Question 1
An inductance of one H carrying a current of two amperes will store the energy of
- A. 2 watts
- B. 2 joules
- C. 4 watts
- D. 4 joules
Answer: B. 2 joules
Explanation: Magnetic energy stored in an inductor W = ½·L·I² = ½×1×2² = 2 J (joules).
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80 questions.
An inductance of one H carrying a current of two amperes will store the energy of
Answer: B. 2 joules
Explanation: Magnetic energy stored in an inductor W = ½·L·I² = ½×1×2² = 2 J (joules).
The square waveform of current has following relation between r.m.s value and average value
Answer: C. r.m.s. value of current is equal to the average value
Explanation: For a symmetrical square wave of amplitude Im, |i(t)| = Im at all times, so Irms = Iavg = Im.
Two numbers of 500 ohms one watt resistors are connected in parallel. Their combined resistance and wattage rating will be
Answer: B. 250 ohms, 2 watts
Explanation: In parallel: Req = 500/2 = 250 Ω. Total power rating is the sum: 1 W + 1 W = 2 W.
In a long uniform coil of inductance 2L and associated resistance 2R ohms is physically cut in to two exact halves which are rewound in parallel. The resistance and inductance of the combination are
Answer: C. R/2 and L/2
Explanation: Cutting a uniform coil of (2L, 2R) into two equal halves gives each half turns N/2 and length l/2, so each has L' = L and R' = R. Connecting them in parallel gives Leq = L/2 and Req = R/2.
For transfer function of a physical two-port network
Answer: D. All the above
Explanation: For physical realizability and stability of a passive two-port network, transmission poles and zeros must lie in the LHP (or on the jω axis as simple poles), satisfying all these conditions.
The starting torque of a three phase induction motor can be increased by
Answer: B. Increasing rotor resistance
Explanation: Starting torque Tst ∝ R2/(R2² + X2²). Increasing rotor resistance R2 shifts the maximum torque toward standstill (s = 1), thereby increasing starting torque.
The capacitor start-capacitor run single phase induction motor is operationally a
Answer: B. Two phase motor
Explanation: Under running conditions, the two windings are displaced 90° in space with currents 90° apart in time, causing the motor to operate virtually as a balanced two-phase motor.
What would happen if the field of a D.C. shunt motor is opened?
Answer: C. Speed will enormously increase damaging the motor
Explanation: For a DC shunt motor, speed N ∝ Eb/Φ. If the field circuit is opened, the flux Φ drops to the residual value (near zero), causing the speed to increase dangerously to destructive levels.
Equalizer rings in D.C. generator with lap windings are used for
Answer: A. Equal distribution of current at brush for sparkless commutation
Explanation: Equalizer rings connect points in a lap-wound armature that are at identical potentials, bypassing circulating currents away from the brushes and ensuring sparkless commutation.
The Buchholz relay is normally used to protect the
Answer: B. Oil immersed transformers against all internal faults
Explanation: Buchholz relay is a gas-actuated protective device installed in the pipe connecting the tank and conservator of oil-immersed transformers to detect incipient and severe internal faults.
The power transformer is a
Answer: D. Pulsating main flux device
Explanation: In a power transformer, the core flux is alternating and pulsating sinusoidally (Φ = Φm sin ωt) to induce alternating EMFs in the windings.
A change of 5% in supply voltage to an induction motor will produce a change of approximately
Answer: C. 10% in the rotor torque
Explanation: Torque developed by an induction motor is proportional to V² (T ∝ V²). A 5% change in voltage produces ΔT/T ≈ 2(ΔV/V) = 2×5% = 10% change in rotor torque.
If the supply frequency to the transformer is increased, the iron loss
Answer: A. Will increase
Explanation: Hysteresis loss Wh ∝ f·Bm^(1.6) and eddy current loss We ∝ f²·Bm². If supply voltage is not fixed or flux is maintained, core losses increase with frequency.
