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ESE Prelims Electrical practice questions

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Question 1Electrical Circuits and Network Theory

A 10 Ω resistor carries a current given by i(t) = 2 + 3sin(100t) A. What is the average power dissipated in the resistor?

  1. 130 W
  2. 40 W
  3. 85 W
  4. 65 W Correct answer.

Explanation

Average power for a periodic current equals R times the mean square of the current. The DC term contributes 2² = 4, and the sinusoidal term contributes 3²/2 = 4.5, giving a mean square of 8.5 A². Multiplying by 10 Ω yields 85 W.

Question 2Electrical Circuits and Network Theory

In a series RLC circuit at resonance, which statement about the voltages is correct?

  1. The voltage across the inductor equals the voltage across the capacitor in magnitude and they are 180° out of phase. Correct answer.
  2. The inductor and capacitor voltages are in phase and add to the supply voltage.
  3. The supply voltage equals the arithmetic sum of the three element voltages.
  4. The voltage across the resistor is zero because the impedance is minimum.

Explanation

At series resonance the inductive and capacitive reactances are equal, so their voltage magnitudes are equal while their phasors oppose each other. They cancel in the phasor sum, leaving the supply voltage entirely across the resistance. This is why element voltages can individually exceed the source voltage.

Question 3Electrical Circuits and Network Theory

The Thevenin equivalent resistance of a two-terminal network is found by

  1. open-circuiting all independent sources and measuring the short-circuit current
  2. short-circuiting all independent voltage sources and open-circuiting all independent current sources, then looking into the terminals Correct answer.
  3. setting all dependent sources to zero and computing the open-circuit voltage
  4. measuring the open-circuit voltage and dividing by the total resistance

Explanation

Thevenin resistance is the resistance seen at the terminals with every independent source deactivated: voltage sources become short circuits and current sources become open circuits. Dependent sources must be retained because they are controlled by circuit variables, and an external test source is used if they are present.

Question 4Electrical Circuits and Network Theory

A 100 V source feeds a load through a line of resistance 2 Ω. Maximum power is transferred to the load when the load resistance is

  1. 2 Ω Correct answer.
  2. 100 Ω
  3. 50 Ω
  4. 0 Ω

Explanation

The maximum power transfer theorem states that a load receives maximum power when its resistance equals the Thevenin resistance of the source network. Here the source resistance is the 2 Ω line resistance, so the matched load is 2 Ω. At this point the efficiency is only 50 percent, which is a key practical limitation.

Question 5Electrical Circuits and Network Theory

For a two-port network, the h-parameters relate which pair of variables?

  1. V1 and I2 as dependent variables in terms of I1 and V2 Correct answer.
  2. V1 and I1 in terms of V2 and I2
  3. I1 and I2 in terms of V1 and V2
  4. V1 and V2 in terms of I1 and I2

Explanation

Hybrid parameters express the input voltage and output current as functions of the input current and output voltage: V1 = h11·I1 + h12·V2 and I2 = h21·I1 + h22·V2. This mixed choice makes them convenient for transistor modelling. The other parameter sets use either both currents or both voltages as independent variables.

Question 6Electrical Circuits and Network Theory

A first-order RC low-pass filter has R = 1 kΩ and C = 1 μF. At what frequency does the output magnitude fall to 1/√2 of its low-frequency value?

  1. 1.59 kHz
  2. 15.9 Hz
  3. 159.2 Hz Correct answer.
  4. 1 kHz

Explanation

The cutoff frequency of a first-order RC filter is f = 1/(2πRC). With R = 1000 Ω and C = 1e-6 F, RC = 1 ms, so f = 1/(2π × 0.001) ≈ 159.2 Hz. At this frequency the output is 0.707 of the passband value, a 3 dB drop.

Question 7Electrical Circuits and Network Theory

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

  1. 231 V Correct answer.
  2. 400 V
  3. 692 V
  4. 133 V

Explanation

In a star connection the line voltage is √3 times the phase voltage, so the phase voltage is 400/√3 ≈ 231 V. Each phase winding sees this lower voltage, which is why star connection suits higher line voltages. In delta connection the line and phase voltages are equal.

Question 8Electrical Circuits and Network Theory

A network has the admittance Y(s) = (s + 2)/((s + 1)(s + 3)). Which statement about its pole-zero plot is correct?

