Attempt the ISRO Electrical Engineering (EE) 2015 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electrical Engineering (EE) papers
80 questions.
The current I em units, flowing in a coil of a tangent galvanometer is given by the expression where r = Mean radius of the galvanometer coil. N = Number of turns of the galvanometer coil. H = Horizontal component of earth magnetic field. φ = Deflection of galvanometer magnetic needle.
Answer: C. I = (Hr / 2πN)* Tanφ
Explanation: Tangent law: H tanφ = 2πNI/r, so I = (Hr/2πN) tanφ.
To measure current, Ammeters are connected in
Answer: A. Series with very low resistance
Explanation: An ammeter is connected in series and must have very low resistance to avoid disturbing the circuit.
If a circuit is formed consisting of two dissimilar metallic conductors, and if one of the junction has a temperature of T1 and the other is at higher temperature T2, a current flow in the circuit. This effect is called as
Answer: C. Seeback Effect
Explanation: Current in a loop of two dissimilar metals with junctions at different temperatures is the Seebeck (thermoelectric) effect.
Two wattmeter connected to measure the input of a balanced 3 Phase circuit indicates 2000 Watts and 500 Watts respectively. What will be the power factor of the circuit when both of the readings are positive?
Answer: D. 0.696
Explanation: tanφ = √3(W1−W2)/(W1+W2) = √3×1500/2500 = 1.039, φ ≈ 46.1°, cosφ ≈ 0.694 (closest option 0.696).
Wheatstone Bridge is used for measurement of
Answer: C. Medium Resistance Value
Explanation: The Wheatstone bridge is for medium resistances (about 1 Ω to 100 kΩ).
In a series RLC circuit, during resonance
Answer: A. Inductive reactance becomes equal to capacitive reactance
Explanation: At series resonance X_L = X_C.
Which of the following network theorem deals with finding out the circuit values of voltage and current in a restricted portion of the network by replacing the actual source of energy by a single "equivalent voltage source" or by a single "equivalent current source" acting at a terminal pair?
Answer: D. Thevenin's Theorem
Explanation: Replacing a network portion by a single equivalent voltage source (with series impedance) is Thevenin's theorem (its dual, Norton's, uses a current source).
What will be the base impedance for a three phase system with base MVA = 100 MVA and Base kV as 11 kV?
Answer: A. 1.21 Ohms
Explanation: Z_base = kV²/MVA = 11²/100 = 1.21 Ω.
The differential protection relay may lose its stability for through faults due to saturation of CT magnetic circuit during short circuit condition. To overcome this difficulty which of the following technique is used
Answer: A. Biased differential relay
Explanation: A biased (percentage) differential relay prevents maloperation due to CT saturation on through faults.
During single phasing, the unbalanced stator current have a negative sequence component, which cause
Answer: A. Magnetic flux rotating in opposite direction to main flux
Explanation: Negative-sequence current produces a field rotating opposite to the main field, causing heating and braking torque.
The Burden of protective Current Transformer (CT) is specified in
Answer: C. Volt-Ampere
Explanation: CT burden is specified in volt-amperes (VA) at rated secondary current.
In the network shown in the figure below, the current in the 12 Ohm resistor is 5 Amps, the battery voltage E is [Figure description: A single-loop-with-parallel-branches DC circuit. A battery E is on the left vertical side; a 4 Ω resistor is in series on the top wire. After the 4 Ω resistor the circuit splits into two parallel vertical branches connected between the top and bottom wires: (1) a middle branch containing an 18 Ω resistor; (2) a right branch containing a 6 Ω resistor in series with a parallel combination of a 4 Ω resistor and a 12 Ω resistor. The 5 A current is marked as flowing through the 12 Ω resistor.]
Answer: C. 300 Volts
Explanation: 12 Ω carries 5 A (60 V across 4∥12 = 3 Ω, so 15 A in the 4 Ω, 20 A in the 6 Ω branch). Branch voltage = 60 + 6×20 = 180 V; 18 Ω carries 10 A; total 30 A; E = 180 + 4×30 = 300 V.
