Pancyprian TS Elec. Eng. III ΘΚ study materials
Pancyprian TS Elec. Eng. III ΘΚ study resources

Pancyprian TS Elec. Eng. III ΘΚ: Electrical Measurements, Instrumentation and Testing

Pancyprian TS Elec. Eng. III ΘΚ · Work through your exam, one topic at a time.

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Electrical Measurements, Instrumentation and Testing

Measurement questions test instrument choice, correct connection, interpretation of readings and the meaning of accuracy. In the ΘΚ they usually appear as a commissioning or fault-finding scenario: which instrument, on which range, connected how, and what does the reading prove? Instrument transformers, the two-wattmeter method and insulation/loop/RCD testing are recurring items.

Core rules

  • Ammeter in series with low resistance; voltmeter in parallel with high resistance; a wattmeter needs both current and voltage connections.
  • Current transformers: ratio 100/5 means a secondary reading multiplied by 20; the secondary must never be opened under load. Voltage transformers must never be short-circuited on the secondary.
  • Two-wattmeter method for a balanced three-phase load: P = W1 + W2, Q = √3(W1 − W2), tanφ = √3(W1 − W2)/(W1 + W2). Keep the sign of W2 when it reads backwards.
  • Accuracy class is a percentage of full-scale deflection, so a 1.5% class instrument on a 100 A range can err by 1.5 A even at low readings.
  • True-RMS meters are needed on distorted waveforms such as VFD outputs; average-responding meters under-read.
  • Insulation resistance at 500 V DC, at least 1 MΩ; earth-fault loop impedance tester; RCD tester at half, one and five times rated residual current; clamp meter for load current; power-quality analyser for THD and unbalance.
  • Calibration must be traceable to national standards, and the meter's safety category (CAT III, CAT IV) must suit the point of measurement.

Worked example

Two wattmeters connected through 100/5 CTs into a balanced 400 V motor circuit read 260 W and 70 W on their own scales. Multiplying by the CT ratio 20 gives W1 = 5.2 kW and W2 = 1.4 kW. Then P = 6.6 kW, Q = √3(5.2 − 1.4) = 6.58 kvar, tanφ = 6.58/6.6 = 0.997 so φ = 44.9° and power factor = 0.71. Apparent power is 9.3 kVA. Had W2 read negative (the pointer driven backwards), the sign would be carried into both formulas, not discarded.

Traps that cost marks

  • Forgetting the CT ratio, which makes every subsequent figure wrong by a factor of 20.
  • Dropping the sign of a negative W2 reading.
  • Testing insulation resistance with sensitive electronics still connected, or on a live circuit.
  • Trusting a cheap non-true-RMS meter on a VFD output.
  • Quoting an accuracy class error against the reading rather than against full scale.