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Vidyut Sahayak (Pareshan) Contractual study materials
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Vidyut Sahayak (Pareshan) Contractual practice questions

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Question 1Electrical Circuits and Network Theory

A 20 Ω resistor in a DC circuit carries a steady current of 0.5 A. What is the power dissipated in the resistor?

  1. 10 W
  2. 20 W
  3. 2.5 W
  4. 5 W Correct answer.

Explanation

Power in a resistor is P = I²R. Substituting 0.5 A and 20 Ω gives 0.25 × 20 = 5 W. The voltage across the resistor is 10 V, and P = VI also gives 5 W.

Question 2Electrical Circuits and Network Theory

Two resistors of 10 Ω and 20 Ω are connected in series across a 30 V DC supply. What is the voltage across the 20 Ω resistor?

  1. 15 V
  2. 20 V Correct answer.
  3. 10 V
  4. 30 V

Explanation

In series the same current flows through both resistors. The total resistance is 30 Ω, so the current is 30 V / 30 Ω = 1 A; the voltage across 20 Ω is 1 A × 20 Ω = 20 V. Voltage division gives 30 × 20/30 = 20 V.

Question 3Electrical Circuits and Network Theory

At a node in a DC network, two currents of 5 A and 3 A enter the node. One outgoing branch carries 6 A. What current must flow in the other outgoing branch?

  1. 14 A
  2. 4 A
  3. 8 A
  4. 2 A Correct answer.

Explanation

Kirchhoff's current law requires the sum of currents entering a node to equal the sum leaving it. The entering current is 5 + 3 = 8 A; with 6 A already leaving, the remaining outgoing current is 8 − 6 = 2 A.

Question 4Electrical Circuits and Network Theory

A 24 V source is connected in series with a 6 Ω resistor, and a 12 Ω resistor is connected across the output terminals. For the circuit seen by a load connected across the 12 Ω resistor, what are the Thevenin voltage and Thevenin resistance?

  1. 16 V, 4 Ω Correct answer.
  2. 16 V, 12 Ω
  3. 12 V, 4 Ω
  4. 24 V, 6 Ω

Explanation

With the load removed, the 6 Ω and 12 Ω resistors form a voltage divider, so Vth = 24 × 12/(6 + 12) = 16 V. Turning off the source shorts it, leaving 6 Ω in parallel with 12 Ω, so Rth = (6 × 12)/(6 + 12) = 4 Ω.

Question 5Electrical Circuits and Network Theory

In a linear network, a 2 A current source alone produces 3 A through a branch, and a 12 V voltage source alone produces 1.5 A through the same branch in the opposite direction. Using superposition, what is the net current in the branch?

  1. 0 A
  2. 1.5 A Correct answer.
  3. 3 A
  4. 4.5 A

Explanation

Superposition adds the branch responses caused by each independent source separately, including their signs. Taking the 3 A response as positive, the opposing 1.5 A response is negative, so the net current is 3 − 1.5 = 1.5 A.

Question 6Electrical Circuits and Network Theory

A 24 V DC source has an internal resistance of 6 Ω and feeds a variable load resistor. What load resistance gives maximum power transfer, and what is that maximum power?

  1. 6 Ω, 48 W
  2. 6 Ω, 24 W Correct answer.
  3. 12 Ω, 12 W
  4. 3 Ω, 24 W

Explanation

Maximum power transfer occurs when the load resistance equals the source resistance, so RL = 6 Ω. The maximum power is V²/(4R) = 24²/(4 × 6) = 576/24 = 24 W.

Question 7Electrical Circuits and Network Theory

An RC circuit has a 10 kΩ resistor and a 100 µF capacitor. What is the time constant of the circuit?

  1. 0.01 s
  2. 0.1 s
  3. 1 s Correct answer.
  4. 10 s

Explanation

The time constant of an RC circuit is τ = RC. Using 10,000 Ω and 100 × 10⁻⁶ F gives τ = 10⁴ × 10⁻⁴ = 1 s, the time to reach about 63.2% of the final capacitor voltage.

