Open-circuit and short-circuit tests

How the OC and SC tests isolate core and copper loss, and how to extract R0, X0, R01, X01 and predict efficiency and regulation from them.

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Why it matters

You cannot load a 500 kVA transformer to full load in a test bay just to find its efficiency. The open-circuit (OC) and short-circuit (SC) tests find all the equivalent-circuit parameters and both kinds of loss while drawing only a few percent of rated power, and from them you predict efficiency and regulation at any load. Reading wattmeter, ammeter and voltmeter data correctly is also a standard lab and viva skill.

Key ideas

Open-circuit (no-load) test — finds the shunt branch and core loss

  • Connection: rated voltage at rated frequency applied to one winding (normally the LV side, because rated LV voltage is easy to supply and the current is measurable), other winding open. Instruments: voltmeter, ammeter, low-power-factor wattmeter.
  • The current is only the no-load current I0 (2–5 % of rated), so copper loss I0²R1 is negligible. The wattmeter reads core loss Pi.
  • Since flux depends on V/f, the core loss measured at rated voltage and frequency is the core loss at every load.
  • The no-load power factor is low (about 0.1–0.3). If a calculation gives a no-load power factor near 1, the data are wrong.
  • From V0, I0 and W0: cos φ0 = W0/(V0·I0), Iw = I0·cos φ0, Iμ = I0·sin φ0, R0 = V0/Iw, X0 = V0/Iμ — referred to the side on which the test was done.

Short-circuit test — finds the series branch and copper loss

  • Connection: one winding (normally the LV side) short-circuited with a thick link; a reduced voltage applied to the other (HV) side and raised from zero until rated current flows. Instruments on the HV side: voltmeter, ammeter, wattmeter.
  • Only 5–10 % of rated voltage is needed (it equals the impedance drop), so the flux and core loss are negligible. The wattmeter reads copper loss at the test current.
  • From Vsc, Isc and Wsc: Z01 = Vsc/Isc, R01 = Wsc/Isc², X01 = √(Z01² − R01²) — referred to the HV side.
  • If the test is done at a current other than rated, full-load copper loss = Wsc × (I_rated/Isc)².
  • Applying rated voltage to a short-circuited transformer would draw 10–20 times rated current and destroy it.

Why two tests? Each isolates one kind of loss because in each test one branch of the approximate equivalent circuit carries negligible current. Together they give R01, X01, R0, X0, Pi and Pcu,FL, which is everything needed for efficiency and regulation. Their limitation: the SC test cannot separate R1 from R2' or X1 from X2' (by convention take X1 = X2'), and the tests give losses at temperatures that may differ from operating temperature.

Other tests you should know by name. Sumpner's (back-to-back) test loads two identical transformers at full current while drawing only losses from the supply — a heat-run test. The polarity test checks terminal marking before parallel operation.

Formulas

  • cos φ0 = W0 / (V0·I0) — no-load power factor; W0 in W, V0 in V, I0 in A.
  • Iw = I0·cos φ0 ; Iμ = I0·sin φ0 — core-loss and magnetising components (A).
  • R0 = V0 / Iw ; X0 = V0 / Iμ — shunt-branch resistance and reactance (Ω), on the test side.
  • Z01 = Vsc / Isc ; R01 = Wsc / Isc² ; X01 = √(Z01² − R01²) — series branch (Ω), on the test side.
  • Pcu,FL = Wsc · (I_rated / Isc)² — full-load copper loss (W).
  • Pi = W0 — core loss (W), constant with load.
  • Z' = a²·Z — move any result from LV to HV side (a = N_HV/N_LV).

Worked examples

Example 1 (standard): finding the equivalent circuit. Given: 10 kVA, 2000/200 V, 50 Hz transformer. OC test (LV side): 200 V, 2.0 A, 100 W. SC test (HV side): 67 V, 5 A, 150 W.

  1. OC power factor: cos φ0 = 100/(200 × 2.0) = 0.25; sin φ0 = 0.968.
  2. Iw = 2.0 × 0.25 = 0.5 A; Iμ = 2.0 × 0.968 = 1.94 A.
  3. R0 = 200/0.5 = 400 Ω, X0 = 200/1.94 = 103 Ω (LV side). Referred to HV (× 100): 40 kΩ and 10.3 kΩ.
  4. SC: Z01 = 67/5 = 13.4 Ω; R01 = 150/5² = 6.0 Ω; X01 = √(13.4² − 6²) = 12.0 Ω (HV side).
  5. Rated HV current is 10000/2000 = 5 A, so the SC test was at full load: Pcu,FL = 150 W, Pi = 100 W.

Example 2 (GATE level): test at reduced current, then efficiency and regulation. Given: 50 kVA, 2200/220 V transformer. OC test (LV): 220 V, 6 A, 360 W. SC test (HV): 50 V, 15 A, 225 W. Find the full-load efficiency and regulation at 0.8 pf lagging.

  1. Rated HV current: I1 = 50000/2200 = 22.73 A.
  2. From SC: Z01 = 50/15 = 3.33 Ω; R01 = 225/15² = 1.0 Ω; X01 = √(3.333² − 1²) = 3.18 Ω.
  3. Full-load copper loss: Pcu,FL = 225 × (22.73/15)² = 516.5 W (same as 22.73² × 1.0).
  4. Core loss Pi = 360 W. Efficiency: η = 40000/(40000 + 360 + 516.5) = 97.86 %.
  5. %R = 22.73 × 1.0/2200 × 100 = 1.03 %; %X = 22.73 × 3.18/2200 × 100 = 3.28 %.
  6. Regulation at 0.8 lag: ε = 1.03 × 0.8 + 3.28 × 0.6 = 2.80 %. (Check: no-load power factor = 360/(220 × 6) = 0.27 — sensible.)

