Electrical Machines MCQ – 65 previous year questions with solutions

In a DC motor, back EMF is generated by armature rotation and opposes the applied supply. Eb=V−IaRa=220−10×1=210 V. Converted electromagnetic power≈EbIa=210×10=2100 W. Key relation: E_b=V−I_aR_a; P_converted≈E_b I_a; T∝ΦI_a.

Exam focus: Higher load → initially lower speed/Eb → higher armature current. Common trap: Do not say Eb rises simply because load current rises at constant supply.

The induced armature emf in a running DC motor is the back emf. By Lenz's law it opposes the applied supply voltage, limiting armature current as the motor gains speed.

A — Zero. Back EMF is proportional to speed; at start the speed is zero, so Eb = 0 and the starting current is very high — hence a starter is needed.

C — Back EMF increases with speed. Eb = kΦN, so for constant flux back EMF is directly proportional to speed.

D — 10 A. V = Eb + IaRa, so Ia = (V − Eb)/Ra = (230 − 220)/1 = 10 A.

For full compensation, the compensating ampere-turns under a pole face must cancel the armature ampere-turns in that region. Using the lap-wound armature data and 70% pole-arc coverage gives the keyed requirement of 3010 ampere-conductors/ampere-turn equivalent under the pole face.

B — Increasing the armature current. For a DC motor T ∝ ΦIa; with constant flux, torque rises directly with armature current.

During the short commutation interval, a coil is effectively short-circuited by the brush and its current must reverse from +I to −I. Interpoles are placed between main poles and connected so their flux produces a commutating emf of the required polarity. Key relation: Reactance voltage is associated with L·di/dt during current reversal; ideal commutation completes the reversal within the brush short-circuit time.

Exam focus: Keywords: commutation = current reversal; interpole = sparkless commutation. Common trap: Confusing the commutator with a source of DC; it is a mechanical switching/rectifying element.

C — Compound DC generator. The series field adds flux as load current rises, compensating the voltage drop of the shunt field, so terminal voltage stays nearly constant automatically.

The generated EMF of a DC machine is proportional to flux per pole and speed. Field flux is controlled by excitation current, armature reaction distorts/weakens the main field, and the commutator provides mechanical rectification so the external generated voltage is unidirectional. Key relation: E_g=PΦZN/(60A).

Exam focus: Remember A=2 for wave winding and A=P for simplex lap winding. Common trap: Do not swap the parallel-path counts for lap and wave windings.

C — Total number of armature conductors. In E = PΦZN/(60A), Z is the total number of armature conductors.

D — Direction of rotation remains unchanged. Torque T ∝ ΦIa; reversing both Φ and Ia leaves the sign of torque the same. Reverse only one of them to reverse rotation.

A — It becomes half. E = PΦZN/(60A), so with constant flux the generated EMF is directly proportional to speed.

The air-gap line is the straight unsaturated reference line on a DC machine magnetization curve. It represents the magnetic behaviour when the reluctance is dominated by the air gap and iron saturation is neglected.

A DC series motor develops very high starting torque because, before saturation, flux is approximately proportional to armature current and torque T∝ΦIa≈Ia². This makes it suitable for traction applications such as electric locomotives. Both assertion and reason are true, and the reason correctly explains the assertion: option A.

A — Fine speed control over whole range. The Ward Leonard system provides smooth DC motor speed variation by adjusting the generator output voltage.

The field test is a regenerative/back-to-back test suited to DC series machines, including traction motors, and allows efficiency to be determined under load conditions without wasting full output power.

The electromagnetic torque of a DC motor is proportional to the product of air-gap flux and armature current. If flux is held constant, torque varies directly with armature current, so doubling current ideally doubles torque. Key relation: T=kΦI_a.

Exam focus: Constant flux converts the general torque equation into T∝I_a. Common trap: Do not generalize constant-flux behavior into 'torque is independent of flux'.

C — Back EMF. As the armature of a running motor rotates in the magnetic field, an EMF is induced that opposes the applied supply voltage.

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.

The angle φ between stator voltage and stator current is the power-factor angle; input power factor = cos φ.

Maximum torque at start (s = 1) needs R₂ = X₂. A cage rotor’s resistance is fixed and low, and no external resistance can be added, so its maximum torque occurs at a slip below 1.

Pull-out (breakdown, maximum) torque is the peak of the torque-speed curve; in the paper’s figure the peak is labelled c.

External rotor resistance can be added through the slip rings, giving high starting torque and limited speed control — so slip-ring motors suit those duties.

For a given supply, the value of maximum torque is approximately independent of rotor resistance, while the slip at which it occurs is proportional to rotor resistance. Adding rotor resistance in a slip-ring motor can improve starting torque and shift maximum torque toward standstill. Key relation: s_max≈R₂/X₂ (referred form, simplified); T∝sR₂/(R₂²+(sX₂)²).

Exam focus: Rotor resistance shifts the torque-slip curve horizontally; it does not materially raise Tmax. Common trap: Maximum torque does not occur at synchronous speed because slip is zero there.

The torque-slip curve changes shape when rotor resistance is changed. The given initial s_max=0.2 is not needed to prove the qualitative relationships. Doubling R2 doubles the approximate slip at maximum torque until model limits; Tmax remains unchanged. Key relation: Approximate condition: s_max ∝ R2/Xeq. At s=1, starting torque rises as rotor resistance is increased toward the optimum R2≈Xeq.

