On-load tap changing allows the ratio of a transformer to be altered
Answer: A
A diverter switch transfers the current through a transition impedance so that neither an open circuit nor a short circuit occurs during the change. Off-circuit tapping links, by contrast, require the transformer to be isolated.
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Tap changing is normally provided on the high-voltage winding because
Answer: B
Correct answer: the current there is smaller and the switchgear cheaper
Ampere-turns balance means the high-voltage winding carries the smaller current, so the tap changer contacts and the switching duty are lighter and the equipment cheaper. The high-voltage winding also has more turns available, giving finer voltage steps. The higher voltage does complicate the insulation, but the current saving outweighs it.
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The inrush current when a transformer is energised can reach
Answer: C
Switching at an unfavourable instant drives the core into deep saturation for part of each cycle, so a large asymmetric current flows until the transient decays. Protection has to be arranged so that the inrush does not trip it.
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The magnetising current of a transformer is non-sinusoidal because
Answer: D
Correct answer: the core's magnetisation curve is nonlinear
Producing a sinusoidal flux in a nonlinear magnetic material requires a non-sinusoidal current, because the relation between flux and magnetomotive force is a curve and not a straight line. The distortion is rich in third harmonic, which is why three-phase connections must provide it a circulating path - the reason a delta or tertiary winding exists.
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A transformer core of 0.02 m² carries a peak flux density of 1.2 T at 50 Hz. A winding of 200 turns upon it develops
Answer: A
Correct answer: about 1,066 V
\[ E = 4.44\,f\,N\,\Phi_m = 4.44 \times 50 \times 200 \times (1.2 \times 0.02) \] \[ = 4.44 \times 50 \times 200 \times 0.024 = 1066\ \text{V.} \] Note that the flux is the flux density times the area, not the density itself; using 1.2 directly gives an answer about fifty times too large.
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A 25 kVA transformer has 300 W of iron loss and 400 W of full-load copper loss. Its efficiency at half load and unity power factor is nearest
Answer: B
At half load the output is 12.5 kW and the copper loss falls to 100 W, so the losses total 400 W and the efficiency is \( 12{,}500/12{,}900 = 0.969 \). The copper loss falls with the square of the loading, which is the key step.
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That transformer reaches its maximum efficiency at a loading of
Answer: C
Correct answer: about 87 per cent of full load
With 300 W of iron loss and 400 W of full-load copper loss, \[ x = \sqrt{\frac{P_i}{P_{cu,fl}}} = \sqrt{\frac{300}{400}} = \sqrt{0.75} = 0.87. \] So the efficiency peaks at about 87 per cent of full load. Because the curve is flat near its maximum, the efficiency at full load is barely below the peak value.
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A 500 kVA transformer supplies 300 kW at 0.8 power factor lagging. The apparent power drawn is
Answer: D
Correct answer: 375 kVA
\[ S = \frac{P}{\cos\phi} = \frac{300}{0.8} = 375\ \text{kVA.} \] The transformer is therefore loaded to 75 per cent of its 500 kVA rating even though it is delivering only 300 kW. This is exactly why the rating is in kVA: the current, which sizes the conductors and produces the copper loss, depends on the apparent power and not on the real power.
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A transformer is rated in kVA rather than kW because
Answer: A
Correct answer: its losses depend on current and voltage rather than on power factor
Copper loss is fixed by the current and core loss by the voltage, and neither depends on the phase angle between them. Since the manufacturer cannot know what power factor the load will have, the rating is expressed as the product of the voltage and current the transformer can sustain - volt-amperes - leaving the real power to the user’s load.
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Consider the following statements about transformer tests.
1. The open circuit test gives the core loss.
2. The short circuit test gives the copper loss.
3. Both require full rated power.
Which are correct?
Answer: B
Correct answer: 1 and 2 only
Statements 1 and 2 are correct. Statement 3 is the reverse of the truth, and it is the reason the tests are used at all: the open circuit test draws rated voltage at only the small no-load current, and the short circuit test rated current at a few per cent of voltage, so between them they consume a tiny fraction of rated power.
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