Common mistakes

  • Treating the SC-test wattmeter reading as full-load copper loss when the test current is not rated current; scale by the square of the current ratio.
  • Mixing sides: OC results belong to the LV side and SC results to the HV side; refer them with a² before combining.
  • Using a no-load power factor near unity, or computing X0 with I0 instead of Iμ.
  • Forgetting that core loss depends on applied voltage: an OC test at reduced voltage does not give rated core loss.
  • Saying the SC test "measures copper loss at rated voltage" — the voltage is only a few percent of rated.
  • Doing the OC test on the HV side in a lab, which needs a high-voltage supply and gives a tiny current that is hard to read.

For GATE IN

Expect numericals that give OC and SC readings and ask for R0, X0, R01, X01, the efficiency at some load and pf, the load for maximum efficiency, or the regulation. Conceptual MCQs ask which winding is shorted or open, what each wattmeter reads and why a low-power-factor wattmeter is used in the OC test. The wattmeter and instrument choices link directly to the Measurements syllabus.

Quick check

  1. In an SC test at half the rated current the wattmeter reads 100 W. What is the full-load copper loss?
  2. Which test gives core loss, and on which side is it usually done?
  3. SC test: 40 V, 10 A, 240 W. Find R01 and X01.
  4. Why is core loss negligible in the SC test?

Answers: 1. 400 W. 2. The OC test, usually on the LV side. 3. R01 = 2.4 Ω, Z01 = 4 Ω, X01 = 3.2 Ω. 4. The applied voltage, and hence the flux, is only a few percent of rated.

Interview questions

All Electrical Machines interview questions

Try answering each one aloud before you open it.

  1. 1.What is an open-circuit test in electrical machines?Concept

    An open-circuit test is performed on electrical machines to determine the core losses and magnetizing current of a transformer. During this test, the secondary winding is left open, and the rated voltage is applied to the primary winding. The test helps in calculating the no-load current and losses, which are essential for efficiency calculations.

  2. 2.What is a short-circuit test in electrical machines?Concept

    A short-circuit test is conducted to determine the copper losses and equivalent impedance of a transformer. In this test, the secondary winding is short-circuited, and a reduced voltage is applied to the primary winding to circulate the rated current. This test helps in calculating the full-load copper losses and the transformer's impedance.

  3. 3.Explain why open-circuit and short-circuit tests are important for transformers.Concept

    Open-circuit and short-circuit tests are crucial for transformers as they help in determining the transformer's efficiency and voltage regulation. The open-circuit test provides information about core losses and magnetizing current, while the short-circuit test gives insights into copper losses and equivalent impedance. Together, these tests allow engineers to design transformers that meet specific performance criteria.

  4. 4.How is the open-circuit test conducted on a transformer?Concept

    Rated voltage at rated frequency is applied to one winding, usually the LV side, with the HV winding left open. A voltmeter, ammeter and low-power-factor wattmeter read V0, I0 and W0. Because I0 is only a few percent of rated current, copper loss is negligible and W0 is the core loss; from cos φ0 = W0/(V0·I0) you get Iw and Iμ and hence R0 = V0/Iw and X0 = V0/Iμ.

  5. 5.How is the short-circuit test conducted on a transformer?Concept

    The LV winding is shorted with a thick link and a variable low voltage is applied to the HV side, raised from zero until rated current flows; this needs only about 5–10 % of rated voltage. Voltmeter, ammeter and wattmeter read Vsc, Isc and Wsc. Since the flux is small, core loss is negligible and Wsc is the copper loss; Z01 = Vsc/Isc, R01 = Wsc/Isc² and X01 = √(Z01² − R01²) referred to the HV side.

  6. 6.Why is a reduced voltage used during the short-circuit test?Application

    A reduced voltage is used during the short-circuit test to ensure that the rated current flows through the windings without causing excessive heating or damage. Since the secondary winding is short-circuited, applying the full rated voltage would result in a very high current, which could damage the transformer.

  7. 7.In an open-circuit test at rated voltage of 230 V, a transformer draws 2 A and 150 W. What is the core loss, and what is the no-load power factor?Numerical

    The wattmeter reading is taken as core loss, 150 W, because the copper loss due to the small no-load current is negligible. The no-load power factor is 150/(230 × 2) = 0.33 lagging, which is in the normal low range for a transformer on no load.

  8. 8.During a short-circuit test a transformer draws 10 A at 50 V with 400 W input. What does the wattmeter reading represent, and what is the equivalent resistance?Numerical

    With only 50 V applied the flux and core loss are negligible, so 400 W is the copper loss at 10 A. Equivalent resistance on the test side is R = 400/10² = 4 Ω, and Z = 50/10 = 5 Ω, so X = 3 Ω. If 10 A is the rated current, 400 W is the full-load copper loss; otherwise scale it by (I_rated/10)².

  9. 9.Explain the significance of measuring no-load current in an open-circuit test.Application

    The no-load current, together with the wattmeter reading, gives the no-load power factor and splits I0 into its core-loss component Iw and magnetising component Iμ. These give the shunt-branch parameters R0 and X0 of the equivalent circuit. A no-load current that is larger than the usual 2–5 % of rated current indicates a poor core, a larger joint air gap or over-fluxing.

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