Exam focus: Rotor resistance shifts the torque-slip curve horizontally, not vertically at Tmax. Common trap: Saying maximum torque occurs at synchronous speed; torque is zero at s=0.

C — Less than 1. In normal stable operation the full-load torque is below the breakdown (maximum) torque, so T_FL/T_max < 1 — typically about 0.4–0.5.

C — Increasing the rotor resistance. Higher rotor resistance shifts the maximum-torque slip upward while the maximum torque magnitude stays about the same, so the torque at full-load slip rises relative to Tmax.

An induction motor is essentially a transformer with a rotating, short-circuited secondary. The stator acts like the primary and the rotor like the secondary, with power crossing the air gap electromagnetically.

D — Induction generator. It generates when driven above synchronous speed and needs no separate DC rotor-excitation supply, although magnetizing reactive power is still required.

C — To reduce the supply voltage. At start the delta-rated winding is connected in star, so each phase gets V_L/√3; starting current and torque fall to one-third of direct-on-line values.

An induction motor requires appreciable magnetizing current because its magnetic circuit includes an air gap, whose reluctance is much larger than the nearly closed iron path of a transformer. Hence its no-load current is generally higher. Both assertion and reason are true, but the reason statement as printed is not the complete/correct explanation expected by the item, leading to option B.

Rotor-current frequency is proportional to slip: at standstill it equals supply frequency, and at synchronous speed it becomes zero. Rotor frequency fr=s f=0.05×50=2.5 Hz. At synchronous speed s=0, fr=0; at standstill s=1, fr=f. Key relation: s=(N_s−N)/N_s; f_r=s f_s.

Exam focus: Standstill: s=1; synchronous speed: s=0. Common trap: Rotor frequency is not independent of slip.

D — Less than 1. In a normal squirrel-cage induction motor the starting torque is lower than the maximum (breakdown) torque, which occurs at a slip well below 1.

Steady torque exists only in synchronism, and for the basic cylindrical-rotor model T ∝ sin δ. A conventional synchronous motor has no inherent starting torque, and maximum torque occurs near δ = 90°, not 180°.

A synchronous motor runs at synchronous speed, has zero inherent starting torque and holds speed as load changes. Torque depends on load angle, not slip, and converter-fed synchronous motors are used in variable-speed drives.

A conventional synchronous motor is not inherently self-starting. By adjusting field excitation it can operate at lagging, unity or leading power factor; over-excited operation is used for reactive-power support. Key relation: N_s=120f/P.

Exam focus: Excitation controls power factor; frequency and pole count set speed. Common trap: Load changes torque angle, not steady synchronous speed.

A — Absence of slip losses. A synchronous motor runs at zero slip, so there are no rotor slip (copper) losses of the induction-motor kind.

GPSC's final key accepts 'All of the mentioned'. A larger air gap reduces synchronous reactance and armature-reaction sensitivity, which improves stability/parallel operation; appropriate air-gap design also helps achieve a smoother magnetic-field distribution.

C — Hunting. Damper windings reduce the rotor oscillations that follow a sudden load change by producing a torque that opposes those oscillations.

For an alternator connected to an infinite bus, bus voltage and frequency are fixed. Increasing prime-mover mechanical input increases torque angle δ and therefore increases the active power delivered to the bus.

C — 100 A approximately. Electrical input is 50/0.90 = 55.56 kW, giving line current 55,555.56/(√3 × 400 × 0.8) = 100.23 A.

A — It increases up to the stability (pull-out) limit. With V and Xs constant, P = (VE/Xs) sinδ, so raising excitation E raises Pmax = VE/Xs.

The final GPSC answer is B. A field winding placed on the stator produces an air-gap field stationary with respect to the stator. A rotor moving at synchronous speed sees that stationary field moving backward at Ns, not at 2Ns.

A generator with more poles produces the same frequency at a lower mechanical speed; for fixed pole count, frequency is directly proportional to speed. f=PN/120. With P=8 and N=750 rpm, f=50 Hz; at 1500 rpm, f=100 Hz. Key relation: f=PN/120; N_s=120f/P.

Exam focus: Use rpm in the 120-factor formula. Common trap: Frequency is not independent of rotor speed in a synchronous generator.

In conventional electromagnetic energy-conversion devices, electrical input establishes magnetic flux in the coupling field, so energy is stored primarily in the magnetic field before being converted into mechanical work and/or returned electrically.

In a conventional doubly excited rotating machine, electromagnetic torque results from interaction between stator and rotor magnetic fields; hence both windings must be excited and carry current for this torque-production mechanism.

D — A resistance with the magnetising reactance. Core loss (hysteresis + eddy current) is a real-power loss, so it is modelled by a resistance (R_c) in the magnetising branch.

At standstill a single-phase stator field can be resolved into equal forward and backward rotating fields. Their starting torques are equal and opposite, giving zero net starting torque; hence the motor is not self-starting. Both assertion and reason are true and the reason correctly explains the assertion: option A.

Distributed and short-pitched AC windings are deliberate design choices that improve waveform quality and suppress selected harmonics. Short pitching reduces the fundamental induced EMF slightly through pitch factor kp; distribution across slots introduces distribution factor kd. Key relation: k_w=k_p k_d; k_p=cos(α/2) for chording angle α.

Exam focus: Winding factor is normally less than 1. Common trap: A concentrated winding does not inherently guarantee a sinusoidal EMF.