The maximum temperature permitted for class A insulation is
Answer: B. 105 degree centigrade
Explanation: Standard insulation temperature limits: Class Y = 90°C, Class A = 105°C, Class E = 120°C, Class B = 130°C, Class F = 155°C, Class H = 180°C, Class C > 180°C.
Two transformers connected in parallel share load in the ratio of their KVA ratings, provided their ohmic impedance are
Answer: B. In direct ratio of their ratings
Explanation: For parallel transformers to share load proportional to their ratings without circulating currents, their per-unit impedances on their own ratings must be equal (load shared in direct ratio of ratings).
The high frequency hum in the transformers is mainly due to
Answer: B. Magnetostriction
Explanation: Transformer hum is caused primarily by magnetostriction—the microscopic periodic change in physical dimensions of ferromagnetic core laminations under an alternating magnetic flux.
A 400/200 volts transformer has pu impedance of 0.05. The HV side voltage required to circulate full load current during short circuit test is
Answer: A. 20 V
Explanation: Full-load impedance drop on the HV side = pu impedance × HV rated voltage = 0.05 × 400 V = 20 V.
Non-loading heat run test on transformers is performed by means of
Answer: D. Sumpner’s test
Explanation: Sumpner's (back-to-back) test allows two identical transformers to be tested at full load for temperature rise without drawing total load power from the mains.
A three phase induction motor is driving full-load torque which is independent of speed. If the line voltage drops to 90% of the rated value, % increase in motor copper losses
Answer: A. 23%
Explanation: Torque T ∝ sV² is constant, so s ∝ 1/V². Motor current I ≈ sV/R2 ∝ 1/V. When voltage drops to 0.9 V, current rises to I/0.9 = 1.111 I, and copper losses (I²R) increase to (1/0.9)² = 1/0.81 = 1.2346 (+23.5% ≈ 23%).
A synchronous generator is feeding power to infinite bus bars at unity power factor. Its excitation is now increased. It will feed
Answer: B. The same power but at a lagging power factor
Explanation: Increasing the excitation of an alternator connected to infinite bus increases its induced EMF E above bus voltage V, delivering reactive power to the bus and causing it to operate at a lagging power factor (overexcited).
The steam input in to a turbogenerator connected to infinite bus is increased. Which of the events will take place?
Answer: B. The generator will feed more real power to bus bar and power angle will increase
Explanation: Increasing steam input supplies more mechanical power (Pm), which increases the active electrical power output (Pe = (EV/X) sin δ) and increases the power/torque angle δ.
In an ideal transformer, which of the following statement is correct?
Answer: D. all the above are correct
Explanation: An ideal transformer has zero winding resistance (no I²R loss), zero leakage flux (no leakage reactance drops), zero core losses (no hysteresis/eddy losses), and infinite permeability (negligible magnetizing mmf).
The voltage applied to a transformer primary is increased keeping v/f fixed. How will core loss and magnetizing current change?
Answer: A. Core loss will increase and magnetizing current remain same
Explanation: Keeping V/f constant keeps peak flux density Bm constant (since Bm ∝ V/f). Magnetizing current remains constant, while core loss increases due to higher frequency (Wh ∝ f, We ∝ f²).
Frequencies in the UHF range normally propagate by means of
Answer: D. Space waves
Explanation: UHF radio waves (300 MHz to 3 GHz) penetrate the ionosphere without reflection and propagate primarily via line-of-sight space waves (tropospheric/direct waves).
Indicate the antenna that is not wideband
Answer: D. Marconi
Explanation: Discone, folded dipole, and helical antennas exhibit wide bandwidth characteristics, whereas a standard Marconi (quarter-wave monopole) antenna is resonant and narrow-band.
The depth of penetration of a wave in a lossy dielectric increases with increasing
Answer: C. Wavelength
Explanation: Skin depth (depth of penetration) δ = 1/√(πfμσ). Since f = c/λ, δ ∝ 1/√f ∝ √λ, so depth of penetration increases with wavelength (lower frequency).
Copper behaves as a
Answer: C. Conductor or dielectric depending on frequency
Explanation: At low and RF frequencies, conduction current dominates (σ >> ωε), so copper acts as a good conductor; at optical and X-ray frequencies above the plasma frequency, displacement current dominates and it behaves as a dielectric.
