6.2 Transformers & Rotating Machines

Ideal transformer ratios

A transformer magnetically couples two windings to change AC voltage levels at (nearly) constant power. Define the turns ratio a = N1/N2, with N1 primary turns and N2 secondary turns. For an ideal (lossless) transformer:

• Voltage: V1/V2 = N1/N2 = a
• Current: I1/I2 = N2/N1 = 1/a
• Power: S1 = S2, so V1·I1 = V2·I2

Because voltage scales by a and current by 1/a, a secondary-side impedance reflects to the primary as Z_ref = a^2·Z_2. This a-squared scaling is a frequent FE answer in impedance-matching and reflected-load problems.

Worked example: reflected impedance and current

A transformer has a = 10 (e.g., 1200 turns to 120 turns). A 4-ohm secondary load reflects to the primary as Z_ref = 10^2·4 = 400 ohms. If the secondary draws 5 A, the primary current is 5/a = 0.5 A, and V1·I1 = V2·I2 confirms power balance.

Transformer losses and efficiency

Real transformers have two loss groups. Core (iron) losses — hysteresis plus eddy currents — depend on voltage and frequency and are essentially constant whenever energized. Copper (winding) losses are I^2R and scale with the square of load current. Efficiency is:

eta = P_out / (P_out + P_core + P_copper)

Maximum efficiency occurs at the load where copper loss equals core loss. Because copper loss tracks load^2, a transformer rated for full-load copper loss P_cu,FL has copper loss x^2·P_cu,FL at fractional load x.

Worked example: efficiency

A 10 kVA transformer at unity-pf full load (P_out = 10 kW) has core loss 120 W and full-load copper loss 180 W.

• eta = 10,000/(10,000 + 120 + 180) = 10,000/10,300 = 0.9709, or 97.1%. At half load, copper loss = 0.5^2·180 = 45 W, P_out = 5 kW, so eta = 5000/(5000+120+45) = 96.8%.

Synchronous speed and synchronous machines

Every AC machine has a rotating stator field whose speed is fixed by frequency and pole count, the synchronous speed:

Ns = 120 · f / P (rpm)

where f is supply frequency (Hz) and P is the (even) number of poles. The factor 120 converts (2f/P) rev/s to rev/min. At 60 Hz: 2-pole = 3600 rpm, 4-pole = 1800 rpm, 6-pole = 1200 rpm, 8-pole = 900 rpm.

A synchronous machine rotor locks to this field and turns at exactly Ns under all loads. As a generator it sets system frequency; as a synchronous condenser, over-excitation makes it supply VARs (leading), correcting plant power factor.

Induction motors, slip, DC machines

The induction motor is industry's workhorse. Its rotor carries no external excitation; the stator field induces rotor currents, which requires the rotor to spin slightly below synchronous speed. The fractional lag is slip:

s = (Ns - N)/Ns

where N is actual rotor rpm. At no load s is near zero; at full load it is typically 2-5%. Rotor electrical frequency equals s·f, and rotor speed N = Ns·(1 - s). If a problem states the rotor turns at exactly Ns, the machine is synchronous, not induction.

Worked example: synchronous speed and slip

A 60 Hz, 6-pole induction motor runs at 1152 rpm.

1. Ns = 120·60/6 = 1200 rpm.
2. s = (1200 - 1152)/1200 = 48/1200 = 0.04, or 4%.
3. Rotor frequency = s·f = 0.04·60 = 2.4 Hz.

For a DC machine, generated EMF E = k·phi·omega (proportional to flux and speed) and torque T = k·phi·I_a (proportional to flux and armature current). Reversing power-flow direction turns the same DC motor into a DC generator.

Machine torque, power, and three-phase connections

Mechanical output ties speed to torque: P_mech = T·omega, where omega = 2·pi·N/60 (rad/s) when N is in rpm. This conversion is a constant trap — a torque problem stated in rpm must convert before multiplying. Useful spot checks: 1 hp = 746 W, and motor efficiency eta = P_mech_out / P_elec_in.

Worked example: motor torque

A motor delivers 10 hp at 1750 rpm. Output power = 10·746 = 7460 W. omega = 2·pi·1750/60 = 183.3 rad/s. Torque T = P/omega = 7460/183.3 = 40.7 N·m.

Three-phase transformer banks combine three single-phase units (or one three-phase core) in delta or wye on each side. Common configurations and their traits:

The delta-wye 30-degree phase shift between primary and secondary line voltages is a frequent conceptual answer. An autotransformer shares one winding (no isolation) and is smaller and cheaper for ratios near 1:1, but provides no galvanic isolation between primary and secondary, so a fault on one side appears directly on the other.