Question 8Electrical Circuits and Network Theory

A series RLC circuit has L = 0.1 H and C = 10 µF. At what frequency will the circuit be in series resonance?

  1. 159 Hz Correct answer.
  2. 1000 Hz
  3. 50 Hz
  4. 1590 Hz

Explanation

Series resonance occurs when XL = XC, giving f = 1/(2π√(LC)). Here LC = 0.1 × 10 × 10⁻⁶ = 10⁻⁶, so √(LC) = 10⁻³ and f = 1/(2π × 10⁻³) ≈ 159 Hz.

Question 9Electrical Circuits and Network Theory

A two-port network is described by z-parameters z11 = 10 Ω, z12 = 5 Ω, z21 = 5 Ω and z22 = 20 Ω. What is the input impedance at port 1 when port 2 is short-circuited?

  1. 6.25 Ω
  2. 15 Ω
  3. 8.75 Ω Correct answer.
  4. 10 Ω

Explanation

With port 2 short-circuited, V2 = 0, so the z-parameter equations give I2 = −z21 I1/z22. Substituting into V1 = z11 I1 + z12 I2 yields Zin = z11 − z12 z21/z22 = 10 − 25/20 = 8.75 Ω.

Question 10Electrical Circuits and Network Theory

Three 30 Ω resistors are connected in delta. What is the equivalent resistance measured between any two of the three terminals?

  1. 10 Ω
  2. 20 Ω Correct answer.
  3. 45 Ω
  4. 30 Ω

Explanation

Between two terminals of a delta, one 30 Ω resistor is in parallel with the series combination of the other two, giving 30 + 30 = 60 Ω in the other path. The equivalent resistance is 30 || 60 = (30 × 60)/(30 + 60) = 20 Ω.

Question 11Electrical Circuits and Network Theory

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 parallel resistors, 1/Req = 1/6 + 1/3 = 1/2, so Req = 2 Ω. The equivalent resistance of a parallel combination is always less than the smallest branch resistance, which here is 3 Ω.

Question 12Electrical Circuits and Network Theory

A 100 µF capacitor is charged to a voltage of 50 V. How much energy is stored in the capacitor?

  1. 0.125 J Correct answer.
  2. 0.25 J
  3. 2.5 J
  4. 0.0125 J

Explanation

The stored energy in a capacitor is W = ½ C V². Substituting C = 100 × 10⁻⁶ F and V = 50 V gives W = 0.5 × 100 × 10⁻⁶ × 2500 = 0.125 J. The factor ½ is essential for capacitor energy.

Question 13Electrical Circuits and Network Theory

The current through a 2 H inductor is increasing at a steady rate of 3 A/s. What is the magnitude of the voltage induced across the inductor?

  1. 1.5 V
  2. 6 V Correct answer.
  3. 12 V
  4. 0.67 V

Explanation

The inductor voltage is v = L di/dt. With L = 2 H and di/dt = 3 A/s, v = 2 × 3 = 6 V. The induced voltage opposes the change in current according to Lenz's law.

Question 14Electrical Circuits and Network Theory

A 12 V ideal DC voltage source is in series with a 4 Ω resistor. Viewed from its output terminals, what are the Norton equivalent current and resistance?

  1. 3 A in parallel with 12 Ω
  2. 0.33 A in parallel with 4 Ω
  3. 3 A in parallel with 4 Ω Correct answer.
  4. 12 A in parallel with 4 Ω

Explanation

The Norton current is the short-circuit current, I_N = V/R = 12/4 = 3 A. The Norton resistance is found by turning off the voltage source, which leaves the 4 Ω resistor as the equivalent resistance.

Question 15Electrical Circuits and Network Theory

A 12 V DC source feeds a 2 Ω resistor in series with a parallel pair of 3 Ω and 6 Ω resistors. What is the voltage across the parallel pair?