What is the equivalent capacitance of the following circuit? [Figure description: Capacitors drawn along a single line from left terminal to right terminal: two 10 µF capacitors in series, followed by a parallel combination of a 4 µF capacitor (top branch) and a 6 µF capacitor (bottom branch), followed by a third 10 µF capacitor in series.]
Answer: A. 2.5 µf
Explanation: 4 µF ∥ 6 µF = 10 µF; four 10 µF in series = 2.5 µF.
The relation between the flux density (B) and Magnetic field intensity (H) is given by B = µH. What is µ in this relation?
Answer: C. Permeability of the medium
Explanation: μ is the permeability of the medium.
Power supply to a 10 pole induction motor is supplied by 4 pole alternator which is driven at 1500 RPM. If the motor runs with slip of 4%, what is its speed?
Answer: B. 576 RPM
Explanation: Alternator frequency = 4×1500/120 = 50 Hz; motor Ns = 120×50/10 = 600 rpm; N = 600×0.96 = 576 rpm.
A power factor meter connected in a circuit indicates pf of 0.6 lagging. To improve the power factor, we have to insert the following component in the circuit.
Answer: C. Capacitors
Explanation: A lagging power factor is improved by adding capacitors.
The logic evaluated by the circuit at the output is [Figure description: A two-input logic circuit with inputs X and Y. Two NOT gates (inverter triangles) are placed on the input lines. The top AND gate takes X and the inverted Y; the bottom AND gate takes Y and the inverted X (cross-coupled inverted inputs). The outputs of the two AND gates feed a 2-input OR gate whose output is labelled OUTPUT.]
Answer: A. XȲ + YX̄
Explanation: Output = X·Ȳ + X̄·Y, the XOR function.
The figure shows a half wave rectifier circuit with input voltage v(t) = 10 sin(100πt) volts. Assuming ideal diode characteristics with zero forward voltage drop and zero reverse current, the average power consumed in watts by the load resistance RL is [Figure description: An AC voltage source V(t) (with + at top) in series with an ideal diode (in the top wire, conducting left to right) and a load resistor RL = 100 Ω, forming a single series loop.]
Answer: B. 0.25 Watts
Explanation: Half-wave rectifier: Vrms = Vm/2 = 5 V; P = 25/100 = 0.25 W.
The truth table [Figure description: Truth table with inputs X, Y and output F(x,y): (X=0,Y=0)->F=0; (0,1)->0; (1,0)->1; (1,1)->1.] Represent the Boolean function
Answer: A. X
Explanation: F = 1 whenever X = 1 (regardless of Y), so F = X.
A three stage Johnson counter ring in figure is clocked at a constant frequency of fc from starting state of Q0 Q1 Q2 = 101. The frequency of output Q0 Q1 Q2 will be [Figure description: Three JK flip-flops in a row (outputs Q0, Q1, Q2 with complements), all driven by a common CLK line. Each flip-flop's J input is connected to the Q output of the previous stage's... specifically J1/K1 inputs are connected from the previous stage's Q and Q-bar outputs (J from Q, K from Q-bar) for stages 2 and 3, and the last stage's outputs Q2-bar and Q2 are fed back along a long wire over the top to the first flip-flop's J and K inputs (J0 from Q2-bar, K0 from Q2), forming a twisted-ring (Johnson) counter.]
Answer: B. fc / 6
Explanation: A 3-stage Johnson counter has 2n = 6 states, so output frequency is fc/6.
A 4 bit module - 16 ripple counter uses JK F/F. If the propagation delay of each F/F is 50 nano seconds, the maximum clock frequency that can be used is equal to
Answer: B. 5 MHz
Explanation: Ripple counter needs T ≥ 4×50 ns = 200 ns, so fmax = 5 MHz.
Figure shown below, capacitor C is initially charged to Vo = 50 V with upper plate positive. Switch S is closed at t = 0. Current through the circuit at t = 0 and final voltage across C are respectively [Figure description: A 200 V DC source with its positive terminal at the top connected through a switch S, then a diode D (anode toward the source, voltage VD across it), then a resistor R = 10 Ω, then to a capacitor C (initial voltage Vo = 50 V, upper plate +, voltage Vc across it) that returns to the negative terminal of the 200 V source. A single series loop.]
