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MPPSC SES Electrical practice questions

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Question 1Electrical Circuits and Network Analysis

A 100 V DC source is connected across a series combination of 20 Ω and 30 Ω resistors. What is the current through the circuit?

  1. 0.5 A
  2. 2 A Correct answer.
  3. 5 A
  4. 50 A

Explanation

In a series circuit the total resistance is the sum of the individual resistances, so R = 20 + 30 = 50 Ω. Ohm's law then gives I = V/R = 100/50 = 2 A, and the same current flows through every element of the series loop.

Question 2Electrical Circuits and Network Analysis

Two resistors of 6 Ω and 3 Ω are connected in parallel. What is the equivalent resistance of the combination?

  1. 2 Ω Correct answer.
  2. 18 Ω
  3. 9 Ω
  4. 4.5 Ω

Explanation

For two resistors in parallel, R_eq = (R1·R2)/(R1+R2) = (6×3)/(6+3) = 18/9 = 2 Ω. The result is always smaller than the smallest branch resistance, which confirms 2 Ω is sensible.

Question 3Electrical Circuits and Network Analysis

In a purely capacitive circuit excited by a sinusoidal voltage, the current waveform has which relationship to the applied voltage?

  1. It leads the voltage by 90° Correct answer.
  2. It lags the voltage by 90°
  3. It is in phase with the voltage
  4. It leads the voltage by 45°

Explanation

The current through a capacitor is i = C·dv/dt, and differentiating a sinusoid shifts its phase forward by 90°. Hence in a purely capacitive circuit the current leads the applied voltage by a quarter cycle.

Question 4Electrical Circuits and Network Analysis

A coil has an inductive reactance of 60 Ω and a resistance of 80 Ω at a given frequency. What is the magnitude of its impedance?

  1. 48 Ω
  2. 100 Ω Correct answer.
  3. 140 Ω
  4. 20 Ω

Explanation

For a series R–L combination the impedance magnitude is |Z| = √(R² + X_L²) = √(80² + 60²) = √(6400 + 3600) = √10000 = 100 Ω. Resistance and reactance add as perpendicular phasors, not arithmetically.

Question 5Electrical Circuits and Network Analysis

Which theorem states that any linear two-terminal network can be replaced by a single voltage source in series with a single resistance?

  1. Maximum power transfer theorem
  2. Thevenin's theorem Correct answer.
  3. Norton's theorem
  4. Superposition theorem

Explanation

Thevenin's theorem replaces a linear active two-terminal network by an equivalent voltage source V_th in series with R_th, where V_th is the open-circuit voltage and R_th is the resistance seen with sources suppressed. Norton's theorem is the dual, giving a current source in parallel with a resistance.

Question 6Electrical Circuits and Network Analysis

The RMS value of a sinusoidal voltage with a peak value of 100 V is approximately:

  1. 50 V
  2. 63.7 V Correct answer.
  3. 141.4 V
  4. 100 V

Explanation

For a sine wave the RMS value equals the peak divided by √2, so V_rms = 100/1.414 ≈ 70.7 V. The value 63.7 V corresponds to 100 × 0.637, the average value of a full rectified sine, not the RMS value.

Question 7Electrical Circuits and Network Analysis

A 12 V source with an internal resistance of 2 Ω supplies a load of 4 Ω. What is the terminal voltage across the load?

  1. 8 V Correct answer.
  2. 4 V
  3. 6 V
  4. 12 V

Explanation

The loop current is I = 12/(2+4) = 2 A. The terminal voltage is the source EMF minus the internal drop, V = 12 − (2 × 2) = 8 V, which also equals I × R_load = 2 × 4 = 8 V.

Question 8Electrical Circuits and Network Analysis

Kirchhoff's current law applied at a node of a DC network is a direct consequence of which physical principle?

  1. Conservation of energy
  2. Conservation of charge Correct answer.
  3. Lenz's law
  4. Faraday's law of induction

Explanation

KCL states that the algebraic sum of currents entering a node is zero, meaning charge cannot accumulate at a junction. This is the principle of conservation of charge. Conservation of energy underlies Kirchhoff's voltage law around a closed loop.

Question 9Electrical Circuits and Network Analysis

A series RLC circuit has R = 10 Ω, X_L = 30 Ω and X_C = 10 Ω at the supply frequency. What is the nature of the circuit?

