Transformers Questions and Answers

Answer: B

The primary needs about 118 turns.

The emf equation of a transformer is \(V = 4.44 f \Phi_{m} N\), so the number of turns for the 110 V primary is

\[ \begin{aligned} N &= \frac{110}{4.44 \times 50 \times 4.2 \times 10^{-3}} \\ &= \frac{110}{0.9324} = 117.97 \approx 118 \end{aligned} \]

220 turns would be needed for the 220 V side, and 120 and 150 do not satisfy the emf equation for the given flux.

Answer: A

The no-load (open circuit) test determines the core loss.

In the open circuit test, rated voltage is applied to one winding with the other open. The current drawn is only the small no-load current, so the copper loss is negligible, while the flux is at its normal value and the full core (iron) loss occurs. The wattmeter reading is therefore the core loss, and the test also gives the magnetising branch parameters.

Copper loss is found from the short circuit test, which is done at reduced voltage with full-load current. A static transformer has no mechanical loss.

Answer: B

The no-load current is the phasor sum of \(I_{m}\) and \(I_{c}\).

On no load the primary draws a small current \(I_{0}\) with two components at right angles: the magnetising current \(I_{m}\), in phase with the flux and lagging the voltage by 90°, which sets up the flux; and the core loss component \(I_{c}\), in phase with the voltage, which supplies the iron loss. Since they are 90° apart they add as phasors: \(I_{0} = \sqrt{I_{m}^{2} + I_{c}^{2}}\).

Adding them algebraically ignores the 90° between them. \(I_{0}\) is itself the no-load current, and \(I_{1}\) is the primary current under load, so options built from them are wrong.

Answer: A

Zero voltage regulation occurs at a leading power factor.

Approximately, the regulation of a transformer is

\[ \text{VR} \approx \frac{I(R\cos\varphi \pm X\sin\varphi)}{V} \]

with + for lagging and − for leading power factor. For a lagging or unity power factor both terms add, so the regulation is always positive. Only with a leading power factor can the reactive term cancel the resistive term, which happens when \(\tan\varphi = R/X\), giving zero regulation. At zero power factor lagging the regulation is close to its maximum.

Answer: C

The condition is \(P_{cu} = \frac{16}{9} P_{i}\).

A transformer is most efficient at the load where the variable copper loss equals the constant iron loss. Copper loss varies as the square of the load fraction x, so at a fraction x of full load it is \(x^{2}P_{cu}\).

\[ \begin{aligned} x^{2} P_{cu} &= P_{i} \\ \left(\frac{3}{4}\right)^{2} P_{cu} &= P_{i} \\ P_{cu} &= \frac{16}{9} P_{i} \end{aligned} \]

9/16 is the same relation turned upside down, and 3/4 and 4/3 forget that copper loss goes with the square of the load, not the load itself.

Answer: C

Silicon steel is preferred because it reduces hysteresis loss.

Adding about 3 to 4 per cent silicon to iron narrows its hysteresis loop, so less energy is lost in reversing the magnetisation fifty times a second. It also raises the resistivity of the steel, which together with lamination cuts the eddy current loss, and it reduces ageing of the core.

The other reasons are wrong. Silicon raises resistivity rather than reducing it, and lower resistivity would increase eddy current loss. Silicon steel keeps a high permeability, which is what a transformer core needs. A change in tensile strength is not a reason to choose a core material; silicon actually makes the steel more brittle.