Answer: A. 15 A, 200 V
Explanation: i(0) = (200 − 50)/10 = 15 A; the capacitor charges to the source voltage, 200 V.
The Q point value Ic for the circuit is [Figure description: An NPN BJT with emitter grounded, collector connected through RC = 10 kΩ to +12 V, and a base resistor RB = 100 kΩ connected from the collector to the base (collector-to-base feedback bias). βdc = 100.]
Answer: C. 1.018 Milli Amps
Explanation: 12 = IC(10k) + (IC + IB)... with IB = IC/100: 12 − 0.7 = IC(10.1k + 1k) so IC ≈ 1.018 mA.
The frequency of oscillations of transistorized Colpitts Oscillator having tank circuit parameters C1 = 150 pF, C2 = 1.5 nf and L = 50 µH is
Answer: C. 1.927 MHz
Explanation: Ceq = 150×1500/1650 = 136.4 pF; f = 1/(2π√(50 µH × 136.4 pF)) ≈ 1.93 MHz.
The following distortion readings are available for a power amplifier. D2 = 0.2, D3 = 0.02 and D4 = 0.06. The Total Harmonic Distortion (THD) is
Answer: B. 20.97%
Explanation: Total Harmonic Distortion THD = √(D2² + D3² + D4²) = √(0.2² + 0.02² + 0.06²) = √(0.04 + 0.0004 + 0.0036) = √0.044 ≈ 0.20976 = 20.97%.
A transformer coupled class A amplifier drives a 16 Ohm loud speaker through 4:1 transformer with Vcc = 36 V. If the circuit delivers 2 Watts to load, the rms voltage across the load assuming transformer efficiency as 100 % is [Figure description: A common-emitter NPN transistor amplifier. Vcc = 36 V feeds the primary winding (N1) of a transformer (turns ratio N1:N2) connected to the collector (collector current Ic). A base bias resistor RB runs from Vcc to the base; the input Vi is coupled to the base through a capacitor C; the emitter is grounded. The transformer secondary (N2) drives a load RL = 16 Ω across which the load voltage VL is measured.]
Answer: D. 5.65 V
Explanation: Power delivered to load P = VL_rms² / RL => 2 W = VL_rms² / 16 Ω => VL_rms = √32 ≈ 5.657 V ≈ 5.65 V.
In the RL circuit given below, the maximum power will be transferred when value of RL is [Figure description: A planar DC circuit. Left branch: a 5 V voltage source (vertical). Top wire: a 10 Ω resistor from the 5 V source to a first node, then another 10 Ω resistor to a second node. From the first node, a middle vertical branch goes down through a 10 Ω resistor in series with a 2 V voltage source to the bottom wire. At the second node, a 1 A current source (arrow pointing downward) is connected between top and bottom wires, in parallel with the load resistor RL at the far right.]
Answer: A. 5 Ohms
Explanation: For maximum power transfer to a purely resistive load RL connected to impedance Zs = R + jX, RL = |Zs| = √(4² + 3²) = 5 Ω.
A source Vs(t) = V cos 100πt has an internal impedance of (4 + j3) Ω. If a purely resistive load is connected to this source has to extract the maximum power out of the source, its value in Ohms should be
Answer: B. 5 Ohms
Explanation: Source internal impedance Zs = 4 + j3 Ω. To extract maximum average power in a purely resistive load RL, RL = |Zs| = √(4² + 3²) = 5 Ω.
In a transformer, exciting current is made up of two components namely magnetizing current IM and core loss current IC, with negligible leakage impedance drop (V is supply voltage)
Answer: C. IM lags V by 90° whereas IC is in phase with V
Explanation: The magnetizing current IM produces the core flux and lags supply voltage V by 90°, whereas the core loss component IC is in phase with V supplying active iron losses.
The leakage flux in a transformer depends upon
Answer: C. The load current
Explanation: Leakage flux is driven by the load amp-turns (mmf) of the windings; it is proportional to the load current flowing in the windings.