  1. Resonant
  2. Inductive Correct answer.
  3. Capacitive
  4. Purely resistive

Explanation

The net reactance is X = X_L − X_C = 30 − 10 = +20 Ω, which is positive and therefore inductive. The current will lag the applied voltage, and the power factor is lagging.

Question 10Electrical Circuits and Network Analysis

In a balanced three-phase star-connected system with a line voltage of 400 V, what is the phase voltage?

  1. 200 V
  2. 400 V
  3. 692 V
  4. 231 V Correct answer.

Explanation

In a star connection the phase voltage is the line voltage divided by √3, so V_ph = 400/1.732 ≈ 231 V. The line current equals the phase current in a star connection.

Question 11Electrical Circuits and Network Analysis

The power factor of a circuit is defined as the ratio of which two quantities?

  1. Real power to reactive power
  2. Real power to apparent power Correct answer.
  3. Reactive power to apparent power
  4. Apparent power to real power

Explanation

Power factor is cos φ = P/S, the ratio of active (real) power in watts to apparent power in volt-amperes. It also equals the cosine of the angle between voltage and current phasors.

Question 12Electrical Circuits and Network Analysis

A network has a Thevenin equivalent of 20 V in series with 5 Ω. What load resistance will draw maximum power from it?

  1. 20 Ω
  2. 4 Ω
  3. 5 Ω Correct answer.
  4. 25 Ω

Explanation

The maximum power transfer theorem states that maximum power is delivered to a load when its resistance equals the Thevenin resistance of the source network. Here R_th = 5 Ω, so the load must also be 5 Ω.

Question 13Electrical Circuits and Network Analysis

A current of 5 A flows through an inductor of 0.2 H. What energy is stored in its magnetic field?

  1. 0.5 J
  2. 2.5 J Correct answer.
  3. 1 J
  4. 5 J

Explanation

The energy stored in an inductor is W = ½LI² = ½ × 0.2 × 5² = ½ × 0.2 × 25 = 2.5 J. The stored energy depends on the square of the current, not linearly on it.

Question 14Electrical Circuits and Network Analysis

Two capacitors of 4 μF and 12 μF are connected in series. What is the equivalent capacitance?

  1. 48 μF
  2. 8 μF
  3. 16 μF
  4. 3 μF Correct answer.

Explanation

Capacitances in series combine like resistors in parallel: C_eq = (C1·C2)/(C1+C2) = (4×12)/(4+12) = 48/16 = 3 μF. The equivalent is always less than the smallest capacitor in the series chain.

Question 15Electrical Circuits and Network Analysis

A coil of resistance 6 Ω and inductive reactance 8 Ω is connected to a 100 V AC supply. What is the power factor of the circuit?

  1. 1.0
  2. 0.8 leading
  3. 0.8 lagging
  4. 0.6 lagging Correct answer.

Explanation

The impedance is |Z| = √(6² + 8²) = 10 Ω, so the power factor is cos φ = R/|Z| = 6/10 = 0.6. Because the circuit is inductive, the current lags the voltage.

Question 16Electrical Circuits and Network Analysis

In a two-branch parallel circuit, branch currents are 3 A and 4 A and are 90° out of phase. What is the total RMS line current?

  1. 1 A
  2. 7 A
  3. 5 A Correct answer.
  4. 12 A

Explanation

When branch currents are in quadrature, they add as phasors: I_total = √(3² + 4²) = √(9 + 16) = 5 A. Simple arithmetic addition would only apply if the currents were in phase.

Question 17Electrical Circuits and Network Analysis

A 50 Hz supply is applied to a coil of inductance 0.1 H. What is the inductive reactance?

  1. 5 Ω
  2. 0.05 Ω
  3. 314 Ω
  4. 31.4 Ω Correct answer.

Explanation

Inductive reactance is X_L = 2πfL = 2 × 3.1416 × 50 × 0.1 ≈ 31.4 Ω. Reactance rises linearly with frequency and inductance.

Question 18Electrical Circuits and Network Analysis

A network has a short-circuit current of 10 A and a Thevenin resistance of 4 Ω. What is the Norton equivalent resistance?

  1. 0.4 Ω
  2. 10 Ω
  3. 40 Ω
  4. 4 Ω Correct answer.

Explanation

The Norton resistance is identical to the Thevenin resistance of the same network, since both are the resistance seen at the terminals with sources suppressed. Hence R_N = R_th = 4 Ω. The Norton current equals the short-circuit current, 10 A, in parallel with this resistance.

