Transformers Questions and Answers

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.

Answer: A

Correct answer: 98.5 per cent

At full load and unity power factor the output is 100 kW and the total loss is \( 500 + 1000 = 1500 \) W: \[ \eta = \frac{100\,000}{100\,000 + 1500} = 0.985 = 98.5\ \text{per cent.} \] Note the loss goes in the denominator with the output, since input equals output plus losses - dividing by the output alone gives a figure above unity.

Answer: B

Because a distribution transformer is energised continuously but loaded for only part of the day, its iron loss runs for twenty-four hours while its copper loss runs for a few. Designing for a low iron loss is what raises the figure.

Answer: C

Iron loss is present the moment the transformer is connected, so on a lightly loaded feeder it dominates the day's energy loss. Power transformers, loaded nearer their rating for most of the time, are optimised differently.

Answer: D

Correct answer: the fall in secondary voltage from no load to full load

Regulation measures how much the secondary voltage sags when load is applied, expressed as a fraction of the no-load value. It is caused by the voltage dropped across the winding resistance and, more importantly, the leakage reactance. A low regulation means a stiff supply, which is why distribution transformers are designed for a few per cent.

Answer: A

Correct answer: a lagging power factor

With a lagging load the current lags the voltage, so the drop across the leakage reactance adds almost directly to the resistive drop and the two combine to give the largest fall. At unity power factor the reactive drop is largely in quadrature and contributes less. A leading load can actually raise the secondary voltage above its no-load value, giving negative regulation.