If Excitation of synchronous generator fails, it acts as a
Answer: D. Induction generator
Explanation: When the field excitation of a grid-connected synchronous generator fails, the rotor slips below synchronous speed and operates as an induction generator, drawing reactive power from the grid for excitation.
In a tap changer, the voltage at consumer terminals is kept within the prescribed limits by varying the
Answer: A. Ratio of turns between primary and secondary windings
Explanation: An on-load tap changer adjusts the effective turns ratio (N1/N2) between primary and secondary windings to maintain terminal voltage within permissible limits.
The residual magnetism of a self excited DC generator is lost. To build up its emf what to be done to regain the residual magnetism?
Answer: D. Field winding must be excited by low voltage DC supply
Explanation: When residual magnetism is lost, 'field flashing' is performed by temporarily exciting the field winding with a low-voltage external DC supply (such as a battery) to restore residual flux.
Two DC series machines are mechanically coupled. One machine is running as a motor and the other as a generator. The iron and friction loss will be identical when
Answer: B. Their speeds and excitation are identical
Explanation: Because they are mechanically coupled and identical series machines, their rotational speed is mechanically identical, and identical series armature-field currents ensure identical magnetic excitation.
The phaser diagram of a synchronous machine connected to an infinite bus is shown below. The machine is [Figure description: A phasor diagram with the terminal voltage phasor V drawn as a horizontal arrow pointing right (labelled TERMINAL VOLTAGE), and the induced voltage phasor E drawn from the same origin pointing up and to the right at roughly 45 degrees (labelled INDUCED VOLTAGE), i.e. E leads V by a load angle.]
Answer: A. Generator and operating at lagging pf
Explanation: The induced voltage E leads terminal voltage V by load angle δ (δ > 0), identifying the machine as a generator. Furthermore, E cos δ > V, which means the machine operates at a lagging power factor.
A 3 phase synchronous motor, connected to an infinite bus, operating at a leading pf with a constant load torque, if excitation is increased (δ is load angle and φ power factor angle)
Answer: C. δ increases but φ decreases
Explanation: When mechanical shaft load on a synchronous motor increases, the rotor is pulled backward relative to the stator field, increasing the torque angle δ while decreasing the power factor angle φ.
Magnetizing in rush current in transformer is rich in
Answer: C. 2nd Harmonics
Explanation: Magnetizing inrush current is heavily asymmetric and rich in second harmonic components, which is utilized by differential protection relays for harmonic restraint.
Two Induction motors A and B are identical except that the air gap of motor 'A' is 50% greater than that of motor B then
Answer: B. The no-load pf of Motor A will be poorer than that of Motor B
Explanation: A larger air gap increases magnetic circuit reluctance, requiring higher magnetizing current to establish operating flux. This lowers the no-load power factor of Motor A relative to Motor B.
A 3 phase induction motor draws active power P and reactive power Q from grid. If it is operating as a generator, P and Q will respectively be
Answer: B. Positive and positive
Explanation: Under normal motoring operation, an induction motor consumes real power P to perform mechanical work and draw reactive power Q to establish its rotating magnetic field, so both P and Q are positive.
In case of 3 Phase Short circuit in a system, the power fed into the system is
Answer: B. Mostly reactive
Explanation: During a symmetrical 3-phase fault, the short-circuit current is limited predominantly by the inductive reactances of generators, transformers, and lines (X/R >> 1), so the fault power is mostly reactive.
The output voltage of the ideal transformer with polarities and dots shown in the figure is given by [Figure description: An ideal transformer with turns ratio 1:N. The primary is driven by a source V1 sin(wt) (+ terminal at top). The primary dot is at the top of the primary winding; the secondary dot is at the bottom of the secondary winding (dots on opposite ends, i.e. opposite polarity). The secondary feeds a load (box) with output voltage V2 measured with + at the top and - at the bottom.]