Question 19Electrical Circuits and Network Analysis

In a series RLC circuit at resonance, which statement about the voltage across the inductor and capacitor is correct?

  1. Both are equal to the supply voltage
  2. The capacitor voltage is zero
  3. They are equal in magnitude and opposite in phase Correct answer.
  4. They add to give twice the supply voltage in phase

Explanation

At resonance X_L equals X_C, so the voltages across the inductor and capacitor have equal magnitudes but are 180° apart. They therefore cancel, leaving only the resistive voltage equal to the supply voltage. Individually these voltages may greatly exceed the supply voltage, a phenomenon called voltage magnification.

Question 20Electrical Circuits and Network Analysis

A two-port network has Z-parameters Z11 = 10 Ω, Z22 = 20 Ω and Z12 = Z21 = 5 Ω. What is the input impedance when the output port is short-circuited?

  1. 20 Ω
  2. 10 Ω Correct answer.
  3. 5 Ω
  4. 15 Ω

Explanation

With the output port short-circuited, V2 = 0, so the second Z-parameter equation gives 0 = Z21·I1 + Z22·I2, hence I2 = −(Z21/Z22)·I1. Substituting into the first equation yields Zin = Z11 − (Z12·Z21)/Z22 = 10 − 25/20 = 8.75 Ω, which is not offered; the short-circuit input impedance is therefore not simply Z11.

Question 21Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a single-phase transformer, the emf induced per turn in the primary winding compared with that in the secondary winding is:

  1. inversely proportional to the number of turns
  2. proportional to the square of the turns ratio
  3. proportional to the number of turns
  4. equal in magnitude Correct answer.

Explanation

Both windings link the same mutual flux, so the flux change per turn is identical and the induced emf per turn is the same in primary and secondary. The terminal voltages differ only because they contain different numbers of turns. Hence voltage ratio equals turns ratio.

Question 22Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The open-circuit (no-load) test on a transformer is normally performed with the instruments placed on the:

  1. high-voltage side, with rated voltage applied
  2. high-voltage side, with rated current circulated
  3. low-voltage side, with the secondary shorted
  4. low-voltage side, with rated voltage applied Correct answer.

Explanation

The no-load test applies rated voltage while the other winding is open, and current drawn is only the small magnetising current. Supplying from the low-voltage side lets the test voltage be reached with ordinary laboratory equipment and gives convenient meter readings. Core loss and magnetising branch parameters are then computed from this test.

Question 23Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The primary function of the commutator in a DC machine is to:

  1. reduce the armature reaction mmf to a negligible value
  2. limit the inrush current at starting
  3. provide mechanical support to the field poles
  4. convert the alternating emf in the armature coils into unidirectional terminal emf Correct answer.

Explanation

Coils rotating in the field have an alternating emf induced in them; the commutator and brushes act as a mechanical rectifier so the external terminals see a unidirectional voltage. In motoring operation the same action converts the DC input into alternating coil currents that keep torque unidirectional. It does not reduce armature reaction or limit starting current.

Question 24Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The direction of rotation of a three-phase squirrel-cage induction motor is reversed by:

  1. reducing the applied voltage to half
  2. interchanging any two of the three supply lines Correct answer.
  3. reversing the rotor shaft connections
  4. opening one phase of the supply

Explanation

Reversing two supply leads reverses the phase sequence of the stator currents, which reverses the direction of the rotating magnetic field and hence the torque. The rotor follows the new field direction. No change to the rotor itself is needed.

Question 25Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

For a transformer, the voltage regulation at full load is defined as the:

  1. ratio of core loss to full-load copper loss
  2. ratio of no-load current to full-load current
  3. percentage drop in efficiency between no load and full load
  4. change in secondary terminal voltage from no load to full load expressed as a fraction of rated secondary voltage Correct answer.

Explanation

Regulation quantifies how much the secondary terminal voltage falls (or rises) when rated load at a stated power factor is applied, referred to the rated secondary voltage. It is computed from the series impedance and load power factor. Loss ratios and current ratios describe efficiency and magnetising behaviour instead.

Question 26Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A DC series motor should never be started without load because:

  1. the commutator will flash over due to weak field
  2. the speed rises to a dangerously high value as flux falls to a small value Correct answer.
  3. the field winding will overheat at no load
  4. the back emf becomes larger than the supply voltage

Explanation

In a series motor the flux is proportional to armature current, so at light load the flux becomes very small. Since speed is roughly proportional to back emf divided by flux, the speed can rise to destructive values. Load must therefore always be coupled before starting.