  1. One zero at s = 2 and two poles at s = 1 and s = 3
  2. One zero at s = -2 and two poles at s = -1 and s = -3 Correct answer.
  3. Two zeros at s = -1 and s = -3 and one pole at s = -2
  4. One pole at s = -2 and two zeros at s = -1 and s = -3

Explanation

Zeros are the roots of the numerator and poles are the roots of the denominator. Setting s + 2 = 0 gives a zero at s = -2, and setting (s + 1)(s + 3) = 0 gives poles at s = -1 and s = -3. All lie in the left half-plane, consistent with a stable network.

Question 9Electrical Circuits and Network Theory

The Laplace transform of the unit step function u(t) is

  1. 1/(s + 1)
  2. 1/s²
  3. s
  4. 1/s Correct answer.

Explanation

The unit step is the integral of the impulse, and since the impulse transform is 1, the step transform is 1/s. This applies for the one-sided transform with region of convergence Re(s) > 0. The ramp function, by contrast, transforms to 1/s².

Question 10Electrical Circuits and Network Theory

A source-free series RC circuit is initially charged to V0 and allowed to discharge. The time constant and the voltage after one time constant are

  1. τ = RC, with voltage equal to V0 at all times
  2. τ = RC, with voltage equal to 0.368 V0 Correct answer.
  3. τ = 1/(RC), with voltage equal to 0.5 V0
  4. τ = R/C, with voltage equal to 0.632 V0

Explanation

For a discharging RC circuit the voltage decays as v(t) = V0·e^(-t/RC), so the time constant is τ = RC. After one time constant the voltage has fallen to e^(-1) ≈ 0.368 of its initial value. The 0.632 figure is the fraction of the final value reached during charging, not the remaining fraction during discharge.

Question 11Electrical Circuits and Network Theory

A current source of 5 A is connected across a 4 Ω resistor. What is the equivalent voltage source and series resistance?

  1. 1.25 V in series with 4 Ω
  2. 5 V in series with 4 Ω
  3. 20 V in parallel with 4 Ω
  4. 20 V in series with 4 Ω Correct answer.

Explanation

Source transformation converts a current source in parallel with a resistance into a voltage source in series with the same resistance. The voltage is the product of the current and resistance, 5 × 4 = 20 V, and the polarity is chosen so the terminal behaviour is unchanged. The resistance value remains 4 Ω.

Question 12Electrical Circuits and Network Theory

A coil has resistance 6 Ω and inductive reactance 8 Ω. Its impedance magnitude and phase angle are

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

Explanation

The impedance is R + jX = 6 + j8, so its magnitude is √(6² + 8²) = 10 Ω. The phase angle is arctan(8/6) ≈ 53.1°, with current lagging voltage. The 3-4-5 triangle pattern makes this a standard result.

Question 13Electrical Circuits and Network Theory

Which condition is necessary and sufficient for a two-port network to be reciprocal?

  1. z11 = z22
  2. h12 = h21
  3. z12 = z21 Correct answer.
  4. y11 = y22

Explanation

A network is reciprocal when the transfer impedance is the same in both directions, so z12 = z21. Equivalently, y12 = y21 and h12 = -h21 for reciprocal networks. Symmetry, by contrast, requires z11 = z22, which is a separate property.

Question 14Electrical Circuits and Network Theory

A series RLC circuit has R = 10 Ω, L = 1 mH and C = 1 μF. The resonant frequency and quality factor are approximately

  1. 5.03 kHz and 3.16 Correct answer.
  2. 503 Hz and 31.6
  3. 5.03 kHz and 31.6
  4. 1.59 kHz and 1.0

Explanation

Series resonance occurs at f = 1/(2π√(LC)). Here √(LC) = √(1e-9) ≈ 3.16e-5 s, giving f ≈ 5.03 kHz. The quality factor is Q = (1/R)√(L/C) = (1/10)√(1e-3/1e-6) = (1/10)(31.6) ≈ 3.16.

Question 15Electrical Circuits and Network Theory

Superposition in a linear circuit applies only when the circuit contains

  1. only reactive elements and no resistors
  2. at least one nonlinear element
  3. dependent sources and nonlinear elements
  4. only independent sources and linear elements Correct answer.

Explanation

Superposition relies on linearity, so it is valid only for circuits built from linear elements and independent sources. Each independent source is considered alone with the others deactivated, and the individual responses are added. Nonlinear devices such as diodes violate the principle.

Question 16Electrical Circuits and Network Theory

A 50 Hz sinusoidal voltage has an RMS value of 230 V. Its peak value and angular frequency are

  1. 325 V and 50 rad/s
  2. 230 V and 50 rad/s
  3. 163 V and 314 rad/s
  4. 325 V and 314 rad/s Correct answer.

