Transformers Questions and Answers

Answer: C

Correct answer: flux that links one winding but not the other

Not every line of flux produced by one winding passes through the other; a small proportion closes through air or through the tank. That leakage flux links only its own winding, so it behaves exactly like a series inductance, and it is what limits the short circuit current and produces most of the voltage regulation. Hysteresis and winding resistance are quite separate effects.

Answer: D

It is normally done from the low-voltage side for convenience, and the wattmeter reads the core loss because the copper loss at no-load current is negligible. The magnetising branch parameters follow from the same readings.

Answer: A

Correct answer: reduced voltage giving rated current in the shorted winding

In the short circuit test one winding is shorted and the applied voltage raised only until rated current flows - a few per cent of rated voltage. The flux, and therefore the core loss, is negligible at that voltage, so the wattmeter reads essentially the full-load copper loss, and the readings also give the equivalent series impedance.

Answer: B

Correct answer: the core loss and the magnetising branch parameters

With the secondary open, the only current drawn is the small no-load current, so the copper loss is negligible and the wattmeter reads the core loss. Voltage, current and power together give the magnetising branch: the conductance representing the core loss and the susceptance representing the magnetising current.

Answer: C

Rated current at a small flux isolates the winding resistance and leakage reactance and excludes the core loss. Together with the open circuit test it completes the equivalent circuit without ever loading the transformer fully.

Answer: D

Correct answer: supplying its full output power

This is the practical value of the two tests. Each draws only a small fraction of rated power - the open circuit test rated voltage at no-load current, the short circuit test rated current at a few per cent of voltage - yet between them they give both loss components, from which the efficiency at any load can be computed. Loading a large transformer to full output for a test would be impractical.

Answer: A

Correct answer: the core flux is fixed by the applied voltage

From \( E = 4.44fN\Phi_m \), the peak flux is set by the applied voltage and frequency, both of which stay constant as load varies. Hysteresis and eddy current loss depend on that flux, so the core loss barely changes from no load to full load. This is why it is treated as a constant loss when efficiency is computed.

Answer: B

Correct answer: the square of the load current

Copper loss is \( I^2R \) in each winding, so it varies as the square of the load current: at half load it is a quarter of the full-load value. That is why the two loss components behave so differently - the constant iron loss dominates at light load and the rising copper loss dominates at heavy load, and the efficiency curve peaks where they cross.

Answer: C

Correct answer: the copper loss equals the iron loss

Efficiency is maximised where the variable loss equals the constant loss. Differentiating the efficiency expression with respect to load gives exactly that condition, and it corresponds to a loading of \( \sqrt{P_i/P_{cu,fl}} \). Transformers expected to run lightly loaded much of the time are therefore designed with the maximum placed below full load.

Answer: D

Correct answer: half of full load

Maximum efficiency occurs where the copper loss equals the iron loss: \[ x^2 P_{cu,fl} = P_i \quad\Rightarrow\quad x = \sqrt{\frac{400}{1600}} = \sqrt{0.25} = 0.5. \] So the peak is at half load. A design like this suits a transformer that spends most of its life lightly loaded, since the peak is placed where it actually operates.