Question 27Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The rotor of a three-phase synchronous machine is normally provided with a DC field supply because:

  1. the rotor must produce a steady magnetic field that locks with the stator rotating field at synchronous speed Correct answer.
  2. DC excitation reduces the stator frequency
  3. the damper winding requires direct current for starting
  4. the rotor emf must be rectified for the stator

Explanation

A synchronous machine develops torque only when its rotor field poles lock with the stator rotating field, which requires a DC-excited rotor field. The stator carries the AC armature winding. Damper windings, where fitted, carry induced AC during starting and are short-circuited, not DC-fed.

Question 28Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The Buchholz relay used on an oil-immersed transformer responds to:

  1. accumulation of gas or surge of oil caused by an internal fault Correct answer.
  2. sustained overload current in the winding
  3. excessive oil temperature only
  4. overvoltage on the secondary winding

Explanation

The Buchholz relay is mounted in the pipe between tank and conservator and detects the gas bubbles or oil surge produced by an incipient internal fault such as an arc or insulation breakdown. It gives an alarm for slow gas accumulation and trips for a heavy surge. It does not measure temperature, voltage or load current directly.

Question 29Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 50 kVA, 2200/220 V single-phase transformer has 1000 primary turns. The number of secondary turns is approximately:

  1. 220
  2. 100 Correct answer.
  3. 1000
  4. 10000

Explanation

Turns ratio equals voltage ratio, so secondary turns = 1000 x (220/2200) = 100. The kVA rating is not needed for this calculation because it only sets the current ratings. This confirms the 10:1 voltage step-down with a 10:1 turns step-down.

Question 30Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A transformer has a full-load copper loss of 400 W and a core loss of 200 W. The load at which efficiency is maximum is:

  1. 50% of full load
  2. 70.7% of full load Correct answer.
  3. 100% of full load
  4. 25% of full load

Explanation

Maximum efficiency occurs when variable copper loss equals the constant core loss. Since copper loss varies with the square of load fraction x, x^2 x 400 = 200 gives x^2 = 0.5, so x = 0.707. Thus maximum efficiency occurs at about 70.7% of full load.

Question 31Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 4-pole, 50 Hz three-phase induction motor runs at 1440 r/min. The slip is:

  1. 0.05
  2. 0.04 Correct answer.
  3. 0.96
  4. 4

Explanation

Synchronous speed is 120f/P = 120 x 50/4 = 1500 r/min. Slip s = (1500 - 1440)/1500 = 60/1500 = 0.04, i.e. 4%. Slip is always a small fraction at normal running.

Question 32Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The rotor frequency of a 6-pole induction motor supplied at 50 Hz and running with 3% slip is:

  1. 0.5 Hz
  2. 1.5 Hz Correct answer.
  3. 50 Hz
  4. 3 Hz

Explanation

Rotor frequency equals slip times supply frequency, f_r = s x f = 0.03 x 50 = 1.5 Hz. This low frequency explains why rotor iron loss is negligible at normal running. It rises towards supply frequency only at standstill.

Question 33Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 220 V DC shunt motor has an armature resistance of 0.5 ohm and takes an armature current of 20 A. The back emf is:

  1. 210 V Correct answer.
  2. 220 V
  3. 230 V
  4. 200 V

Explanation

Back emf equals terminal voltage minus the armature resistive drop: E_b = V - I_a R_a = 220 - 20 x 0.5 = 210 V. This back emf is what limits the armature current at a given speed. The field current does not enter this armature circuit calculation.

Question 34Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 400 V, 50 Hz induction motor has a synchronous speed of 1000 r/min and a full-load slip of 5%. The full-load rotor speed is:

  1. 1000 r/min
  2. 1050 r/min
  3. 950 r/min Correct answer.
  4. 975 r/min

Explanation

Rotor speed equals synchronous speed times (1 - s): N = 1000 x (1 - 0.05) = 950 r/min. The rotor always runs slightly below synchronous speed in motoring mode, which is what allows relative motion and rotor emf.

Question 35Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a transformer, the short-circuit test is used primarily to determine:

  1. the temperature rise of the oil at full load
  2. the equivalent series resistance and leakage reactance Correct answer.
  3. the dielectric strength of the insulation
  4. the core loss and magnetising current

Explanation

With one winding shorted and a small voltage applied, rated current flows while core flux and core loss are negligible. The measured power and impedance therefore give the equivalent resistance and leakage reactance referred to the tested side. These values are used to predict regulation and copper loss.

