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Open circuit and short circuit tests

Short answer

Open circuit test: apply rated voltage to LV with HV open, measures core loss and shunt parameters R0, X0. Short circuit test: short LV, apply reduced voltage to HV to get rated current, gives full‑load copper loss and series parameters Req, Xeq. Both tests determine efficiency and regulation without full loading.

Describe the open circuit test and short circuit test for a single phase transformer.

The open circuit (OC) and short circuit (SC) tests find the equivalent circuit parameters, the losses and the efficiency of a transformer without loading it fully. They need very little power, which is only the loss of the transformer.

Open circuit test (no-load test)

Purpose: to find the core loss PiP_i, no-load current I0I_0 and the shunt branch R0R_0, X0X_0.

        W   A
 AC o---(W)--(A)----+----------+
 supply            (V)   HV winding open
 o-----------------+----------+
        LV winding (rated V)

Procedure: the HV winding is left open, and rated voltage V0V_0 at rated frequency is applied to the LV winding (this is safer and needs lower-range instruments). A wattmeter, ammeter and voltmeter are connected on the LV side.

Calculations:

P0=V0I0cos⁡ϕ0,Iw=I0cos⁡ϕ0,Iμ=I0sin⁡ϕ0P_0 = V_0I_0\cos\phi_0,\quad I_w = I_0\cos\phi_0,\quad I_\mu = I_0\sin\phi_0 R0=V0Iw,X0=V0IμR_0 = \frac{V_0}{I_w},\qquad X_0 = \frac{V_0}{I_\mu}

Since I0I_0 is small (2 to 5% of rated), the copper loss I02R1I_0^2R_1 is negligible. So the wattmeter reading P0P_0 is the core (iron) loss, constant at all loads.

Short circuit test

Purpose: to find the full-load copper loss PcuP_{cu} and the equivalent resistance, reactance and impedance ReqR_{eq}, XeqX_{eq}, ZeqZ_{eq}.

        W   A
 AC o---(W)--(A)----+----------+
 supply (low V,    (V)   LV winding
 via variac)             short-circuited
 o-----------------+----------+
        HV winding

Procedure: the LV winding is short-circuited with a thick conductor. A low voltage VscV_{sc} (about 5 to 10% of rated) is applied to the HV winding and slowly increased by a variac until the rated full-load current IscI_{sc} flows. Wattmeter, ammeter and voltmeter are on the HV side.

Calculations:

Zeq=VscIsc,Req=PscIsc2,Xeq=Zeq2−Req2Z_{eq} = \frac{V_{sc}}{I_{sc}},\qquad R_{eq} = \frac{P_{sc}}{I_{sc}^2},\qquad X_{eq} = \sqrt{Z_{eq}^2 - R_{eq}^2}

Because the applied voltage is low, the flux and hence the core loss are very small (core loss varies as B2B^2). So the wattmeter reading PscP_{sc} is the full-load copper loss.

Uses of the results

  • Equivalent circuit parameters, from which regulation is found: VR≈I(Reqcos⁡ϕ±Xeqsin⁡ϕ)V\text{VR} \approx \dfrac{I(R_{eq}\cos\phi \pm X_{eq}\sin\phi)}{V}.
  • Efficiency at any load: η=xScos⁡ϕxScos⁡ϕ+Pi+x2Pcu\eta = \dfrac{xS\cos\phi}{xS\cos\phi + P_i + x^2P_{cu}}.
  • The maximum efficiency load x=Pi/Pcux = \sqrt{P_i/P_{cu}}.
  • Together, the two tests give the transformer's full performance while drawing only the loss power from the supply. This is their advantage over a direct load test.

✓ Answer checked by Claude Opus 5.5 (high effort)

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