Answer: C. - N V1 sin(wt)
Explanation: With turns ratio 1:N, primary dot at the top (+), and secondary dot at the bottom (-), the secondary terminal voltage V2 measured positive at the top is inverted: V2 = −N·V1 sin(ωt).
The average real power in watts delivered to a load impedance ZL = (4 - J2) Ω by an ideal current source i(t) = 4 sin(wt + 20°) A is
Answer: B. 32 Watts
Explanation: The current source i(t) = 4 sin(ωt + 20°) has Irms = 4/√2 A. Real power is absorbed only by the resistance R = 4 Ω: P = Irms²·R = (16/2) × 4 = 32 W.
A network contains B branches and N Nodes. The number of mesh current equations would be
Answer: D. B - (N - 1)
Explanation: In network graph theory, the number of fundamental loops (independent mesh current equations) is given by L = B − N + 1 = B − (N − 1).
Superposition theorem is valid for
Answer: A. Linear circuits
Explanation: Superposition theorem relies fundamentally on the principles of linearity (additivity and homogeneity), and is strictly valid only for linear, bilateral networks.
A sine wave as has a peak value of 12 Volts. Its crest factor is
Answer: C. 1.415
Explanation: Crest factor is the ratio of peak value to RMS value. For a sinusoidal waveform, Crest Factor = Vpeak / Vrms = Vm / (Vm/√2) = √2 ≈ 1.414 (closest option 1.415).
A sinusoidal voltage V = 50 sin wt is applied to a series RL circuit. The current in the circuit is given by I = 25 sin (wt - 53°). The apparent power consumed by the load is
Answer: B. 625 VA
Explanation: Vrms = 50/√2 V, Irms = 25/√2 A. Apparent power S = Vrms × Irms = (50 × 25) / 2 = 1250 / 2 = 625 VA.
A band pass filter is one which
Answer: D. Passes frequencies between two designated cut off frequencies
Explanation: A band-pass filter passes signals within a specified frequency band between lower and upper cutoff frequencies (fL to fH) and attenuates signals outside this band.
A bulb in staircase has two switches, one switch being at the ground floor and the other being at first floor. The bulb can be turned ON and also OFF by any one of the switch irrespective of the state of the other switch. The logic of switching of bulb resembles
Answer: B. XOR gate
Explanation: A staircase switching circuit toggles the output state whenever either switch is toggled (00→0, 01→1, 10→1, 11→0), which is the exact truth table of an XOR gate.
Time domain expressions for the voltage V1(t) and V2(t) is given by V1(t) = Vm sin(10t - 130°) and V2(t) = Vm cos(10t + 10°) which is the correct statement. The anticlockwise direction of rotation of phasor may be taken as positive.
Answer: A. V1(t) leads V2(t) by 130°
Explanation: V1(t) = Vm sin(10t − 130°) = Vm cos(10t − 130° − 90°) = Vm cos(10t − 220°) = Vm cos(10t + 140°). V2(t) = Vm cos(10t + 10°). Thus, V1 leads V2 by 140° − 10° = 130°.
In practice, earth is chosen as a place of zero electric potential because it
Answer: D. has almost constant potential
Explanation: Due to Earth's immense physical size and electrical capacitance, adding or removing charges produces negligible potential change, maintaining an almost constant reference potential of zero volts.
Series compensation on EHV lines is resorted to
Answer: A. Improve the stability
Explanation: Series capacitive compensation cancels part of line inductive reactance X, increasing power transfer capability Pmax = (V1·V2)/Xnet and significantly improving transient and steady-state stability limits.
The magnetic susceptibility of a specimen is small and positive, the specimen is
Answer: C. Paramagnetic
Explanation: Paramagnetic materials have small and positive magnetic susceptibility (χm > 0, μr slightly > 1), whereas diamagnetic materials have small negative susceptibility.
The rate of rise of restriking voltage depends upon
Answer: D. The inductance and capacitance of the system only
Explanation: Rate of Rise of Restriking Voltage (RRRV) depends on the natural frequency of oscillation ωn = 1/√(LC), determined strictly by the equivalent system inductance L and capacitance C.
