Transformers MCQ [Free PDF] – 16 Objective Questions with Answers

Transformer core loss is approximately constant at fixed voltage/frequency, while copper loss is I²R and therefore varies with the square of load. The competition between these losses creates a definite maximum-efficiency load point. Key relation: At x p.u. load: Pcu=x²Pcu,FL. Maximum η when x²Pcu,FL = Pi. All-day η = output energy (kWh)/input energy (kWh).

Exam focus: Remember: power transformer → ordinary efficiency; distribution transformer → all-day efficiency is frequently tested. Common trap: Saying efficiency is constant because transformer has no moving parts.

At x p.u. load, transformer output is proportional to x while copper loss is proportional to x² and iron loss remains approximately constant. The given losses make the maximum-efficiency load x=sqrt(500/500)=1 p.u. Every statement follows numerically, so B is correct. Key relation: Pcu(x)=x²Pcu,FL. Here Pcu(0.5)=0.25×500=125 W. Maximum η: x²Pcu,FL=Pi → x=1.

Exam focus: If core loss equals full-load copper loss, maximum efficiency is at full load—an instant result. Common trap: Scaling copper loss linearly with load instead of with load squared.

C — Copper loss = iron loss. Maximum efficiency occurs at the load where the variable copper loss equals the constant iron (core) loss.

The given transformer ratio is 200/1000 V and the emf per turn is 10 V. For the 200 V winding, N = E/(emf per turn) = 200/10 = 20 turns.

D — Maximum magnetic flux in the core. In E = 4.44 f N φm, φm is the peak value of the sinusoidal core flux.

The two specified loads are 1050 kVA and 180 kVA. The total apparent power supplied to the loads is therefore 1050 + 180 = 1230 kVA, matching option A.

Dry-type transformers use solid insulation and air/natural/forced-air cooling rather than flammable insulating oil. Their lower fire and environmental risk makes them suitable for indoor/commercial installations. Key relation: Cooling classes include AN/AF for dry-type designs.

Exam focus: Indoor fire-risk questions often favor dry-type transformers. Common trap: Oil maintenance belongs to oil-filled transformers, not dry-type units.

An ideal transformer conserves apparent power and relates voltages, turns and currents inversely. Turns ratio a=230/115=2. Output power=115×4=460 W. Key relation: V1/V2=N1/N2=a; I1/I2=N2/N1=1/a; ideal V1I1=V2I2.

Exam focus: Voltage ratio follows turns ratio; current ratio is inverse. Common trap: In an ideal transformer input power equals output power.

A — 1 : 1. A usual isolation transformer has equal primary and secondary turns so that it transfers approximately the same AC voltage while providing galvanic separation between the circuits.

In the usual transformer tests, the open-circuit test is performed by energizing the LV side and leaving the HV side open. The short-circuit test is usually performed by shorting the LV side and applying a reduced voltage to the HV side. This matches option A.

If two transformers have different impedance magnitudes but identical X/R ratios, their impedance angles are equal. They may share unequal magnitudes of load, but the load-current phase angles are the same, so both operate at the same power factor.

A — Dd1. A conventional delta–delta transformer connection does not provide the 30° displacement required by clock number 1.

D — It provides a magnetic path for flux linkage between the coils. The core gives a low-reluctance closed path to the alternating flux produced by the primary so that it links the secondary.

D — High permeability. A high-permeability core provides a low-reluctance path and establishes the required flux with comparatively little magnetizing current.

OC and SC tests separate transformer loss components without loading the transformer fully. OC emphasizes shunt/core behavior; SC emphasizes series/leakage behavior. Key relation: OC: P0 ≈ Pi(core). SC at rated current: Psc ≈ Pcu,FL; Zeq = Vsc/Isc; Req=Psc/Isc².

Exam focus: OC → rated voltage, small current, core loss. SC → rated current, small voltage, copper loss. Common trap: Confusing “rated voltage” in OC with “rated current”; the OC current is deliberately small.

A — Less than the input voltage. In an ideal transformer V2/V1 = N2/N1; with N1 > N2 it is a step-down transformer.