Assuming constant transmission efficiency, if voltage is increased ‘n’ times, the size of the conductor would be
Answer: A. Reduced to 1/n^2 that of the original
Explanation: For given power P, distance l, and percentage loss, conductor volume is inversely proportional to V² (Volume ∝ 1/V²). If voltage increases by a factor of n, conductor volume reduces to 1/n².
Shunt conductance in overhead power transmission lines is primarily due to
Answer: A. Leakage over the insulators
Explanation: Shunt conductance G represents the real leakage current flowing across the surface of overhead line insulators to the supporting towers and ground.
The dielectric strength of SF6 gas (used in circuit breakers) is approximately
Answer: B. 2 to 3 times more than air
Explanation: Sulphur hexafluoride (SF6) is an electronegative gas whose dielectric breakdown strength is roughly 2.5 to 3 times that of air at atmospheric pressure.
Ferranti effect states that under certain conditions, the sending end voltage is
Answer: A. Less than receiving end voltage
Explanation: Ferranti effect occurs on lightly loaded or open-circuited long transmission lines where line charging current through line inductance causes receiving end voltage VR to exceed sending end voltage VS (i.e. VS < VR).
The equal area criterion of stability is applicable to
Answer: B. One machine system and infinite bus bars
Explanation: Equal area criterion is a graphical method based on the swing equation, applicable strictly to a single machine connected to an infinite bus (or a two-machine equivalent system).
The power transmission capacity of the transmission line is
Answer: D. Proportional to the square of the operating voltage
Explanation: Maximum power transmission capacity Pmax = (VS·VR)/X. Since VS and VR are proportional to nominal voltage V, Pmax ∝ V² (proportional to the square of operating voltage).
If the torque angle ‘delta’ increases indefinitely, the system indicates
Answer: C. Instability
Explanation: Stable operation requires dPe/dδ > 0. If the power angle δ increases beyond 90° indefinitely, synchronizing torque vanishes and the system loses synchronism (instability).
Resistance switching is normally employed in
Answer: D. Air-blast circuit breaker
Explanation: Resistance switching is universally employed in air-blast circuit breakers to damp transient recovery voltage (TRV) oscillations and suppress current chopping.
Series capacitors are used to
Answer: B. Compensate for line inductive reactance
Explanation: Series capacitors inserted directly in the line cancel part of the inductive reactance (Xnet = XL − Xc), reducing line impedance, enhancing power transfer capability, and improving voltage regulation.
Over fluxing protection is normally recommended for
Answer: A. Generator transformers in power stations
Explanation: Generator step-up (GSU) transformers in generating stations are vulnerable to overfluxing (V/f ratio exceeding limits) during generator startup, shutdown, or load rejection.
In general AC distribution systems, arc interruption in vacuum circuit breakers are designed to function in a time period of
Answer: A. One cycle
Explanation: In vacuum circuit breakers, the arc diffuses rapidly into the high-dielectric vacuum and is cleanly extinguished at the very first natural current zero (within one cycle).
Systems for Protection from negative sequence current is provided normally for
Answer: B. Generators
Explanation: Negative-sequence currents produce a reverse-rotating magnetic field at twice synchronous speed relative to the rotor, inducing double-frequency rotor eddy currents and severe overheating in synchronous generators.
Bundled conductors are mainly used in high voltage overhead transmission lines to
Answer: C. Reduce corona
Explanation: Bundling conductors increases the effective conductor radius (GMR), lowering surface electrical field gradient, which significantly reduces corona loss and audible/radio interference.
Surge impedance of overhead transmission line is normally in the order of
Answer: C. 300 – 500 ohms
Explanation: Surge impedance Zs = √(L/C) is typically 400 Ω (range 300–500 Ω) for overhead transmission lines, compared to 30–50 Ω for underground cables.
