Transformers Questions and Answers

Answer: A

Correct answer: \( 4.44fN\Phi_m \)

The induced emf is the rate of change of flux linkage, and for sinusoidal flux the rms value works out as \[ E = 4.44\,f\,N\,\Phi_m, \] where 4.44 is \( 2\pi/\sqrt{2} \). Everything follows from it: the flux is fixed by the applied voltage and frequency, which is why core loss is nearly constant with load, and why a transformer run at reduced frequency saturates.

Answer: B

Induction requires a changing flux linkage, and direct current gives a steady flux once the transient has died away. The primary then behaves as a low resistance across the supply, which is why applying dc to a transformer destroys it.

Answer: C

Correct answer: the turns ratio of its windings

Both windings link the same core flux, so each develops \( 4.44fN\Phi_m \) and the ratio of the two voltages is simply the ratio of the turns. That is exactly why the flux term cancels and the relation holds regardless of frequency, core material or size - it is the defining property of the transformer.

Answer: D

Correct answer: the inverse of the turns ratio

Neglecting the small magnetising current, the ampere-turns of the two windings must balance, so \( N_1I_1 = N_2I_2 \) and the current ratio is the inverse of the turns ratio. The winding with more turns carries less current, which is why the high-voltage winding is wound in finer wire and why tap changers are placed there.

Answer: A

Correct answer: the square of the turns ratio

Voltage transforms by the turns ratio and current by its inverse, so impedance, being their quotient, transforms by the square: \( Z_1 = (N_1/N_2)^2 Z_2 \). This is the basis of impedance matching, and it is why per unit quantities are so convenient - expressed in per unit, an impedance has the same value on either side.

Answer: B

Correct answer: 100 V

\[ V_2 = V_1\frac{N_2}{N_1} = 400 \times \frac{250}{1000} = 100\ \text{V.} \] The transformer steps down by four to one, so the secondary voltage is a quarter of the primary. The distractor of 1600 V is what you get by inverting the ratio, which is the usual slip when the direction of transformation is not fixed in mind first.

Answer: C

Correct answer: 5 A

Ampere-turns balance, so the current transforms by the inverse of the turns ratio: \[ I_1 = I_2\frac{N_2}{N_1} = 20 \times \frac{250}{1000} = 5\ \text{A.} \] Check it against the power: the secondary delivers \( 100 \times 20 = 2000 \) VA and the primary draws \( 400 \times 5 = 2000 \) VA, as it must if losses are neglected.

Answer: D

Correct answer: 64 Ω

Impedance transforms by the square of the turns ratio: \[ Z_1 = \left(\frac{N_1}{N_2}\right)^2 Z_2 = 4^2 \times 4 = 64\ \Omega. \] Forgetting the square gives 16, which is one of the distractors. A step-down transformer therefore makes a low impedance look large from the primary side, which is precisely what impedance matching exploits.

Answer: A

The applied voltage and the frequency fix the flux through the emf equation, and the load's ampere-turns are balanced by an equal primary contribution. That constancy is why the iron loss of a transformer hardly changes with load.

Answer: B

One part in quadrature with the voltage establishes the flux, and a small part in phase with it supplies the hysteresis and eddy current loss. Together they amount to a few per cent of rated current in a power transformer.