unequal-load calculations.

Do not assign balanced branch voltages automatically to an unequal floating star.

Check Wattmeter Signs and Ranges

Use the labelled A/AB and C/CB arrangement.

Keep original signed readings.

Restore a documented polarity reversal before applying the original formulas.

Use the algebraic sum for active power.

Use the difference-derived Q and PF only under balanced sinusoidal conditions.

Check line voltage against each pressure-coil range and line current against each current-coil range.

Independent Numerical Checks

Final Calculation Habit

State the model, connection, balance assumption, RMS convention and references.

Write the governing relationships before substituting numbers.

Retain useful intermediate precision and report units.

Interpret current phase, reactive-power sign and original wattmeter signs.

A complete answer should agree when checked from both phase and line viewpoints.

Related B.Tech First Semester Study Materials

CSE 1st Year All Subject PYQs → CSE 1st Year All Subject Notes → CSE 1st Year All Subject Syllabus →