Short notes

Transformer Equivalent Circuit Short Notes

When you do an open circuit (OC) test on a 5 kVA, 230/115 V transformer — applying rated voltage to the LV side, leaving HV open — the wattmeter, voltmeter, and ammeter readings directly give you the shunt branch parameters R0 and X0 of the equivalent circuit. These represent core loss and magnetising reactance without ever having to dismantle the machine.

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How it works

The equivalent circuit referred to the primary replaces the actual transformer with a single series branch (R1 + a²R2) + j(X1 + a²X2) and a shunt branch R0 ∥ jX0 across the input. Short-circuit (SC) test supplies reduced voltage (typically 5–8% of rated) to the HV side while LV is shorted; at rated current the wattmeter gives total copper loss Psc. For a 10 kVA unit the SC test might give Vsc = 18 V, Isc = 43.5 A, Psc = 450 W — from these, Zeq = 18/43.5 = 0.414 Ω, Req = 450/43.5² = 0.238 Ω, Xeq = 0.331 Ω.

Key points to remember

The approximate equivalent circuit neglects R0 and X0, moving the shunt branch to the input terminals — valid because no-load current is only 2–5% of full-load current. Voltage regulation VR = (V2NL − V2FL)/V2FL × 100%; for a resistive load regulation is low, for a lagging load it is highest, and for a leading load it can be negative. Regulation formula: VR ≈ εR cosφ + εX sinφ where εR = per-unit resistance drop and εX = per-unit reactance drop. Examiners routinely ask for regulation at 0.8 pf lagging.

Exam tip

Every Anna University exam includes a numerical where you calculate regulation and efficiency from OC and SC test data — practice the full procedure of finding Req, Xeq from SC test and R0, X0 from OC test before your exam night.

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