Explanation

The peak value equals RMS multiplied by √2, so 230 × 1.414 ≈ 325 V. The angular frequency is ω = 2πf = 2π × 50 ≈ 314 rad/s. Confusing frequency in hertz with angular frequency in rad/s is a common error.

Question 17Electrical Circuits and Network Theory

A two-branch parallel circuit has branch currents of 6∠0° A and 8∠90° A. The total current magnitude is

  1. 2 A
  2. 48 A
  3. 10 A Correct answer.
  4. 14 A

Explanation

The branch currents are phasors 90° apart, so they add vectorially: 6 + j8. The magnitude is √(6² + 8²) = 10 A. Adding the magnitudes arithmetically would be valid only if the currents were exactly in phase.

Question 18Electrical Circuits and Network Theory

A linear time-invariant circuit has the transfer function H(s) = 1/(s² + 2s + 5). The natural response is

  1. an undamped oscillation with damping factor 2
  2. a damped sinusoid with damping factor 1 and damped frequency 2 rad/s Correct answer.
  3. a pure sinusoid with frequency 5 rad/s
  4. two real exponentials with time constants 1 s and 5 s

Explanation

The poles satisfy s² + 2s + 5 = 0, giving s = -1 ± j2. The real part -1 sets the exponential decay rate and the imaginary part 2 rad/s sets the damped oscillation frequency. A complex-conjugate pole pair therefore produces a decaying sinusoid.

Question 19Electrical Circuits and Network Theory

In nodal analysis, the number of independent node equations for a circuit with N nodes (including the reference node) is

  1. N
  2. N + 1
  3. 2N
  4. N - 1 Correct answer.

Explanation

One node is chosen as the reference, and Kirchhoff's current law is written at each of the remaining nodes. This gives N - 1 independent equations, matching the number of unknown node voltages. Writing an equation at the reference node adds no new information.

Question 20Electrical Circuits and Network Theory

A balanced delta-connected load of impedance 30∠30° Ω per phase is supplied from a 400 V three-phase line. The line current magnitude is

  1. 40 A
  2. 13.3 A
  3. 7.7 A
  4. 23.1 A Correct answer.

Explanation

In a delta load the phase voltage equals the line voltage, so the phase current is 400/30 ≈ 13.3 A. The line current is √3 times the phase current, giving 13.3 × 1.732 ≈ 23.1 A. The impedance angle affects power factor but not the current magnitude.

Question 21Electrical Circuits and Network Theory

A circuit has the impedance Z = 3 - j4 Ω at a given frequency. The power factor and its nature are

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

Explanation

The impedance magnitude is √(3² + 4²) = 5 Ω, so the power factor is R/|Z| = 3/5 = 0.6. The negative reactance indicates a capacitive circuit, so the current leads the voltage. Hence the power factor is 0.6 leading.

Question 22Electrical Circuits and Network Theory

For a series RC circuit driven by a step voltage, the capacitor voltage rises towards the source value. The initial rate of rise of capacitor voltage is

  1. V/R
  2. V/(RC) Correct answer.
  3. V·RC
  4. zero

Explanation

At the instant of switching the capacitor voltage is zero, so the full source voltage appears across R and the initial current is V/R. The capacitor voltage rises at dV_c/dt = i/C, giving an initial slope of V/(RC). The slope decreases exponentially as the capacitor charges.

Question 23Electrical Circuits and Network Theory

A network function has a pole exactly on the imaginary axis at s = jω0, with all other poles in the left half-plane. The circuit response to a bounded input is

  1. an exponentially decaying response
  2. an impulse that decays to zero immediately
  3. a sustained oscillation at frequency ω0 Correct answer.
  4. a constant DC offset with no oscillation

Explanation

A pole on the imaginary axis corresponds to a marginally stable system with an undamped natural mode at the pole frequency. With no damping term, the response oscillates at ω0 without decaying. All other poles in the left half-plane contribute only decaying terms.

Question 24Electrical Circuits and Network Theory

In a purely inductive AC circuit, the average real power consumed over a full cycle is

  1. zero Correct answer.
  2. equal to V·I/2
  3. equal to I²X_L
  4. equal to V·I

Explanation

In a pure inductor the current lags the voltage by exactly 90°, so the power factor is cos 90° = 0. Energy is stored in the magnetic field and returned to the source each half cycle, giving zero net real power. The quantity I²X_L represents reactive power in VAR, not average real power in watts.