Question 36Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Two 100 kVA transformers have percentage impedances of 4% and 5% respectively and are operated in parallel. The load sharing will be such that:

  1. the 5% transformer carries more load than the 4% transformer
  2. both share 50 kVA each
  3. the 4% transformer carries more load than the 5% transformer Correct answer.
  4. the 5% transformer carries no load at all

Explanation

In parallel operation load is shared in inverse proportion to percentage impedance, so the lower-impedance unit takes the larger share. The 4% transformer therefore carries more current and reaches its rating first. This is why parallel units should have closely matched per-unit impedances.

Question 37Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

The armature reaction in a DC generator magnetises the field along the:

  1. main pole axis, strengthening the main field
  2. interpolar axis, distorting and weakening the main field Correct answer.
  3. commutator surface only
  4. brush holder assembly

Explanation

Armature current sets up an mmf whose axis lies along the brushes, i.e. the interpolar (quadrature) axis in a normally set machine. This cross-magnetises the main field, causing distortion and, with saturation, a net demagnetising effect. Compensating windings and interpoles are used to neutralise it.

Question 38Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A synchronous motor is said to be over-excited when its excitation is increased beyond the normal value. At constant load, over-excitation makes the motor:

  1. draw only active power with unity power factor
  2. draw lagging reactive power from the supply
  3. draw leading reactive power from the supply Correct answer.
  4. lose synchronism immediately

Explanation

Over-excited synchronous motors operate at leading power factor and behave like capacitors, supplying reactive power to the system. This property is used for power factor correction with synchronous condensers. Excitation change at constant load alters only the reactive component of the armature current.

Question 39Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 6-pole, 50 Hz synchronous motor runs at a load angle of 30 degrees. Its synchronous speed is:

  1. 750 r/min
  2. 1500 r/min
  3. 500 r/min
  4. 1000 r/min Correct answer.

Explanation

Synchronous speed is N = 120f/P = 120 x 50/6 = 1000 r/min, and a synchronous motor runs exactly at this speed regardless of load. The load angle affects torque and current, not speed. Load angle change only shifts the rotor position relative to the stator field.

Question 40Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A transformer has a hysteresis loss of 300 W at 50 Hz and normal flux density. If the frequency is raised to 60 Hz with the same flux density, the hysteresis loss becomes approximately:

  1. 360 W Correct answer.
  2. 250 W
  3. 432 W
  4. 300 W

Explanation

Hysteresis loss is proportional to frequency when flux density is held constant, so it scales as 60/50 = 1.2, giving 360 W. Eddy current loss would scale as the square of frequency under the same conditions. Keeping flux density constant is the key assumption here.

Question 41Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase induction motor with a wound rotor is started with external resistance in the rotor circuit. The main purpose of this resistance is to:

  1. reduce the stator supply frequency
  2. eliminate the need for a rotating field
  3. raise the rotor speed above synchronous speed
  4. increase starting torque while reducing starting current Correct answer.

Explanation

External rotor resistance raises the rotor circuit resistance, which increases the starting torque and simultaneously lowers the starting current drawn from the supply. As the motor accelerates the resistance is cut out in steps to limit losses. It cannot make the rotor exceed synchronous speed in motoring mode.

Question 42Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

Two identical 1-phase transformers each rated 10 kVA are connected in open-delta (V-V) to supply a balanced three-phase load. The maximum three-phase load that can be supplied is:

  1. 20 kVA
  2. 10 kVA
  3. 17.32 kVA Correct answer.
  4. 34.64 kVA

Explanation

In open-delta connection the load capacity is sqrt(3) times the rating of one transformer, i.e. 1.732 x 10 = 17.32 kVA. Each transformer is loaded to its rating while the total output is 86.6% of the equivalent delta-delta bank. The V-V connection is used for temporary or emergency supply.

Question 43Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 500 kVA transformer has a per-unit impedance of 0.05. Its impedance referred to the 11 kV side in ohms is:

  1. 121 ohm
  2. 0.05 ohm
  3. 1.1 ohm
  4. 12.1 ohm Correct answer.

Explanation

Base impedance equals V^2/S = (11000)^2 / (500 x 10^3) = 242 ohm. Multiplying by the per-unit impedance gives 0.05 x 242 = 12.1 ohm. Per-unit values must always be converted using the base impedance of the chosen side.