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

Explanation

The parallel pair has equivalent resistance 3 Ω || 6 Ω = (3 × 6)/(3 + 6) = 2 Ω. The total series resistance is 2 + 2 = 4 Ω, so the source current is 12/4 = 3 A. The voltage across the parallel pair is then 3 A × 2 Ω = 6 V.

Question 16Electrical Circuits and Network Theory

A series RL circuit has a resistance of 2 kΩ and an inductance of 4 H. What is the time constant of the circuit?

  1. 2 µs
  2. 0.5 ms
  3. 2 ms Correct answer.
  4. 8 s

Explanation

For a series RL circuit, the time constant is τ = L/R. Substituting L = 4 H and R = 2000 Ω gives τ = 4/2000 = 0.002 s, which is 2 ms.

Question 17Electrical Circuits and Network Theory

A single-phase load connected to a 230 V, 50 Hz supply draws a current of 10 A and consumes 2 kW. What is the magnitude of its power factor?

  1. 1.00
  2. 0.50
  3. 0.75
  4. 0.87 Correct answer.

Explanation

The power factor magnitude is P/(V I) = 2000/(230 × 10) = 0.8696, which rounds to 0.87. Power factor is the cosine of the phase angle between voltage and current.

Question 18Electrical Circuits and Network Theory

A series RLC circuit has R = 10 Ω, L = 100 mH and C = 1 µF. What is the quality factor at resonance?

  1. 100
  2. 31.6 Correct answer.
  3. 10
  4. 3.16

Explanation

At resonance the quality factor of a series RLC circuit is Q = (1/R)√(L/C). Here Q = (1/10)√(0.1/1 × 10⁻⁶) = (1/10)√100000 ≈ 31.6. This Q compares stored reactive energy with energy lost per cycle.

Question 19Electrical Circuits and Network Theory

An infinite ladder network is built from series resistors of 1 Ω and shunt resistors of 2 Ω, repeated indefinitely. What is the input resistance measured at the first pair of terminals?

  1. 4 Ω
  2. 1 Ω
  3. 2 Ω Correct answer.
  4. 3 Ω

Explanation

Let R be the input resistance. Because the ladder is infinite, adding one section does not change R, so R = 1 + (2 || R). Solving R = 1 + 2R/(2 + R) gives R² − R − 2 = 0, whose positive root is R = 2 Ω.

Question 20Electrical Circuits and Network Theory

A series RLC circuit has R = 2 Ω, L = 1 H and C = 0.01 F. What is the damping ratio and the nature of the response?

  1. ζ = 0.1, overdamped
  2. ζ = 10, underdamped
  3. ζ = 1, critically damped
  4. ζ = 0.1, underdamped Correct answer.

Explanation

For a series RLC circuit, α = R/(2L) = 2/(2 × 1) = 1 s⁻¹ and ω₀ = 1/√(LC) = 1/√(0.01) = 10 rad/s. The damping ratio is ζ = α/ω₀ = 1/10 = 0.1. Since ζ < 1, the circuit is underdamped.

Question 21Electrical Circuits and Network Theory

In a DC network, the reciprocal of resistance is known as which quantity?

  1. Reactance
  2. Conductance Correct answer.
  3. Capacitance
  4. Inductance

Explanation

Conductance is defined as the reciprocal of resistance, so G = 1/R, and it is measured in siemens. Capacitance and inductance store energy rather than directly describe conduction, while reactance is the opposition offered by reactive elements to alternating current.

Question 22Electrical Circuits and Network Theory

A connected network used in nodal analysis has six nodes, one of which is chosen as the reference node. How many independent node-voltage equations are required?

  1. Four equations
  2. Six equations
  3. Seven equations
  4. Five equations Correct answer.

Explanation

For a connected network with N nodes, one node is taken as the reference, leaving N - 1 independent node-voltage equations. With six nodes, the number required is therefore five.

Question 23Electrical Circuits and Network Theory

A sinusoidal voltage has a peak value of 100 V. What is its RMS value for a pure sine wave?

