Question

(a) Explain how the efficiency and regulation of a transformer can be assessed by open circuit and short circuit tests? (8)

(b) What is meant by equivalent resistance? (4)

(c) What is meant by all day transformer efficiency? (4)

Answer 1

Part (a)

Open-circuit (no-load) test: the low-voltage winding is supplied at rated voltage with the high-voltage side open. The wattmeter measures the no-load power, which is almost entirely the iron (core) loss (hysteresis and eddy current losses), since the no-load copper loss is negligible. The no-load current and power give the iron loss and the magnetising and core-loss components of the exciting current.
Short-circuit (impedance) test: the low-voltage winding is short-circuited and a reduced voltage is applied to the high-voltage side so that rated current flows. The wattmeter measures the full-load copper loss (I^2 R), and the voltmeter and ammeter give the equivalent impedance. From this the equivalent resistance and reactance are found.
Efficiency: total losses = iron loss (from OC test) + copper loss (from SC test, scaled by the square of the load current). Efficiency = output/(output + losses). At any load, copper loss = full-load copper loss x (load fraction)^2.
Regulation: from the SC test, the equivalent resistance Req and reactance Xeq referred to one side are found. The percentage voltage regulation = (I (Req cos phi +/- Xeq sin phi)/V) x 100, where the sign depends on lagging or leading power factor. This gives the voltage drop from no-load to full-load.

Part (b)

The equivalent resistance of a transformer is the total resistance of the primary and secondary windings referred to one side (primary or secondary). It is the resistance which, when carrying the current on that side, produces the same copper loss as the actual primary and secondary resistances.
Referred to the primary: Req1 = R1 + R2' where R2' = R2 (N1/N2)^2. Referred to the secondary: Req2 = R2 + R1' where R1' = R1 (N2/N1)^2.
It is used to calculate the copper loss and the voltage regulation of the transformer.

Part (c)

The all-day efficiency is the ratio of the energy output to the energy input over a full day (24 hours), taking into account the varying load during the day.
It is important for distribution transformers which operate at light load for much of the day. The iron loss is constant (occurs all day), while the copper loss varies with the square of the load.
All-day efficiency = (energy output in 24 h) / (energy output + energy losses in 24 h).
It is lower than the ordinary efficiency because the iron loss is incurred even when the transformer is lightly loaded. It is used to select a transformer with the lowest total energy loss over the day.

✓ Verified Model Answer (Text Solution)