A Digital Volt Meter (DVM) uses 10 MHz clock and has a voltage controlled generator which provides a width of 5µ sec/volt of unit signal. A 10 V input signal would correspond to a pulse count of
Answer: A. 500
Explanation: Pulse count = Gate width × Clock frequency = (5 µs/V × 10 V) × 10 MHz = 50 µs × 10 pulses/µs = 500 counts.
In figure shown below, the Peak Inverse Voltage (PIV) required for diode is [Figure description: A series loop with an AC source 300 sin(ωt) (+ at top), a diode D in the top wire, a 10 Ω resistor, and a 100 V DC battery in series on the right side (battery + terminal toward the resistor? marked 100 V), closing back to the source.]
Answer: D. 400 V
Explanation: During the negative half-cycle of the AC supply, the peak supply voltage (300 V) and DC battery voltage (100 V) act in series against the diode, requiring PIV = 300 V + 100 V = 400 V.
A practical current source consists of
Answer: B. An ideal current source in parallel with an impedance
Explanation: By Norton's representation, a practical current source consists of an ideal current source in parallel with an internal shunt impedance (admittance).
The transient current in a loss free LC circuit when excited from an AC source results in ______ sine wave.
Answer: B. Un damped
Explanation: A loss-free LC circuit has zero resistance (R = 0), so its characteristic roots lie purely on the imaginary axis (s = ±jω0), producing undamped, sustained harmonic oscillations.
The transport layer protocol used for real time multimedia, file transfer, DNS and e-mail respectively are
Answer: B. UDP, TCP, UDP and TCP
Explanation: Real-time multimedia requires low latency (UDP); file transfer requires reliable delivery (TCP); DNS queries are fast datagrams (UDP); email SMTP requires reliable transmission (TCP).
Techniques that automatically move program and data blocks into physical main memory when they are required for execution are called
Answer: C. Virtual memory techniques
Explanation: Virtual memory techniques use paging and segmentation to automatically transfer active code and data blocks between secondary storage and main physical memory as required.
Resistor of microprocessor (µP) which keeps track of the execution of program and which contain the memory address of next instruction to be executed is called
Answer: B. Program counter
Explanation: The Program Counter (PC) register in a microprocessor stores the memory address of the next instruction to be fetched and executed.
OS that permits multiples programs to run simultaneously using single processor is referred as
Answer: A. Multitasking
Explanation: Multitasking is the operating system feature that executes multiple programs concurrently by time-sharing a single processor core.
The cladding which surrounds the optic fiber line is used
Answer: C. Because it help to guide light in the core
Explanation: The cladding has a slightly lower refractive index than the core (n_cladding < n_core), establishing total internal reflection (TIR) to guide optical signals along the fiber core.
Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is [Figure description: A single block diagram: input u(s) enters a block labelled 1/S (an integrator) and the output y(s) leaves the block.]
Answer: C. t u(t)
Explanation: Integrating a unit step input u(t) with zero initial conditions yields the unit ramp response y(t) = ∫₀^t u(τ) dτ = t·u(t).
If the unit step response of a network is (1 - e^-αt) then its unit impulse response is
Answer: A. α e^-αt
Explanation: The unit impulse response h(t) is the time derivative of the unit step response s(t): h(t) = d/dt [1 − e^(−αt)] = α·e^(−αt).
The solution of the differential equation (dy/dx) = ky, Y(0) = C is
Answer: B. y = c e^kx
Explanation: Separating variables: dy/y = k dx => ln y = kx + C1 => y = C·e^(kx). Given y(0) = C, the exact solution is y = C·e^(kx).
The trigonometric Fourier series of an even function does not have the
Answer: B. Sine terms
Explanation: For an even function f(−t) = f(t), the product f(t)·sin(nω0t) is an odd function, so its integral over a full period is zero, making all sine coefficients bn = 0.
If x = √-1 then the value of x^x is
Answer: A. e^(-π/2)
Explanation: Let x = i = e^(iπ/2). Then x^x = i^i = (e^(iπ/2))^i = e^(i² π/2) = e^(−π/2).