  1. 70.7 V Correct answer.
  2. 141.4 V
  3. 50.0 V
  4. 100 V

Explanation

For a pure sine wave, the RMS value is the peak value divided by the square root of two. Thus Vrms = 100 / 1.414 = 70.7 V, which is the value that produces the same heating effect as a DC voltage of that magnitude.

Question 24Electrical Circuits and Network Theory

A 30 V ideal DC source feeds a 10 Ω series resistor connected to a 40 Ω resistor, and a load is connected across the 40 Ω resistor. What are the Thevenin voltage and Thevenin resistance seen by the load?

  1. 24 V and 50 Ω
  2. 12 V and 8 Ω
  3. 24 V and 8 Ω Correct answer.
  4. 30 V and 10 Ω

Explanation

The open-circuit voltage across the 40 Ω resistor is found by voltage division: 30 x 40 / (10 + 40) = 24 V. With the source replaced by a short, the 10 Ω and 40 Ω resistors are in parallel, giving RTh = 10 || 40 = 8 Ω.

Question 25Electrical Circuits and Network Theory

A Wheatstone bridge has adjacent ratio arms of 120 Ω and 80 Ω, and a third arm of 240 Ω. What value of the fourth arm balances the bridge?

  1. 240 Ω
  2. 160 Ω Correct answer.
  3. 80 Ω
  4. 360 Ω

Explanation

The bridge balance condition is R1 / R2 = R3 / R4 for the chosen adjacent arms. Rearranging gives R4 = R2 R3 / R1 = 80 x 240 / 120 = 160 Ω, so the bridge has no current through the detector at balance.

Question 26Electrical Circuits and Network Theory

A coil has a resistance of 8 Ω and an inductive reactance of 6 Ω at 50 Hz. What are its impedance magnitude and phase angle by which current lags voltage?

  1. 14 Ω and 36.9° lagging
  2. 10 Ω and 36.9° lagging Correct answer.
  3. 10 Ω and 53.1° lagging
  4. 8 Ω and 6.0° lagging

Explanation

The impedance magnitude is Z = sqrt(R^2 + X^2) = sqrt(8^2 + 6^2) = 10 Ω. The phase angle is arctan(X/R) = arctan(6/8) = 36.9°, and current lags voltage in an inductive coil.

Question 27Electrical Circuits and Network Theory

A 50 µF capacitor is connected across a 230 V, 50 Hz supply. What are the capacitive reactance and the RMS current?

  1. 63.7 Ω and 3.61 A Correct answer.
  2. 31.8 Ω and 7.23 A
  3. 159 Ω and 1.45 A
  4. 63.7 Ω and 2.30 A

Explanation

Capacitive reactance is Xc = 1 / (2πfC) = 1 / (2π x 50 x 50 x 10^-6) = 63.7 Ω. The RMS current is I = V / Xc = 230 / 63.7 = 3.61 A.

Question 28Electrical Circuits and Network Theory

A two-port network has h21 = 50. With the output port short-circuited, an input current of 2 mA is applied. What is the short-circuit output current magnitude?

  1. 25 mA
  2. 2 mA
  3. 100 mA Correct answer.
  4. 0.04 mA

Explanation

The hybrid parameter h21 is the forward current transfer ratio measured with the output short-circuited: h21 = I2 / I1. Therefore I2 = 50 x 2 mA = 100 mA, with the direction given by the defined port current convention.

Question 29Electrical Circuits and Network Theory

A series RLC circuit has R = 5 Ω, L = 20 mH and C = 50 µF. What are the resonant frequency and the bandwidth?

  1. 159 Hz and 39.8 Hz Correct answer.
  2. 318 Hz and 39.8 Hz
  3. 159 Hz and 79.6 Hz
  4. 159 Hz and 5.0 Hz

Explanation

The resonant frequency is f0 = 1 / (2π sqrt(LC)) = 1 / (2π sqrt(20 x 10^-3 x 50 x 10^-6)) = 159 Hz. For a series RLC circuit, bandwidth is BW = R / (2πL) = 5 / (2π x 0.02) = 39.8 Hz.