The bode diagram approach is applied to
Answer: B. Minimum phase network
Explanation: Bode diagram analysis directly relates magnitude and phase uniquely only for minimum-phase systems (all poles and zeros in the left half s-plane).
Which of the following methods is the strongest tool to determine the stability and the transient response of the system?
Answer: D. Root locus
Explanation: Root locus traces the exact trajectories of closed-loop poles in the s-plane as gain varies, providing complete information on both absolute/relative stability and transient response (damping ratio, natural frequency).
If the gain of a critically damped system is increased, it will become
Answer: A. Under damped system
Explanation: For a standard second-order system, increasing forward-path loop gain increases the natural frequency ωn without increasing the damping factor, reducing the damping ratio ζ below 1 (underdamped).
The frequency domain and time domain are related through
Answer: A. Laplace transform
Explanation: The Laplace transform (and Fourier transform) provides the mathematical mapping converting time-domain differential equations into algebraic frequency-domain transfer functions.
A system with characteristic equation s^4 + 2s^3 + 11s^2 + 18s + 18 = 0 will have closed loop poles such that
Answer: C. Two poles lie symmetrically on the imaginary axis of the s-plane.
Explanation: Routh array: s⁴: 1, 11, 18; s³: 2, 18; s²: 2, 18; s¹: row of zeros! Auxiliary polynomial A(s) = 2s² + 18 = 0 gives s = ±j3, so two poles lie symmetrically on the imaginary axis.
If the gain of an open loop system is doubled, the gain margin
Answer: C. Becomes half
Explanation: Gain margin GM = 1/|G(jωpc)|. If open-loop gain is doubled, |G(jωpc)| doubles, so the gain margin becomes half.
An all pass network imparts only
Answer: A. Negative phase to the input
Explanation: An all-pass filter has constant unity gain (|H(jω)| = 1) across all frequencies and introduces pure negative phase lag (delay) to the input signal.
An ON-OFF controller is
Answer: C. Non-linear controller
Explanation: An ON-OFF (two-position) controller is inherently non-linear due to its discontinuous switching characteristic and hysteresis/deadband.
The transfer function of a phase lead controller is (1 + 3Ts) / (1 + Ts). The maximum value of phase provided by this controller is
Answer: D. 30
Explanation: Phase lead transfer function G(s) = (1 + 3Ts)/(1 + Ts) has α = 1/3. Maximum phase lead φm = arcsin((1 − α)/(1 + α)) = arcsin((2/3)/(4/3)) = arcsin(0.5) = 30°.
The thermal time constant is the time
Answer: A. To reach the final steady temperature if the initial rate of increase of temperature were maintained constant
Explanation: Thermal time constant τ = H/A is classically defined as the time required to reach final steady-state temperature if the initial rate of temperature rise were maintained constant (also the time to reach 63.2% of steady temperature rise).
Which of the following circuit will have no transients?
Answer: A. Pure resistive circuit
Explanation: Transients arise from energy storage in magnetic (inductor L) or electric (capacitor C) fields. A purely resistive circuit has no energy storage and responds instantaneously with zero transients.
A conductor of length 100 cm moves right angles to a magnetic field of flux density of 2 wb/sq.m. with the velocity of 25 m/second. The induced emf in the conductor will be
Answer: B. 50 volts
Explanation: Induced EMF e = B·l·v·sin θ = 2 Wb/m² × 1.0 m × 25 m/s × sin 90° = 50 V.
Which of the following devices are required to measure three phase balanced power?
Answer: A. One watt meter
Explanation: For a balanced 3-phase load, total power can be measured using a single wattmeter by connecting its current coil in one line and creating a star/artificial neutral connection (Ptotal = 3·W).
The sensitivity of an instrument is
Answer: C. Ratio of the change in the magnitude of the output to the corresponding change in the magnitude of the input
Explanation: Static sensitivity is defined as the ratio of the incremental change in instrument output magnitude (deflection) to the corresponding incremental change in input measured quantity.
Potentiometer sensitivity can be increased by
Answer: B. Increasing the length of the potentiometer wire
Explanation: Potentiometer sensitivity S = 1/k where potential gradient k = V/L. Increasing slide wire length L decreases the voltage gradient per unit length, allowing finer voltage resolution and higher sensitivity.