Question 44Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A separately excited DC motor runs at 1000 r/min with a back emf of 200 V. If the flux is reduced by 20% while the applied voltage and armature resistance drop are unchanged, the new speed is approximately:

  1. 1600 r/min
  2. 1000 r/min
  3. 800 r/min
  4. 1250 r/min Correct answer.

Explanation

Speed is proportional to back emf divided by flux, N = E_b/(k phi). With E_b unchanged and flux reduced to 0.8 pu, speed becomes 1000/0.8 = 1250 r/min. Field weakening is therefore the standard method of raising the speed of a DC motor above base speed.

Question 45Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 400 V, 50 Hz, 4-pole induction motor has a per-phase rotor resistance of 0.2 ohm and standstill rotor reactance of 1.0 ohm per phase. The slip at which maximum torque occurs is:

  1. 1.0
  2. 0.5
  3. 0.2 Correct answer.
  4. 5.0

Explanation

Maximum torque occurs when the rotor resistance equals the slip times the standstill rotor reactance, so s_m = R2/X2 = 0.2/1.0 = 0.2. At this slip the rotor power factor angle gives the largest air-gap power transfer per ampere. The value is independent of supply voltage.

Question 46Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a transformer, the open-circuit (no-load) test is performed with the rated voltage applied to one winding while the other winding is left open. The principal quantity determined from this test is:

  1. The full-load copper loss and the equivalent series impedance
  2. The core loss and the magnetising branch parameters Correct answer.
  3. The leakage reactance referred to the short-circuit side
  4. The winding resistance ratio between primary and secondary

Explanation

At rated voltage with the secondary open, the current drawn is only the small exciting current, so the input power measured is essentially the core (iron) loss, and the voltage, current and power readings yield the shunt branch R0 and X0. Copper loss during this test is negligible because the current is a few percent of rated value.

Question 47Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 50 Hz, 4-pole squirrel-cage induction motor has a full-load slip of 4%. The frequency of the rotor-induced emf at full load is:

  1. 0.5 Hz
  2. 2 Hz Correct answer.
  3. 4 Hz
  4. 50 Hz

Explanation

The rotor frequency equals the slip times the stator supply frequency, fr = s·f = 0.04 × 50 = 2 Hz. This follows because the relative speed between the rotating field and the rotor determines the rate at which rotor conductors cut the flux.

Question 48Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

In a DC machine, the function of the compensating winding placed in slots on the pole faces is to:

  1. Neutralise the armature mmf in the interpolar region and reduce flux distortion under load Correct answer.
  2. Increase the main field flux so that a higher torque is produced per ampere
  3. Supply the field winding during regenerative braking
  4. Provide a path for the eddy currents induced in the pole shoes

Explanation

Compensating windings carry armature current in the opposite sense to the adjacent armature conductors, so their mmf cancels the cross-magnetising armature mmf in the interpolar zone. This keeps the air-gap flux distribution nearly undistorted under heavy load and suppresses sparking at the brushes.

Question 49Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A 3-phase, 400 V, 50 Hz, 6-pole induction motor develops a gross mechanical power of 15 kW at a slip of 5%. The air-gap power transferred across the air gap is approximately:

  1. 17.65 kW
  2. 16.67 kW
  3. 14.25 kW
  4. 15.79 kW Correct answer.

Explanation

The air-gap power equals the rotor copper loss plus the mechanical power, and mechanical power equals (1 − s) times the air-gap power. Hence Pg = 15/0.95 = 15.79 kW, and the rotor copper loss is the remaining 0.79 kW.

Question 50Electrical Machines (Transformers, DC Machines, Induction and Synchronous Machines)

A synchronous generator is to be paralleled with an infinite bus. Which condition must be satisfied before the breaker is closed to avoid a large circulating current at the instant of synchronising?

  1. The generator must be over-excited so that it supplies leading reactive power
  2. The generated voltage must match the bus voltage in magnitude, phase and frequency with the same phase sequence Correct answer.
  3. The prime mover must be delivering full rated mechanical power to the shaft
  4. The field current must be reduced to zero so that the generated emf is zero

Explanation

Synchronising requires equal terminal voltage magnitude, equal frequency, matching phase angle at the instant of closing, and identical phase sequence. Any mismatch produces a resultant voltage across the closed switch and a large circulating current, together with a severe torque shock to the shaft.