Question 25Electrical Machines

In a DC shunt generator running at constant speed with no load, the terminal voltage builds up only if the field winding is connected so that its mmf aids the residual magnetism. What is this condition called?

  1. Self-excitation with correct polarity Correct answer.
  2. Separately excited operation
  3. Cumulative compounding
  4. Differential compounding

Explanation

Voltage build-up in a shunt generator depends on residual flux inducing a small emf that drives current in a direction that strengthens that flux. If the field terminals are reversed, the field mmf opposes the residual flux and the machine fails to build up. This process of using its own output to supply the field is self-excitation with correct polarity.

Question 26Electrical Machines

A 4-pole, 50 Hz, 3-phase induction motor runs at 1440 rpm. What is the slip?

  1. 0.10
  2. 0.96
  3. 0.06
  4. 0.04 Correct answer.

Explanation

Synchronous speed is 120f/P = 120×50/4 = 1500 rpm. Slip s = (Ns − N)/Ns = (1500 − 1440)/1500 = 60/1500 = 0.04, i.e. 4 percent.

Question 27Electrical Machines

The emf equation of a transformer gives E = 4.44 f N Φm. What does the factor 4.44 represent?

  1. The leakage flux coefficient
  2. The form factor of a sinusoidal flux wave multiplied by 2π/√2 Correct answer.
  3. The ratio of rms to peak current
  4. The winding factor for a distributed winding

Explanation

For a sinusoidal flux, the rms emf per turn is 2πfΦm/√2 = 4.44fΦm. The constant 4.44 combines the form factor 1.11 of a sine wave with the 4fΦm term from Faraday's law. Hence it is tied to the sinusoidal waveform of the flux.

Question 28Electrical Machines

A 3-phase, 6-pole, 50 Hz induction motor has a full-load slip of 3 percent. At what speed does the rotor mmf rotate with respect to the stator?

  1. 970 rpm
  2. 940 rpm
  3. 1000 rpm Correct answer.
  4. 1030 rpm

Explanation

Synchronous speed is 120×50/6 = 1000 rpm. The rotor mmf always rotates at synchronous speed relative to the stator regardless of slip, because rotor frequency adjusts to keep the air-gap field synchronous. The 970 rpm figure is the shaft speed, not the rotor field speed.

Question 29Electrical Machines

In a transformer, which quantity is the same on the primary and secondary sides of an ideal transformer?

  1. Current
  2. Apparent power per winding resistance
  3. Voltage per turn Correct answer.
  4. Impedance

Explanation

In an ideal transformer the same flux links each turn, so the induced emf per turn is identical on both windings. This is why the voltage ratio equals the turns ratio. Current and impedance transform inversely and directly with turns ratio respectively.

Question 30Electrical Machines

A separately excited DC motor has armature resistance 0.5 Ω and is supplied at 220 V. At a certain load the back emf is 210 V. What is the armature current?

  1. 420 A
  2. 20 A Correct answer.
  3. 10 A
  4. 5 A

Explanation

Armature current is (V − Eb)/Ra = (220 − 210)/0.5 = 10/0.5 = 20 A. Back emf opposes the supply, so only the difference drives current through the armature resistance.

Question 31Electrical Machines

Why is the core of a power transformer laminated?

  1. To increase the mechanical strength of the core
  2. To reduce eddy current loss Correct answer.
  3. To reduce hysteresis loss
  4. To increase the saturation flux density

Explanation

Laminations interrupt the paths available for eddy currents, increasing the effective resistance and reducing eddy current loss, which varies with the square of lamination thickness. Hysteresis loss depends on the material's B-H loop, not on lamination thickness.

Question 32Electrical Machines

A 3-phase induction motor develops maximum torque when the rotor resistance per phase equals which quantity?

  1. The stator resistance per phase
  2. The synchronous reactance
  3. The magnetizing reactance
  4. The standstill rotor reactance per phase Correct answer.

Explanation

Maximum torque occurs at the slip where rotor resistance equals rotor standstill reactance, giving s = R2/X2. Adding external rotor resistance shifts the maximum-torque slip toward higher values without changing the peak torque value.

Question 33Electrical Machines

A transformer has 500 primary turns and 100 secondary turns. If the primary is connected to a 230 V, 50 Hz supply, what is the approximate peak flux in the core?

  1. 1.04 mWb
  2. 0.46 mWb
  3. 2.07 mWb Correct answer.
  4. 4.14 mWb

Explanation

Using E = 4.44 f N Φm with the primary: Φm = 230/(4.44×50×500) = 230/111000 ≈ 2.07×10⁻³ Wb. The flux is set by the applied voltage, frequency, and primary turns only.