Low resistance is measured with
Answer: C. Kelvin’s double bridge
Explanation: Kelvin's double bridge eliminates contact and lead wire resistance errors, making it the standard precision instrument for measuring low resistances (< 1 Ω).
Which bridge is used to determine frequency?
Answer: C. Wien bridge
Explanation: Wien bridge is a frequency-selective AC bridge widely used for audio frequency measurement (f = 1/(2πRC)) and harmonic distortion analyzers.
The core of a moving iron instrument is made up of permalloy to
Answer: C. Reduce effect of stray magnetic field
Explanation: Permalloy (high nickel-iron alloy) has extremely high magnetic permeability and negligible hysteresis loss, shielding moving-iron vanes from stray magnetic fields and minimizing hysteresis errors.
Accuracy is defined as
Answer: B. Closeness with which an instrument reading approaches the true value of the quantity being measured
Explanation: Accuracy denotes the degree of closeness between the measured instrument reading and the true accepted value of the quantity being measured.
If three amplifiers having the same bandwidth are cascaded, the bandwidth of the resulting amplifier will be
Answer: B. Worse than that of each stage
Explanation: Cascading n identical non-interacting amplifier stages reduces overall bandwidth: BWcascade = BWstage × √(2^(1/n) − 1). For n = 3, BWcascade ≈ 0.51 BWstage (worse than each stage).
An element is said to have negative resistance when
Answer: B. The current / voltage curve has negative slope
Explanation: Dynamic negative resistance is characterized by a region in the I-V curve where incremental resistance dV/dI < 0 (negative slope), as seen in tunnel diodes.
The operational amplifiers are seldom used for differentiation because
Answer: D. The noises are amplified and this can be significant in the output.
Explanation: A differentiator has gain magnitude |H(jω)| = ωRC proportional to frequency, which excessively amplifies high-frequency noise and tends toward instability.
Which of the following multivibrator is called the flip flop?
Answer: C. Bistable multivibrator
Explanation: A bistable multivibrator has two stable states and requires an external trigger to switch between them, which is the operational definition of a digital flip-flop.
The amount of feedback applied to an amplifier reduces the gain by a factor of 10. The bandwidth
Answer: B. Increases by a factor of 10
Explanation: Negative feedback preserves the gain-bandwidth product (GBW = Af·BWf = A·BW). If closed-loop gain Af decreases by a factor of 10, the bandwidth increases by a factor of 10.
In a negative feedback amplifier, the output impedance is decreased
Answer: A. If the signal sampled is a voltage
Explanation: Voltage sampling (shunt connection at output) stabilizes output voltage and reduces closed-loop output impedance by (1 + Aβ), whereas current sampling increases it.
The ROM consists of
Answer: A. A decoder followed by an encoder
Explanation: A Read-Only Memory (ROM) architecture consists of an input decoder (generating all 2^n minterms) followed by a programmable/fixed OR encoder array.
An ideal rectifier should have transformer utilization factor (TUF) of 1. If the actual TUF is 3.5, it shows that
Answer: B. The transformer must be 3.5 times larger
Explanation: TUF = Pdc / (Transformer VA rating). A lower TUF means a larger transformer is required for the same DC output; the transformer must be oversized proportionally.
A power MOSFET is a
Answer: A. Voltage controlled device
Explanation: Power MOSFETs are unipolar, voltage-controlled devices where drain current is controlled by the electric field generated by gate-to-source voltage (Vgs).
When transistors are used in series or parallel, a snubber circuit is used to
Answer: D. All of these
Explanation: Snubber networks (R-C-diode) suppress transient overvoltage spikes, limit dv/dt and di/dt during switching, and assist static and dynamic voltage equalization across series/parallel devices.
lim (x -> infinity) of ((x + sin x)/x) equals to
Answer: C. 1
Explanation: lim(x→∞) (x + sin x)/x = lim(x→∞) [1 + (sin x)/x]. Since −1 ≤ sin x ≤ 1, (sin x)/x → 0 as x → ∞, giving 1 + 0 = 1.
