Transformers Questions and Answers

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.

Answer: B

At unity power factor the reactive drop is in quadrature and contributes only to the second order, so the regulation is nearly the resistive component. It is a useful check on whether a regulation calculation has been done correctly.

Answer: C

The reciprocal of the per unit impedance gives the multiple, so a low impedance means a stiff supply and a large fault current. Switchgear has to be rated for that current, which is one reason impedance is not made smaller than it needs to be.

Answer: D

Correct answer: 30 degrees between primary and secondary line voltages

In a delta-star connection the primary line voltage is a phase voltage of the delta while the secondary line voltage is the difference of two star phase voltages, and the two differ in phase by 30 degrees. The shift matters when transformers are paralleled: two units must belong to the same vector group or a large circulating current will flow.