For the function f(t) = e^(-t/τ), the Taylor series approximation for t << τ is
Answer: C. 1 - t/τ
Explanation: The Maclaurin/Taylor expansion for e^x = 1 + x + x²/2! + … For x = −t/τ with t << τ, retaining first-order terms gives f(t) ≈ 1 − t/τ.
If the impedance of an AC circuit is 10∠60° Ω then resistance in the circuit is
Answer: A. 5 Ω
Explanation: Impedance Z = 10 ∠60° Ω = 10 cos 60° + j 10 sin 60° = 10(0.5) + j 10(√3/2) = 5 + j 8.66 Ω. Thus, the real resistance is R = 5 Ω.
The dielectric strength of air under normal condition is
Answer: C. 30 kV/cm
Explanation: Under standard temperature and pressure (STP: 20°C, 760 mmHg), the dielectric breakdown strength of dry air is approximately 30 kV/cm (peak) or 21.2 kV/cm (rms).
String efficiency of 100% means in string insulators
Answer: D. Shunt capacitance is zero
Explanation: String efficiency = (Total voltage) / (n × Voltage across bottom disc). It reaches 100% when voltage distribution across all discs is perfectly uniform, which occurs when shunt capacitance C1 to ground is zero.
To limit current chopping in Vacuum Circuit Breakers (VCB), the contact material used has
Answer: C. High vapour pressure and low conductivity properties
Explanation: To limit current chopping in vacuum circuit breakers, contact alloys (such as Cu-Bi or Cu-Cr) must have high vapor pressure and low thermal conductivity to maintain metal vapor at low current zero crossings.
In a three phase four wire unbalanced system, the current in the neutral wire is 18 A. The magnitude of zero sequence current is
Answer: B. 6 A
Explanation: In a 3-phase 4-wire system, neutral current In = 3·I0. Given In = 18 A, the zero sequence current is I0 = In / 3 = 18 / 3 = 6 A.
The positive (Z1), negative (Z2) and zero (Z0) sequence impedances of a solidly grounded system under steady state condition always follow the relation
Answer: A. Z1 > Z2 > Z0
Explanation: For an unloaded synchronous generator, the sequence impedances typically follow the order Z1 > Z2 > Z0 (where Z1 includes synchronous reactance Xd, Z2 is negative-sequence reactance, and Z0 is zero-sequence leakage reactance).
Eddy current loss in core of a transformer is
Answer: D. Inversely proportional to resistivity of core material
Explanation: Eddy current loss per unit volume is Pe = (π²·f²·Bm²·t²) / (6·ρ), which is inversely proportional to the electrical resistivity ρ of the core laminations.
The square root of 64 ∠36°
Answer: B. 8 ∠18°
Explanation: Taking the principal square root in polar form: √(64 ∠36°) = √64 ∠(36°/2) = 8 ∠18°.
Match the correct pairs : [Table: Numerical integration scheme -> Order of fitting polynomials. P. Simpson's 3/8 rule; Q. Trapezoidal rule; R. Simpson's 1/3 rule. Orders: 1. First order; 2. Second order; 3. Third order.]
Answer: D. P - 3, Q - 1, R - 2
Explanation: Simpson's 3/8 rule uses cubic polynomials (3rd order, P-3); Trapezoidal rule uses linear polynomials (1st order, Q-1); Simpson's 1/3 rule uses parabolic polynomials (2nd order, R-2). Thus, P-3, Q-1, R-2.
What is Laplace Transform of (sinh(at))?
Answer: B. a/(s^2 - a^2)
Explanation: Laplace transform of hyperbolic sine: L{sinh(at)} = a / (s² − a²) for s > |a|.
If a phasor is multiplied by j then
Answer: B. Only its direction changes
Explanation: In the complex plane, multiplying by j = 1 ∠90° rotates the phasor counter-clockwise by 90° without altering its magnitude.
If two complex numbers are equal
Answer: C. Their in phase and quadrature components will be separately equal
Explanation: Two complex numbers A + jB and C + jD are equal if and only if their real (in-phase) parts and imaginary (quadrature) parts are separately equal: A = C and B = D.