Each tab has a short explanation, one or more live labs, and activities. Move any slider and every number and graph updates. Activities with a number to find have a Load these settings button so everyone starts from the same circuit, and a Check button that compares your answer with the model. Nothing is sent anywhere; your answers stay in this page, and the last tab gives you a summary you can copy.

Suggested path: follow the tabs in order. Magnetic circuits and induction explain inductors and transformers; AC power and three-phase give the language of the grid; the two transformer tabs combine everything; the “Putting it together” tab connects it to converters.

The last equation is “Ohm’s law for magnetics”. Reluctances in series add, just like resistances. Air has µ_r = 1 while iron may have 2 500 or more, so a thin air gap can dominate the total reluctance.

Inductance from the magnetic point of view

Inductance is flux linkage per amp: L = N φ/i. Substituting φ = N i/ℛ gives

Inductance grows with the square of the turns, with core area, and with permeability, and falls with path length and with any air gap.

Saturation

Iron cannot carry unlimited flux. Above roughly 1.5 to 2 T for silicon steel (0.3 to 0.5 T for ferrite), µ collapses toward µ₀, ℛ jumps, and the inductance falls sharply. In a converter that means the current spikes. The lab includes saturation with B_sat = 1.6 T.

Change the turns, current, core size, material, and gap. Watch where the MMF is spent: a small gap takes most of it.

The voltage induced in a coil equals the turns times the rate of change of the flux through it. The minus sign is Lenz’s law: the induced voltage would drive a current whose flux opposes the change. Only a changing flux induces voltage; constant flux induces nothing.

For a sinusoidal flux φ = Φ_m sin ωt, the voltage is e = N ω Φ_m cos ωt, a cosine that leads the flux by 90°. Its RMS value is

A wire of length l carrying current i in a field B feels a force

where θ is the angle between the wire and the field. The force is largest when the wire is perpendicular to the field, and its direction is given by the right-hand rule (index finger along the current, middle finger along B, thumb along F). A motor is a set of such wires on a rotor, so the force becomes torque.

Instantaneous power is always p = v i. If the current lags the voltage by θ, then p = V I cos θ + V I cos(2ωt − θ). The first term is constant; the second swings at twice the line frequency and averages to zero. Splitting it further:

An inductive load consumes both P and Q. A capacitive load consumes P and supplies Q. Because cos(+30°) = cos(−30°), a power factor alone does not say whether the load is inductive or capacitive; always state “lagging” or “leading”.

More power for the same copper. A three-phase system delivers more power per unit weight of conductor than a single-phase system, so lines, generators, and transformers are more compact.

Constant power. In single phase the instantaneous power pulses at twice the line frequency. In a balanced three-phase load the three pulsations add to a constant. This gives smooth torque in motors and a smooth DC output when the voltages are rectified.

Balanced sets and phase sequence

A balanced source has three voltages of equal magnitude and 120° apart; a balanced load has three equal impedances. Then the neutral current is zero, and the neutral wire is not needed. The phase sequence is the order in which the voltages reach their peaks. At the instant shown in the lab below, reading clockwise from A gives sequence ABC. Swapping two phases gives ACB, which reverses a motor’s direction.

Each phasor rotates counter-clockwise at the line frequency. The instantaneous voltage is its projection on the horizontal axis. Drag the angle, or press play, and notice that the three values always add to zero.

Phase quantities are across or through one winding or load element. Line quantities are the line-to-line voltage and the current in a line conductor. For a balanced system:

Total power does not depend on the connection when written with line quantities:

where θ is the angle between the phase voltage and the phase current. Because the three phases are identical, you can analyze one phase (a per-phase circuit) and multiply by three.

A four-wire wye system with resistive loads. Make phase C draw less current and watch the neutral current appear and the total power start to ripple.

Connect the same three impedances in wye or delta on the same line voltage. The phasor diagram shows the √3 and the 30° shift.

A load Z_L on the secondary looks, from the primary, like

This is why a transformer can “match” a load to a source, and why, in analysis, you move impedances across the transformer by multiplying or dividing by a² and then delete the ideal transformer.

Dot convention

Dots tell you the polarity without looking at the winding direction. A voltage that is positive at the dotted end of one winding is positive at the dotted end of the other. A current entering the dot on the primary leaves the dot on the secondary, so N₁i₁ − N₂i₂ = 0.

Open-circuit test. Leave the secondary open and apply rated voltage to the other winding. Almost all the current is magnetizing current, and the power measured is the core loss. This gives R_c and X_m.

Short-circuit test. Short the secondary and raise the primary voltage from zero until rated current flows. The voltage is small, so the shunt branch is negligible, and the power measured is the copper loss. This gives R_eq and X_eq.

The transformer is 10 kVA, 2400/240 V. The open-circuit test is made on the 240 V side at rated voltage; the short-circuit test on the 2400 V side at its rated current of 4.167 A.

A smaller VR means a stiffer output. A lagging (inductive) load makes the output droop. A leading (capacitive) load can make the secondary voltage rise, giving a negative VR. Efficiency peaks at the load where the copper loss equals the core loss.

The primary is held at rated voltage. Impedances are in percent (per unit × 100) on the transformer’s own rating.

Instead of volts, amps, and ohms, engineers often divide each quantity by a base value. With a base power S_b and base voltage V_b:

A transformer’s per-unit impedance is the same whether you refer it to the primary or the secondary, so the ratio disappears from the calculation. Per-unit values also show at a glance whether an impedance is large or small. For a three-phase system use the three-phase S_b and the line-to-line V_b.

The voltage rating protects the winding insulation, and also keeps the core out of saturation, so the voltage must be reduced if the frequency is reduced (“derating”). The current rating limits the I²R copper loss, which becomes heat. That is why transformers are rated in kVA: voltage limits the core, current limits the copper.

A three-phase unit is usually three windings on one core inside a tank of insulating oil. The oil insulates the windings from the case, carries heat away, and keeps out moisture. Compared with three separate single-phase units, it is lighter, cheaper, smaller, and slightly more efficient.

The primary and secondary can each be connected in Y or Δ, giving four combinations. Within one transformer the ratio of phase voltages is the turns ratio a, but the ratio of line voltages depends on the connections:

Faraday’s law for a sinusoidal supply says V = 4.44 f N B_m A_c. Solving for the core area:

For a fixed voltage, turns, and flux density, the core area is inversely proportional to frequency. A 60 Hz iron transformer and a 100 kHz ferrite transformer for the same voltage differ enormously in size. The catch: eddy and hysteresis losses rise with frequency, and ferrite saturates at a lower B_m, so the gain is limited.

A six-pulse diode bridge connects the highest line-to-line voltage to the output at every instant. The average output is

so the transformer connection sets V_L,sec, and that sets the DC bus feeding the converter. The transformer also provides isolation from the grid.

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