A fair (unbiased) coin was tossed four times in succession and resulted in the following outcomes: (i) Head (ii) Head (iii) Head (iv) Head. The probability of getting a ‘Tail’ when the coin is tossed again is
Answer: B. 1/2
Explanation: Each toss of a fair coin is an independent Bernoulli trial. Previous outcomes have no memory effect; the probability of getting a Tail on the next toss is strictly 1/2.
What are the eigen values of the following 2 × 2 matrix [[2, -1], [-4, 5]] (rows: (2, -1) and (-4, 5))?
Answer: B. 1 and 6
Explanation: Characteristic equation det(A − λI) = (2 − λ)(5 − λ) − (−1)(−4) = λ² − 7λ + 10 − 4 = λ² − 7λ + 6 = (λ − 1)(λ − 6) = 0, so eigenvalues are 1 and 6.
The double integral from 0 to π/2 and 0 to π/2 of sin(x + y) dx dy is
Answer: D. 2
Explanation: ∫₀^(π/2) ∫₀^(π/2) sin(x + y) dx dy = ∫₀^(π/2) [−cos(x + y)]₀^(π/2) dy = ∫₀^(π/2) [cos y + sin y] dy = [sin y − cos y]₀^(π/2) = (1 − 0) − (0 − 1) = 2.
In the Taylor series expansion of e^x + sin x about the point x = π, the coefficient
Answer: B. 0.5 e^π
Explanation: The Taylor series coefficient for (x − π)² is f''(π)/2!. f'(x) = e^x + cos x, f''(x) = e^x − sin x. At x = π, f''(π) = e^π − sin π = e^π. Thus, coefficient = e^π/2! = 0.5 e^π.
A deck of five cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed from the deck, one at a time. What is the probability that the two cards are selected with the number of the first card being one higher than the number on the second card?
Answer: A. 1/5
Explanation: Total ordered pairs of cards selected without replacement from {1,2,3,4,5} is 5 × 4 = 20. Favorable pairs (first = second + 1) are (2,1), (3,2), (4,3), (5,4), giving 4 pairs. Probability = 4/20 = 1/5.
Using trapezoidal rule for the table given below, the integral from 4 to 5.2 of ln x dx will be. [Table: x: 4, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2; Ln x: 1.39, 1.44, 1.48, 1.53, 1.57, 1.61, 1.65]
Answer: A. 1.8277
Explanation: Trapezoidal rule with step h = 0.2: I ≈ (h/2)·[y0 + yn + 2·Σ y_mid] = 0.1·[1.39 + 1.65 + 2·(1.44 + 1.48 + 1.53 + 1.57 + 1.61)] = 0.1·[3.04 + 15.26] = 0.1 × 18.30 = 1.83 (closest to 1.8277).
Find the z transform of (n + 1)^2
Answer: A. z^2(2z + 1)/(z - 1)^3
Explanation: Z{(n + 1)²} = Z{n² + 2n + 1}. With Z{1} = z/(z−1), Z{n} = z/(z−1)², Z{n²} = z(z+1)/(z−1)³, combining yields z²(2z + 1)/(z − 1)³.
Consider the following system of equations in three real variables x1, x2 and x3: 2x1 - x2 + 3x3 = 1; 3x1 - 2x2 + 5x3 = 2; -x1 - 4x2 + x3 = 3. The system of equations has :
Answer: B. A unique solution
Explanation: Coefficient matrix determinant det = 2(−2·1 − 5·(−4)) − (−1)(3·1 − 5·(−1)) + 3(3·(−4) − (−2)(−1)) = 2(18) + 1(8) + 3(−14) = 36 + 8 − 42 = 2 ≠ 0. Since det ≠ 0, the linear system has a unique solution.
Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0?
Answer: A. sin(x^3)
Explanation: sin(x³) = x³ − x⁹/6 + … contains only odd powers; the others contain even powers.