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Transformers: board exam questions with solutions

On the REE exam under Electrical Engineering Professional Subjects and the RME exam under Technical Subjects. Try each question before you open its solution, then practise with fresh numbers below: 3 computed question types on this topic.

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Build it in the lab

Basic transformer · Linear supply: transformer, bridge, capacitor, regulator: the same circuits on a live bench, with meters you can move.

Questions

1. An ideal step-down transformer has a turns ratio of 10:1. If the primary current is 5 A, what is the secondary current?

  1. 50 A
  2. 0.5 A
  3. 5 A
  4. 500 A
Show the answer and solution

Answer: A. 50 A

In an ideal transformer I2/I1 = N1/N2, so I2 = 5 × 10 = 50 A. Current steps up when voltage steps down.

2. What is the rated full-load secondary current of a 50 kVA, 2400/240 V single-phase transformer?

  1. 20.8 A
  2. 208.3 A
  3. 120.3 A
  4. 166.7 A
Show the answer and solution

Answer: B. 208.3 A

I2 = S/V2 = 50 000/240 = 208.3 A.

3. A 50 kVA transformer has a core loss of 300 W and a full-load copper loss of 800 W. What is its efficiency at full load and 0.8 power factor?

  1. 97.85%
  2. 98.04%
  3. 97.32%
  4. 99.26%
Show the answer and solution

Answer: C. 97.32%

Output = 50 × 0.8 = 40 kW; losses = 300 + 800 = 1100 W. η = 40 000/(40 000 + 1100) = 97.32%.

4. For the same transformer (core loss 300 W, full-load copper loss 800 W), at what fraction of rated load does maximum efficiency occur?

  1. 37.5%
  2. 100.0%
  3. 50.0%
  4. 61.2%
Show the answer and solution

Answer: D. 61.2%

Maximum efficiency occurs when copper loss equals core loss: x² × 800 = 300, so x = √(300/800) = 0.612 = 61.2% of rated load.

5. A transformer's secondary voltage is 245 V at no load and 235 V at full load. What is its voltage regulation?

  1. 4.26%
  2. 4.08%
  3. 10.00%
  4. 6.52%
Show the answer and solution

Answer: A. 4.26%

VR = (V_NL − V_FL)/V_FL × 100 = (245 − 235)/235 × 100 = 4.26%.

6. The open-circuit (no-load) test of a transformer is mainly used to determine which quantity?

  1. Full-load copper losses
  2. Core (iron) losses
  3. Equivalent series impedance
  4. Polarity of the windings
Show the answer and solution

Answer: B. Core (iron) losses

With the secondary open, the primary current is small, so copper loss is negligible and the wattmeter reads essentially the core (hysteresis plus eddy-current) loss.

7. Why is a transformer core built from thin insulated laminations rather than solid steel?

  1. To reduce hysteresis losses
  2. To reduce copper losses
  3. To reduce eddy-current losses
  4. To increase the turns ratio
Show the answer and solution

Answer: C. To reduce eddy-current losses

Laminations break up the paths of circulating eddy currents, greatly reducing eddy-current loss. Hysteresis loss is reduced by the steel grade (e.g. silicon steel), not by laminating.

8. Two 50 kVA single-phase transformers are connected open-delta (V-V) to supply a balanced three-phase load. What is the maximum three-phase load they can carry without overload?

  1. 100.0 kVA
  2. 150.0 kVA
  3. 75.0 kVA
  4. 86.6 kVA
Show the answer and solution

Answer: D. 86.6 kVA

An open-delta bank can supply √3 × (rating of one unit) = 1.732 × 50 = 86.6 kVA, i.e. 57.7% of the 150 kVA a full delta of three units would carry.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. An ideal 13200/24 V transformer supplies 5 A on its secondary. What current does the primary draw?

  1. 2.75 kA
  2. 5 A
  3. 12.9 mA
  4. 9.09 mA
Show the answer and solution

Answer: D. 9.09 mA

Ideal transformer: V₁I₁ = V₂I₂, so I₁ = 5 × 24/13200 = 9.09 mA.

10. What is the rated secondary current of a 100 kVA single-phase transformer at 230 V?

  1. 217 A
  2. 348 A
  3. 251 A
  4. 435 A
Show the answer and solution

Answer: D. 435 A

Single-phase rated current = S/V = 100 kVA/230 V = 435 A.

11. A 100 kVA transformer has 200 W core loss and 900 W full-load copper loss. What is its efficiency at 100% load and 1 pf?

  1. 148.4%
  2. 98.91%
  3. 49.46%
  4. 98.9%
Show the answer and solution

Answer: B. 98.91%

Copper loss scales with load²: 1² × 900 = 900 W; core 200 W. η = 100000/(100000 + 1100) = 98.91%.

Other topics in the same subjects

DC circuits · AC circuits · Three-phase systems · Transients (RC, RL, RLC) · Semiconductors and diodes · Power electronics · Instrumentation and measurement · Control systems · DC machines · Induction motors · Synchronous machines · Power systems and transmission · Protection: fuses, breakers, relays · Illumination · Power plant engineering · PEC: branch circuits, ampacity and protection · Tools and test equipment · Motors, starters and control circuits · RA 7920 and ethics (electrical)

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