Question 34Electrical Machines

In a DC series motor, why must the load never be removed completely?

  1. The armature reaction would reverse the field
  2. The speed would rise to a dangerously high value Correct answer.
  3. The commutator would flash over at low current
  4. The field current would exceed its rating

Explanation

In a series motor the flux is proportional to armature current, so as load and current fall, flux falls too. Since speed is roughly proportional to Eb/Φ, a very small flux makes the speed rise to a destructive value. Hence a series motor must always be started with a mechanical load coupled.

Question 35Electrical Machines

The open-circuit test on a transformer is normally performed with the rated voltage applied to one winding and the other left open. What does this test primarily determine?

  1. Core losses and magnetizing branch parameters Correct answer.
  2. Leakage reactance and winding resistance
  3. Copper losses and equivalent resistance
  4. Turns ratio under load

Explanation

With the secondary open, only the small no-load current flows, so copper loss is negligible and the measured power is essentially core loss. The results yield the shunt branch of the equivalent circuit, namely the magnetizing reactance and core-loss resistance.

Question 36Electrical Machines

A 3-phase, 400 V, 50 Hz star-connected induction motor has a per-phase stator resistance of 0.5 Ω and draws a line current of 10 A at a power factor of 0.8. What is the total stator copper loss?

  1. 150 W Correct answer.
  2. 120 W
  3. 50 W
  4. 400 W

Explanation

In a star connection the phase current equals the line current, so per-phase copper loss is I²R = 10²×0.5 = 50 W. For three phases the total is 3×50 = 150 W. Power factor is irrelevant to copper loss.

Question 37Electrical Machines

A transformer has an equivalent resistance of 0.02 pu and an equivalent reactance of 0.06 pu. At full-load current with 0.8 power factor lagging, what is the approximate per-unit voltage regulation?

  1. 0.020 pu
  2. 0.060 pu
  3. 0.076 pu
  4. 0.052 pu Correct answer.

Explanation

Regulation ≈ I(R cosφ + X sinφ) = 1×(0.02×0.8 + 0.06×0.6) = 0.016 + 0.036 = 0.052 pu. The second-order term is small and is neglected in this approximation.

Question 38Electrical Machines

A 6-pole, 50 Hz, 3-phase induction motor has a rotor resistance of 0.2 Ω and standstill reactance of 0.8 Ω per phase. At what slip does maximum torque occur?

  1. 0.25 Correct answer.
  2. 0.20
  3. 0.80
  4. 4.00

Explanation

Maximum torque occurs at s = R2/X2 = 0.2/0.8 = 0.25. This is the slip at which the reflected rotor resistance equals the reflected rotor reactance in the equivalent circuit.

Question 39Electrical Machines

In a transformer, the no-load current is 2 A at a power factor of 0.2 lagging when the rated voltage is applied. What is the approximate magnetizing component of the no-load current?

  1. 1.96 A Correct answer.
  2. 2.00 A
  3. 0.40 A
  4. 1.60 A

Explanation

The core-loss component is I0 cosφ0 = 2×0.2 = 0.4 A. The magnetizing component is I0 sinφ0 = 2×sin(78.5°) ≈ 2×0.98 = 1.96 A. The magnetizing component dominates because the no-load current is nearly reactive.

Question 40Electrical Machines

A 4-pole DC lap-wound generator has 400 armature conductors and runs at 1000 rpm. If the flux per pole is 0.02 Wb, what is the generated emf?

  1. 533.3 V
  2. 133.3 V Correct answer.
  3. 66.7 V
  4. 266.7 V

Explanation

For a lap winding the number of parallel paths equals the number of poles, so A = P. Emf = ΦZNP/(60A) = 0.02×400×1000×4/(60×4) = 32000/240 ≈ 133.3 V. The pole number cancels in the lap case.

Question 41Electrical Machines

A transformer has a full-load copper loss of 400 W and a core loss of 200 W. At what fraction of full load is the efficiency maximum?

  1. 0.50
  2. 1.00
  3. 0.25
  4. 0.71 Correct answer.

Explanation

Efficiency is maximum when variable copper loss equals constant core loss, i.e. x²Pcu = Pcore. Thus x = √(200/400) = √0.5 ≈ 0.71 of full load.

Question 42Electrical Machines

In a 3-phase induction motor, the air-gap power is 10 kW and the rotor copper loss is 400 W. What is the mechanical power developed?