Question 30Electrical Circuits and Network Theory

A two-port network has z11 = 12 Ω, z12 = 4 Ω, z21 = 4 Ω and z22 = 8 Ω. What is the input impedance when the output port is short-circuited?

  1. 2 Ω
  2. 10 Ω Correct answer.
  3. 12 Ω
  4. 8 Ω

Explanation

With the output short-circuited, V2 = 0, and the z-parameter equations give Zin = z11 - z12 z21 / z22. Substituting the values gives 12 - (4 x 4) / 8 = 12 - 2 = 10 Ω.

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

The RMS value of the EMF induced in a transformer winding having N series turns and a peak core flux of Φmax, when supplied at frequency f, is given by:

  1. 1.11 f N Φmax
  2. 4.00 f N Φmax
  3. 2.22 f N Φmax
  4. 4.44 f N Φmax Correct answer.

Explanation

The induced EMF is E = 4.44 f N Φmax, where the constant 4.44 equals 2π/√2 for a sinusoidal flux wave. The factor comes from the form factor of a sine wave (1.11) multiplied by the average EMF per turn, 4 f Φmax.

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

In a DC machine, the primary function of the commutator and brushes is that:

  1. It converts the alternating EMF induced in the armature into a unidirectional voltage. Correct answer.
  2. It provides mechanical support and rigidity to the field poles and the yoke.
  3. It keeps the brush pressure on the commutator constant at all speeds and loads.
  4. It limits the eddy-current loss in the laminated armature core to a very small value.

Explanation

The commutator with its brushes acts as a mechanical rectifier: it reverses the connection of each armature coil as the coil crosses the brush axis, so the voltage appearing at the terminals is unidirectional. In motoring duty the same action reverses the coil currents to keep the developed torque unidirectional.

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

A three-phase, 4-pole, 50 Hz induction motor runs at 1440 rpm at full load. Its per-unit slip is:

  1. 0.06
  2. 0.05
  3. 0.02
  4. 0.04 Correct answer.

Explanation

Synchronous speed is ns = 120f/P = 120 × 50/4 = 1500 rpm. Slip is s = (ns − n)/ns = (1500 − 1440)/1500 = 0.04, that is 4 per cent.

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

In a loaded DC generator, the effect of armature reaction is that:

  1. It shifts the magnetic neutral axis along rotation and lowers the generated EMF. Correct answer.
  2. It reverses the residual magnetism of the field poles when the load is removed.
  3. It removes the need for interpoles or compensating windings in the machine.
  4. It strengthens the main field flux and raises the generated EMF above the no-load value.

Explanation

Armature reaction is the magnetising effect of the armature MMF on the main field. In a generator it distorts the flux so that the magnetic neutral axis moves in the direction of rotation, and the cross-magnetising and demagnetising components slightly reduce the generated EMF. Interpoles and compensating windings are used to neutralise this effect, not to make them unnecessary.

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

A 220 V DC shunt motor drives a constant-torque load at 1000 rpm. Resistance is now inserted in the field circuit while the supply voltage and the load torque stay constant. The motor speed will:

  1. decrease
  2. remain unchanged
  3. increase Correct answer.
  4. oscillate about 1000 rpm

Explanation

Adding field resistance reduces the field current and therefore the flux. Since torque T = kΦIa must stay constant, the armature current rises, and from N ∝ Eb/Φ with Eb = V − IaRa the speed increases. If the field is weakened too far a shunt motor may run away, which is why field circuits are never opened on load.

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

External resistance is added to the rotor circuit of a three-phase slip-ring induction motor. Which statement about its torque–slip characteristic is correct?

  1. Maximum torque is unchanged, but it occurs at a higher slip and starting torque rises. Correct answer.
  2. The maximum torque falls steadily while the slip at maximum torque remains constant.
  3. The maximum torque increases in direct proportion to the value of the added rotor resistance.
  4. Both the maximum torque and the slip at which it occurs decrease as external resistance is added.

