Transformers Questions and Answers

Answer: C

Only the leakage impedance limits the current, so with five per cent impedance it reaches twenty times rated and the winding forces are four hundred times normal. Protection must clear such a fault within a fraction of a second.

Answer: D

Two conductors carrying current attract or repel in proportion to the product of their currents, which for a single winding means the square. That is why a transformer's mechanical bracing is designed for the fault case rather than the normal one.

Answer: A

Correct answer: Friction and windage loss

Friction and windage are the losses of a rotating machine, and a transformer has no moving parts at all. Its losses are hysteresis and eddy current loss in the core, together making up the iron loss, and \( I^2R \) loss in the two windings, the copper loss. There is also a small stray loss from leakage flux in the tank and clamps.

Answer: B

An autotransformer has part of its winding common to the primary and the secondary, so it needs less copper and has lower winding resistance and leakage reactance. Lower resistance and reactance mean a smaller voltage drop under load, so the voltage regulation is better, and lower copper loss means the efficiency is higher than that of a two winding transformer of the same rating.

The options with poor regulation or less efficiency state the opposite of the effect. The answer is better regulation and more efficiency, option B.

Answer: A

A resistance on the secondary side is referred to the primary side by dividing by the square of the turns ratio. The ratio of transformation is \(K = 2\).

\[ \begin{aligned} R_2' = \frac{R_2}{K^2} = \frac{0.48}{2^2} \\ &= \frac{0.48}{4} = 0.12\ \Omega \end{aligned} \]

The value 0.24 ohm comes from dividing by \(K\) instead of \(K^2\), and 0.06 ohm from dividing by \(2K^2\). The value 0.18 has no basis. The answer is 0.12 ohm, option A.

Answer: A

A two winding transformer of 2000/200 V has a high voltage winding of 2000 V and a low voltage winding of 200 V. If the two windings are joined in series so that their voltages add (additive connection), the autotransformer has a primary of 2000 V and a secondary of \(2000 + 200 = 2200\) V, which is the step up connection in option A.

The same two windings can also be connected to subtract, giving 1800 V, but that is not listed. The value 1500 V and 220 V do not arise from any series connection of the two windings, and 200 V is just the low voltage winding on its own, not an autotransformer rating. The answer is option A.

Answer: B

When a two winding transformer is reconnected as an autotransformer, the kVA it can handle is larger than its nameplate rating, because only part of the power is transformed (the rest is conducted directly). The two windings are 240 V and 120 V, and to feed a load at 360 V they are connected in series (additive).

\[ \begin{aligned} \text{Auto kVA} \\ &= \text{kVA}_{2\text{-winding}} \times \frac{V_{high}}{V_{high} - V_{low\,side}} \\ &= 12 \times \frac{360}{360 - 240} = 12 \times 3 \\ &= 36\ \text{kVA} \end{aligned} \]

The rating in kVA does not depend on the load power factor, which only decides how many kW the 36 kVA can deliver (about 30.6 kW at 0.85), and the efficiency data is not needed. The answer is 36 kVA, option B.

Answer: A

The rated high side current is the kVA divided by the high side voltage, \(I = \dfrac{S}{V}\). The high side voltage is 2400 V.

\[ I_1 = \frac{75 \times 10^3}{2400} = 31.25\ \text{A} \]

\[ I_2 = \frac{200 \times 10^3}{2400} = 83.33\ \text{A} \]

The other options are 10 or 100 times too large or too small, which would come from taking the kVA as VA or the voltage as 240 V. The answer is 31.25 A and 83.33 A, option A.

Answer: C

The primary is star connected on a 6600 V line, so the primary phase voltage is

\[ V_{1ph} = \frac{6600}{\sqrt{3}} = 3810.5\ \text{V} \]

The turns ratio of each single phase transformer is 12, so the secondary phase voltage is

\[ V_{2ph} = \frac{3810.5}{12} = 317.5\ \text{V} \]

In a delta secondary the line voltage equals the phase voltage, so the secondary line voltage is 317.5 V, which is the 317.55 V of the options. The primary current of 10 A is not needed for the voltage. The other values are 10, 100 and 1000 times off. The answer is option C.

Answer: B

The currents are 7.58 A, 62.5 A.

For a single-phase transformer, current is the rating divided by the voltage on each side:

\[ \begin{aligned} I_{1} &= \frac{25000}{3300} \approx 7.58\ \text{A} \\ I_{2} &= \frac{25000}{400} = 62.5\ \text{A} \end{aligned} \]

As a check, the currents are in the inverse ratio of the voltages: \(62.5/7.58 \approx 8.25 = 3300/400\).

6.25 A, 75.8 A reuses the same digits, but neither value is 25 kVA divided by 3300 V or by 400 V; 16 A, 132 A and 13.2 A, 160 A do not come from these figures either.