14.2 Losses, Efficiency, Load/No-Load & Turns Ratio

14.2 Losses, Efficiency, Load/No-Load & Turns Ratio

Quick Answer: Vp/Vs = Np/Ns = Is/Ip. Ideal power: Vp Ip = Vs Is. Copper losses = I²R in windings (load-dependent). Iron losses = hysteresis + eddy in the core (present when energised). η = Pout/(Pout+losses).

Section 14.1 built the physical picture. Syllabus 3.15 expects fluent turns-ratio arithmetic, clear separation of copper vs iron losses, and the difference between no-load and loaded operation — all staples of CAAS Module 3 MCQs.

Turns Ratio — Voltage, Current, Power

Define:

Ideal transformer:

Vp / Vs = Np / Ns = Is / Ip

Equivalently:

Vs = Vp × (Ns / Np) and Ip = Is × (Ns / Np) or Is = Ip × (Np / Ns)

Voltage ratio equals turns ratio; current ratio is the inverse. Step-down voltage → step-up current (and vice versa) so that, ideally:

Pin = Pout → Vp × Ip = Vs × Is

(assuming unity power factor on both sides for the basic Module 3 model).

Worked example 1 — step-down voltage

Aircraft bus Vp = 115 V, turns Np : Ns = 5 : 1.

Vs = 115 × (1/5) = 23 V.

Worked example 2 — primary current from secondary load

Same transformer; secondary load draws Is = 10 A (ideal).

Ip = Is × (Ns/Np) = 10 × (1/5) = 2 A.

Check power: Pin = 115 × 2 = 230 W; Pout = 23 × 10 = 230 W.

Worked example 3 — find turns

Instrument transformer: Vp = 115 V, Vs = 28 V, Np = 960 turns. Find Ns.

Ns = Np × (Vs/Vp) = 960 × (28/115) ≈ 233.7 → 234 turns (nearest whole turn).

Worked example 4 — step-up

Np = 200, Ns = 800, Vp = 28 V.

Vs = 28 × (800/200) = 112 V. If Is = 0.5 A, Ip = 0.5 × (800/200) = 2 A.

Impedance reflection (recognition-level)

Load Zs on the secondary appears at the primary approximately as:

Zp ≈ Zs × (Np/Ns)²

Module 3 may only require recognising that impedance scales with the square of the turns ratio — useful in coupling/matching discussions.

Load vs No-Load

No-load (secondary open)

• Secondary current Is ≈ 0 (no load power).
• Primary draws no-load current I0: magnetising component (creates flux) + small core-loss component.
• I0 is typically a few percent of full-load primary current on efficient units.
• Core flux is set mainly by applied Vp and frequency (V/f roughly related to flux).
• Iron losses continue whenever the primary is energised at rated voltage/frequency.
• Copper losses in the secondary are zero; primary copper loss is small (I0² Rp).

On load

• Secondary delivers Is to the load.
• Secondary ampere-turns are opposed by additional primary ampere-turns → Ip rises.
• Ideal relation Ip ≈ Is × (Ns/Np) plus the smaller magnetising component.
• Copper losses in both windings increase roughly with the square of current.
• Terminal voltage Vs may droop slightly on a real transformer because of winding resistance and leakage reactance (voltage regulation) — ideal maths ignore this unless the stem gives regulation data.

Transformer Losses

Real efficiency is less than 100%. Loss categories Module 3 emphasises:

1. Copper losses (winding / I²R losses)

Pcu = Ip² Rp + Is² Rs (or treated as a combined full-load copper-loss figure).

Cause: resistance of copper (or aluminium) windings. Mitigation: larger conductor cross-section, high-conductivity copper, shorter mean turn length. High-current (usually LV) windings need thicker wire.

Copper loss is load-dependent — approximately proportional to load current squared.

2. Iron / core losses

Present whenever the core is magnetised by rated AC voltage (approximately constant with load in the basic model if V and f stay fixed).

(a) Hysteresis loss — energy spent reorienting magnetic domains each cycle; area of the B–H loop each second. Worse at higher frequency and with “harder” magnetic materials. Mitigation: soft silicon steel with a narrow hysteresis loop (low coercivity).

(b) Eddy-current loss — circulating currents induced in the conducting core by alternating flux; heat as I²R inside the iron. Mitigation: thin insulated laminations parallel to flux; silicon to raise core resistivity. Solid cores are unacceptable for AC power transformers.

3. Flux leakage (coupling imperfection)

Some primary flux fails to link the secondary (and vice versa). Effects: leakage inductance, voltage drop under load, slightly less than ideal coupling. Mitigation: shell-type cores, interleaved/concentric windings, short leakage paths.

Overcoming / reducing losses — exam table

Efficiency

η = (Pout / Pin) × 100%

Pin = Pout + Pcu + Piron (+ other small losses)

So:

η = Pout / (Pout + losses) × 100%

Aircraft transformers often exceed ~95% at sensible loads; Module 3 still expects you to compute from given loss watts.

Worked example 5 — efficiency

Rated output 2500 W (unity pf). Copper loss 45 W, iron loss 35 W.

Total losses = 80 W. Pin = 2500 + 80 = 2580 W. η = 2500/2580 × 100% ≈ 96.9%.

Worked example 6 — find Pout from η

Pin = 1200 W, η = 96%.

Pout = 0.96 × 1200 = 1152 W. Losses = 48 W.

Worked example 7 — ideal current then add losses (concept)

Ideal Vp=115 V, Vs=23 V, Is=10 A → Ip=2 A, Pout=230 W. If total losses at that load are 10 W, Pin=240 W, η=230/240≈95.8%. Do not use ideal Ip alone when a stem already gives Pin or losses.

Power Transfer Recap

• Ideal: energy conserved magnetically — voltage up, current down (or reverse).
• Real: a few percent becomes heat (copper + iron).
• kVA rating is based on voltage × current capability (heating/insulation), not on a guaranteed wattage at any power factor.

Common Traps

1. Using Vp/Vs = Ip/Is (forgetting current is inverse).
2. Calling laminations a fix for hysteresis (they primarily cut eddy paths; soft steel fixes hysteresis).
3. Saying iron loss is zero at no-load (opposite: iron loss continues; copper is minimal).
4. Mixing peak and RMS in turns-ratio arithmetic — use consistent RMS nameplate values.

Drill Vp/Vs = Np/Ns = Is/Ip and the copper/iron mitigation table until both are automatic. Section 14.3 extends the same ideas to three-phase banks and autotransformers.