Explanation

The breakdown torque of an induction motor is set by the standstill rotor reactance and the supply voltage, not by rotor resistance, so its magnitude is unchanged. The slip at which it occurs equals R2/X2, so extra resistance pushes the peak towards higher slip and therefore raises the starting torque; the resistance is cut out progressively as the motor accelerates.

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

A three-phase alternator delivers its rated current to a load at 0.8 power factor lagging. Compared with unity power factor at the same current, its voltage regulation is:

  1. larger and positive Correct answer.
  2. smaller and may become negative
  3. zero at all lagging power factors
  4. the same at all power factors

Explanation

Voltage regulation becomes larger and positive as the load power factor moves towards lagging, because the armature reaction of a lagging current is demagnetising and the synchronous reactance drop adds to the internal EMF. At a sufficiently leading power factor the regulation can fall to zero or become negative.

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

Two three-phase distribution transformers are to be operated in parallel. Which condition must be met so that no circulating current flows when the load is disconnected?

  1. Identical percentage impedance and identical X/R ratio for the two units
  2. The same winding material, copper or aluminium, in both of the units
  3. Identical kVA ratings and identical cooling classes for both transformers
  4. The same voltage ratio and matching polarity and phase sequence Correct answer.

Explanation

For parallel operation without circulating current, the voltage ratios must be equal and the polarities (and phase sequence, for three-phase units) must match, so that no resultant EMF appears across the open paralleling switch. Mismatched percentage impedance does not cause a no-load circulating current; it causes unequal load sharing, and kVA ratings may differ.

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

A double-cage squirrel-cage induction motor develops a high starting torque mainly because:

  1. The inner cage is made of high-resistance alloy and dominates the rotor current at all speeds.
  2. Both cages have the same resistance and reactance, so the current divides equally between them.
  3. The rotor leakage reactance is deliberately made very large both at standstill and at running speed.
  4. The outer cage has high resistance and low reactance, so it carries most current at standstill. Correct answer.

Explanation

At standstill the rotor frequency equals the supply frequency, so the leakage reactance of the inner, deeply embedded cage is high and forces most of the current into the outer cage, which has high resistance and is closely coupled to the air gap. High rotor resistance at start gives high starting torque, while at running slip the reactance falls and the low-resistance inner cage takes over, giving good running efficiency.

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

Damper windings fitted in the pole faces of a synchronous machine are intended to:

  1. Increase the excitation voltage that the shaft-mounted exciter can deliver to the field.
  2. Improve the voltage regulation by reducing the armature leakage reactance and armature reaction.
  3. Damp rotor oscillations (hunting) by developing a torque that opposes changes in speed. Correct answer.
  4. Allow the machine to run stably at any speed below synchronous speed under load.

Explanation

Damper windings are short-circuited copper or brass bars in the pole shoes that behave like a squirrel cage. When the rotor speed departs from synchronous speed they carry induced currents whose torque opposes the oscillation, so hunting is suppressed; the same bars supply the starting torque of a synchronous motor.

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

In a power transformer, the core is built from thin laminated steel sheets primarily to reduce which loss?

  1. Copper loss
  2. Stray load loss
  3. Hysteresis loss
  4. Eddy-current loss Correct answer.

Explanation

Laminations interrupt the eddy-current paths in the core, increasing their effective resistance and thereby reducing eddy-current loss. Hysteresis loss is controlled mainly by the core material, such as grain-oriented silicon steel, not by lamination thickness.

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

A DC series motor must never be started without a mechanical load because, as the load approaches zero, the:

  1. flux falls to a low value and speed becomes dangerously high Correct answer.
  2. armature current reverses and the motor runs as a generator
  3. commutator voltage rises above the supply and blocks the armature current
  4. field current falls and the motor stalls before reaching rated speed

Explanation

In a series motor the field flux is produced by the armature current. As the mechanical load falls, armature current and flux fall, and the speed rises sharply; at no load the speed can reach a destructive value.