  1. 10.4 kW
  2. 9.2 kW
  3. 9.6 kW Correct answer.
  4. 0.4 kW

Explanation

The air-gap power divides into rotor copper loss and mechanical power, so mechanical power is 10 − 0.4 = 9.6 kW. This also equals (1 − s) times air-gap power, with slip here being 0.04.

Question 43Electrical Machines

A 3-phase, 50 Hz, 4-pole induction motor operates with a slip of 5 percent. What is the frequency of the rotor emf?

  1. 5 Hz
  2. 50 Hz
  3. 2.5 Hz Correct answer.
  4. 0.5 Hz

Explanation

Rotor frequency equals slip times supply frequency, fr = s·f = 0.05×50 = 2.5 Hz. At standstill slip is one and the rotor frequency equals the supply frequency.

Question 44Electrical Machines

A DC shunt motor has a constant field flux and is connected to a fixed supply. If the load torque is halved while field flux remains constant, what happens to the armature current?

  1. It doubles
  2. It halves Correct answer.
  3. It remains the same
  4. It becomes four times

Explanation

For a DC motor, torque is proportional to flux times armature current. With flux constant, torque is proportional to armature current, so halving the torque halves the current. The speed rises slightly to reduce the emf difference accordingly.

Question 45Electrical Machines

In a transformer, the short-circuit test is performed with reduced voltage applied to one winding while the other is shorted. Why is the applied voltage kept low?

  1. To measure core loss accurately
  2. To eliminate leakage flux
  3. To avoid saturating the core
  4. To limit the current to rated value Correct answer.

Explanation

With the secondary shorted, only the small leakage impedance limits current, so rated current would flow at a few percent of rated voltage. Keeping the applied voltage low limits the current to its rated value and gives the series-branch parameters without overheating the windings.

Question 46Electrical Machines

A 3-phase induction motor is started by a star-delta starter. What is the ratio of starting torque with star-delta starting to the starting torque with direct-on-line starting?

  1. 1/2
  2. 1/3 Correct answer.
  3. 1/√3
  4. 1/9

Explanation

In star connection the phase voltage is 1/√3 of the delta phase voltage, so starting current and torque both fall to one third. Since torque varies with the square of voltage, (1/√3)² = 1/3.

Question 47Electrical Machines

Which condition must be satisfied for two 3-phase transformers to be operated in parallel?

  1. Equal kVA ratings and identical winding resistance
  2. Equal per-unit impedance and equal core loss
  3. Same voltage ratio, same polarity, and same phase-sequence and phase-angle group Correct answer.
  4. Identical core material and identical no-load current

Explanation

Parallel operation requires matching voltage ratios to avoid circulating current, correct polarity to prevent a short circuit, and the same phase-angle group so secondary voltages are in phase. Impedance values should be similar for proper load sharing but need not be equal.

Question 48Electrical Machines

In a DC machine, what is the function of the commutator and brushes?

  1. To increase the generated emf by adding turns
  2. To reduce eddy currents in the armature core
  3. To convert the alternating emf in the armature coils into direct voltage at the terminals Correct answer.
  4. To provide the magnetizing flux for the field

Explanation

The emf induced in rotating armature coils is alternating; the commutator and brushes act as a mechanical rectifier so that the external terminal voltage is unidirectional. In a motor they perform the reverse function, converting DC supply into alternating current in the coils.

Question 49Power Systems

In a 3-phase, 50 Hz transmission line, the phase voltage is 132 kV and the line current is 400 A at a power factor of 0.9 lagging. What is the total 3-phase active power delivered to the load?

  1. 142.6 MW
  2. 82.3 MW Correct answer.
  3. 95.0 MW
  4. 47.5 MW

Explanation

Three-phase active power is P = √3 × V_L × I_L × cos φ, using the line voltage of 132 kV and line current of 400 A. This gives √3 × 132 × 10³ × 400 × 0.9 ≈ 82.3 MW. Multiplying the phase voltage by 3 would be needed for a per-phase sum, not the line-voltage formula.

Question 50Power Systems

A power system has a base of 100 MVA and 220 kV. A transmission line of reactance 40 Ω is to be represented in per unit. What is its per-unit reactance on these bases?

  1. 0.018 pu
  2. 0.083 pu Correct answer.
  3. 0.826 pu
  4. 0.182 pu

Explanation

The base impedance is Z_base = V_base² / S_base = (220×10³)² / (100×10⁶) = 484 Ω. Dividing the actual reactance by the base impedance gives 40 / 484 ≈ 0.083 pu.