Engineering & Technology · Electrical Power, Conversion & Machines

Transformers, three-phase systems and network models

A transformer can increase voltage while decreasing current for the same ideal transferred power. A balanced three-phase network can deliver constant total instantaneous power while each phase varies. Both useful results depend on relationships between terminals, phases and reference conventions.

Electrical Circuits supplies current/voltage, power, rms, phase and ideal storage relations. Engineering Thermodynamics supplies energy balances, losses and system boundaries. This chapter introduces magnetic flux and pole count before using them, derives the simple machine/converter relations and labels every invented parameter. All diagrams and numerical cases are original teaching models. They do not specify equipment construction, protective settings, installation, switching or qualification.

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After reading, you can

Use ideal transformer and balanced three-phase relationships while stating rms, connection and balance assumptions.

Transform terminal relationships without creating power

For an ideal transformer, a turns ratio a=N1/N2 gives voltage magnitude ratio V1/V2=a. Current magnitudes have the inverse ratio when power flows through to the load. If both current arrows are defined into the device’s positive terminals, one signed current is opposite the other; the sum of ideal absorbed port powers is zero.

A load impedance seen through an ideal transformer is multiplied by a². This is a network equivalence, not an increase in actual load power. Real transformers also need magnetising current and have winding resistance, leakage, core losses and limits associated with flux, frequency and temperature. A turns ratio alone does not predict these.

An ordinary transformer transfers through changing flux. Extending an ideal terminal ratio to arbitrary sustained DC excitation would omit the magnetic and winding behaviour that determines the actual result. Isolation and protective suitability likewise depend on construction and the complete system, not the ratio alone.

Specify a balanced sinusoidal set

Three balanced phase voltages have equal magnitude, equal frequency and 120° separation. With matching balanced sinusoidal loads, their currents also have equal magnitudes and equal phase displacement relative to their respective voltages. The ideal phase currents sum to zero. Unbalance or harmonic content can invalidate that conclusion for a real neutral.

The original figure plots a balanced voltage set and compares one phase’s instantaneous power with the three-phase sum. Individual phase power varies at twice the electrical frequency. The corresponding variations cancel in the balanced sum, giving total real power 3 Vphase,rms Iphase,rms cosφ. Here φ is the displacement between each phase voltage and current.

For a balanced wye connection, line voltage magnitude is √3 times phase-to-neutral voltage and line current equals phase current. In a balanced delta connection, the line/branch relationships differ. State which voltage and current are being used; “three-phase voltage” alone is insufficient for substitution.

Horizontal electrical angle 0–360 degrees at 1.5 units/degree. Upper normalised voltages sinθ,sin(θ−120°),sin(θ+120°), vertical 95 units per normalised peak. Actual phase rms 100 V. Lower one-phase instantaneous power 1000 sinθ sin(θ−φ) W withφ=acos 0.8, phase current 5 A rms; it ranges−100 to 900 W. All three corresponding powers sum to 1200 W, green horizontal line. Power scale 0.14 units/W with 0 at 590; plotted extent includes negative one-phase power. Constructed balanced sinusoidal case, not a waveform measurement.

Choose the level of network detail

A one-line network drawing compresses phase detail to show major sources, buses, transformers and loads. A per-phase equivalent can simplify a balanced problem, but faults or unequal loads may need a less symmetric model. An omitted conductor in an abstraction is not evidence that the physical system needs no such connection.

For nonsinusoidal conditions, total power factor is real power divided by apparent power using the chosen rms definitions; it need not equal only cosφ. The simple balanced sinusoidal formulas here do not automatically quantify harmonic heating, fault current or insulation stress.

Check a transformer and a separate three-phase load

First, use an original ideal single-phase transformer with a=5, input 100 V rms and a resistive secondary drawing 3 A at 20 V rms. Output is 60 W and primary current magnitude is 0.6 A, giving the same 60 W. The secondary resistance is 20/3≈6.667 Ω and its primary equivalent is 25 times larger, about 166.7 Ω.

Separately, let a balanced wye load have phase-to-neutral voltage 100 V rms, phase current 5 A rms and displacement factor cosφ=0.8. Total real power is 3×100×5×0.8=1200 W. Line voltage is √3×100≈173.2 V rms and apparent power is 1500 VA.

Using √3 Vline Iline cosφ reproduces 1200 W. Substituting 100 V as though it were this case’s line voltage would give a different, incorrect result. The cases specify mathematical terminal values, not actual supply or transformer ratings.

Try the reasoning

Can an ideal step-up transformer increase both voltage and current at unchanged input power?
Show the answer and reasoning
No. Its inverse current relationship preserves transferred power.
Why is the neutral-current cancellation conditional?
Show the answer and reasoning
It uses a balanced sinusoidal phase set and loads; unequal conditions or harmonics require another account.
Is cosφ always the complete power factor of a distorted waveform?
Show the answer and reasoning
No. Distortion affects the rms/apparent-power relationship; the sinusoidal displacement formula has limited scope.

What to carry forward

Label the ports, rms quantities and connection. Use balance only where its assumptions have been established.

References

Teaching resources and research supporting this lesson’s explanations and examples.