Instrument Transformers Questions and Answers

Answer: C

Its secondary sees only the low impedance of an ammeter and its leads, so the flux and hence the error are small. A potential transformer works with a nearly open secondary, since a voltmeter draws almost nothing. Everything else about their behaviour follows from that difference.

Answer: D

It is a series device inserted in a circuit whose current is fixed by the load elsewhere. This is the reverse of a power transformer, where the primary current follows the secondary load, and it is the reason an open secondary is so dangerous.

Answer: A

The errors are computed from measured quantities such as the exciting current, the winding impedances and the burden, without reference to any other transformer. It is laborious, which is why comparison against a standard is the usual practice.

Answer: B

The transformer under test and a standard of the same nominal ratio carry the same primary current, and the difference between their secondary currents is measured in magnitude and phase. Both a deflection and a null version of the method exist.

Answer: C

The exciting current is not proportional to the primary current, so the error expressed as a percentage is quite different at ten per cent of rated current from what it is at full load. Accuracy classes accordingly state limits at 5, 20, 100 and 120 per cent of rated current.

Answer: D

Correct answer: 400 A

The ratio is \( 500/5 = 100 \), so \[ 4 \times 100 = 400\ \mathrm{A}. \] The secondary reading is multiplied by the ratio, never by the primary rating alone - reading 500 A off a full-scale assumption is the usual error, and the 20 A offered comes from dividing instead of multiplying.

Answer: A

Correct answer: 10.5 kV

\[ 105 \times \frac{11000}{110} = 105 \times 100 = 10{,}500\ \mathrm{V} = 10.5\ \mathrm{kV}. \] A potential transformer works exactly as a current transformer does in this respect: the instrument reading times the ratio. The system is running slightly below nominal, which is what the 105 V rather than 110 V is telling the operator.

Answer: B

Correct answer: 288 kW

Both ratios multiply the wattmeter reading, since power is the product of current and voltage: \[ 120 \times \frac{200}{5} \times \frac{6600}{110} = 120 \times 40 \times 60 = 288{,}000\ \mathrm{W} = 288\ \mathrm{kW}. \] Applying only one of the two ratios gives 4.8 kW or 7.2 kW, both of which are offered here.

Answer: C

Correct answer: 2000/5

A current transformer is a step-up in turns and a step-down in current, so 400 secondary turns against a single primary turn gives a current ratio of 400 to 1. With 5 A in the secondary the primary carries \( 400\times5 = 2000 \) A, and the nominal ratio is written 2000/5. Reading the turns count as the primary rating gives the 400/5 offered.

Answer: D

Roughly, the component of the exciting current in phase with the primary is lost from the secondary, so the secondary reproduces only 198 of the 200 amperes and the ratio error is about one per cent negative. This is why turns compensation is applied.