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

The no-load current drawn by a transformer from the supply consists mainly of:

  1. magnetizing current and core-loss current Correct answer.
  2. circulating current between primary and secondary windings
  3. winding resistance drop current only
  4. full-load copper-loss current and secondary load current

Explanation

At no load the secondary is open, so the primary current is the exciting current. It has a magnetizing component that sets up the mutual flux and a core-loss component that supplies hysteresis and eddy-current losses.

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

A 100 kVA distribution transformer has an iron loss of 1 kW and a full-load copper loss of 2 kW. It supplies 50 kVA at 0.8 power factor for 12 hours and carries no load for the remaining 12 hours. Its all-day efficiency is closest to:

  1. 92.3%
  2. 97.6%
  3. 94.1% Correct answer.
  4. 96.0%

Explanation

Output energy is 50 kVA × 0.8 × 12 h = 480 kWh. Iron loss is 1 kW × 24 h = 24 kWh, and copper loss at half load is 2 kW × (0.5)² × 12 h = 6 kWh. All-day efficiency is therefore 480/(480 + 24 + 6) = 94.1%.

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

An off-circuit tap changer on the HV winding of a transformer is moved from the nominal tap to a tap having 5% more HV turns. If rated voltage is applied to the HV winding, the no-load secondary voltage will be:

  1. 95.2% of nominal Correct answer.
  2. 105% of nominal
  3. 100% of nominal
  4. 90.5% of nominal

Explanation

The secondary voltage is proportional to the turns ratio Ns/Np. Increasing the HV turns by 5% reduces the volts per turn and gives 1/1.05 = 0.952 of the original secondary voltage.

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

A self-excited DC shunt generator is driven at rated speed but fails to build up its terminal voltage. If the machine has normal residual magnetism, the most probable cause is that:

  1. the brushes are seated correctly on a clean commutator
  2. the external load resistance is higher than rated
  3. field resistance is much lower than the normal value
  4. field resistance exceeds the critical field resistance Correct answer.

Explanation

A shunt generator builds up only if the field resistance line intersects the magnetization curve. If field resistance is above the critical value, the generated voltage cannot rise enough to supply its own field current.

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

A three-phase squirrel-cage induction motor develops a locked-rotor torque of 150% of rated torque at rated voltage. If it is started at 80% of rated voltage, the locked-rotor torque becomes:

  1. 120% of rated torque
  2. 64% of rated torque
  3. 96% of rated torque Correct answer.
  4. 150% of rated torque

Explanation

Induction-motor torque at a given slip is proportional to the square of the applied voltage. Therefore locked-rotor torque becomes 150% × (0.8)² = 96% of rated torque.

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

A synchronous motor is connected to an infinite bus and drives a constant mechanical load. If its field excitation is increased from the value giving unity power factor, the armature current will:

  1. increase and the power factor becomes leading Correct answer.
  2. remain unchanged and the power factor becomes unity
  3. decrease and the power factor becomes lagging
  4. decrease and the power factor becomes leading

Explanation

At constant load the power input is fixed, so increasing excitation beyond unity power factor makes the motor draw a leading current. The armature current rises again after passing its minimum at unity power factor.

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

When a separately excited DC motor is plugging-braked, the armature terminals are reversed while the field is left unchanged. The back emf now aids the supply, so excessive armature current is avoided only if:

  1. a capacitor is connected across the armature terminals
  2. the field current is reduced to zero during braking
  3. series resistance is added in the armature circuit Correct answer.
  4. the armature terminals are short-circuited through a resistor

Explanation

In plugging, reversing the armature terminals makes the back emf aid the supply voltage, so the net armature voltage is almost twice normal. Series resistance is inserted to limit the resulting braking current.

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

A transformer has a percentage impedance of 5% on its own base. If its LV terminals are short-circuited and rated voltage is applied to the HV side, the steady-state short-circuit current on the LV side will be approximately:

  1. 10 times rated current
  2. 50 times rated current
  3. 5 times rated current
  4. 20 times rated current Correct answer.

Explanation

Percentage impedance is the per-unit voltage that causes rated current to flow in a short circuit. With 5% impedance, the short-circuit current is 1/0.05 = 20 times the rated current.