Attempt the ISRO Electrical Engineering (EE) 2013 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 energy stored in the magnetic field in a solenoid of length 30cm and diameter 3cm wound with 1000 turns of wire & carrying a current of 10A is
Answer: B. 0.15 joules.
Explanation: L = μ0N²A/l = 4π×10⁻⁷×10⁶×(π×0.015²)/0.3 ≈ 2.96 mH; W = ½LI² = ½×2.96×10⁻³×100 ≈ 0.15 J.
A network is said to be linear, if and only if
Answer: D. The principles of superposition and homogeneity apply.
Explanation: A linear network must satisfy both superposition (additivity) and homogeneity (scaling).
Kirchoff's law fails in the case of
Answer: D. Distributed parameter networks
Explanation: KCL/KVL assume lumped elements; they fail for distributed-parameter networks (e.g. transmission lines at high frequency).
In a four branch parallel circuit, 50mA current flows in each branch. If one of the branches opens, the current in other branches
Answer: C. Unaffected.
Explanation: In a parallel circuit each branch is across the same voltage, so the other branches' currents are unaffected.
The wave length of a wave in a waveguide is
Answer: A. is greater than in free space
Explanation: Guide wavelength λg = λ0/√(1−(λ0/λc)²) is always greater than the free-space wavelength.
Characteristic impedance of a quarter wave transformer connected in between a load of 100 ohm and a transmission line of characteristic impedance 225 ohms is
Answer: D. 150 ohm
Explanation: Z = √(Z_L·Z0) = √(100×225) = 150 Ω.
A transverse electromagnetic wave with circular polarization is received by a dipole antenna due to polarization mismatch. The power transfer efficiency from the wave to the antenna is reduced to about
Answer: A. 50%
Explanation: A linear (dipole) antenna receiving a circularly polarised wave picks up only one linear component: 50 % (3 dB polarisation loss).
The unit of displacement density of a magnetic circuit is
Answer: B. Coulomb / sq. metre
Explanation: Electric displacement density D is charge per unit area: C/m².
The derivative of an ideal step function is
Answer: A. an impulse function
Explanation: d/dt u(t) = δ(t).
An impulse function consists of
Answer: A. entire frequency range with same relative phase
Explanation: The Fourier transform of δ(t) is 1: flat magnitude over all frequencies with zero (same) phase.
The discrete time system described by y(n) = x(n^2) is
Answer: C. non-causal, linear and time invariant
Explanation: y(n)=x(n²) is non-causal (needs future samples for n>1) and linear. It is actually time-VARYING, so no listed option is exactly right; c is closest to the intended key. Verify against the official key.
What does the transfer function of a system describe for the system?
Answer: B. only zero state response
Explanation: H(s) = Y(s)/X(s) with zero initial conditions, so it describes only the zero-state response.
Which of the following measures cannot be effective in reducing the noise?
Answer: C. increase in channel bandwidth
Explanation: Increasing channel bandwidth admits more noise power (N = N0·B); the other measures reduce effective noise.
Which among the following type of transformer have smallest size with same electrical specification ?
Answer: D. OFWF type transformer.
Explanation: OFWF (oil-forced, water-forced) cooling is the most effective, so the transformer is the smallest for the same rating.
Two transformers operating in parallel will share the load depending upon their ?
Answer: D. Per unit impendence.
Explanation: Parallel transformers share load in proportion to rating only if their per-unit impedances are equal; otherwise sharing depends on per-unit impedance.
Transformer core is laminated,
Answer: C. to avoid eddy current and hysteresis losses.
Explanation: Lamination reduces eddy-current loss (key answer lists eddy and hysteresis).
The eddy current losses in the transformer will be reduced if ?
Answer: D. The laminations are thin.
Explanation: Eddy loss ∝ t² (lamination thickness squared); thinner laminations reduce it.
The Buchholz relay is used to protect the ?
Answer: B. Oil immersed transformers against all internal faults.
Explanation: Buchholz relay is a gas-actuated relay for oil-immersed transformers (internal faults).
Why are transformer stamping annealed before being used for the building?
Answer: B. to reduce hysteresis loss due to burning of edges
Explanation: Punching strains the grain and raises hysteresis loss at the cut edges; annealing relieves this.
As compared to Δ – Δ bank, the capacity of the V –V bank of transformers is ---- percent.
Answer: A. 57.7
Explanation: V-V capacity is 1/√3 = 57.7 % of the Δ-Δ bank rating.
A transformer on no-load is switched on to a source of voltage. It will draw a current ---
Answer: B. which is several times the steady-state magnetizing current, depending upon the initial state of the residual flux in the transformer core.
Explanation: Switching-on inrush is several times the magnetising current and depends on the residual flux and the switching instant.
On the two sides of a star/delta transformer ----
Answer: B. the voltage and current both differ in phase by 30°
Explanation: Star/delta introduces a 30° phase shift between both line voltages and line currents.
In a Scott-connected transformer the number of primary and teaser turns respectively are ----
Answer: D. N, √3N/2
Explanation: Main transformer has N turns on primary; teaser has (√3/2)N turns.
The use of higher flux density in the transformer design ---
Answer: A. decreases the total weight / kVA
Explanation: Higher flux density permits a smaller core and winding, reducing total weight per kVA.
The applied voltage of a certain transformer is increased by 50% while the frequency is reduced by 50%. The maximum core flux density will become ---
Answer: A. three times
Explanation: Bm ∝ V/f = 1.5/0.5 = 3 times.
Power input to a transformer on no-load at rated voltage comprises predominantly of
Answer: C. Core loss
Explanation: No-load input supplies core (iron) loss; copper loss is negligible.
Distribution transformers have core loss
Answer: C. Less than full load copper loss
Explanation: Distribution transformers are on load for only part of the day, so they are designed with core loss less than full-load copper loss (max efficiency at part load).
Non loading heat run test on transformers is performed by means of -
Answer: D. Sumpner's test.
Explanation: Sumpner's (back-to-back) test loads both transformers thermally with only the losses drawn from the supply (heat run test).
In power lines, series capacitors are used to----
Answer: B. Compensate inductive reactance.
Explanation: Series capacitor cancels part of the line's inductive reactance (X_net = X_L − X_C).
The starting torque of a slip ring induction motor can be increased by
Answer: A. Adding external resistance to rotor
Explanation: Adding rotor resistance moves the peak torque toward standstill and raises the starting torque.
The synchronous speed of a four-pole induction motor operating at 50Hz is
Answer: A. 25rps
Explanation: Ns = 120×50/4 = 1500 rpm = 25 rps.
A shunt generator has a critical field resistance of 200 ohm at a speed of 800r.p.m. If the speed of the generator is increased to 1000r.p.m., what is the change in critical field resistance of the generator?
Answer: C. Increases to 250 ohm
Explanation: Critical field resistance is proportional to speed: 200×1000/800 = 250 Ω.
A three-phase slip ring induction motor is fed from the rotor side with stator winding short circuited. The frequency of currents flowing in short circuited stator is
Answer: A. Slip frequency
Explanation: The stator is the secondary of a rotor-fed machine, so its currents are at slip frequency relative to the field. Medium confidence.
When the supply voltage to an induction motor is reduced by 10%, the maximum torque decreased by approximately
Answer: C. 20%
Explanation: Tmax ∝ V²: 0.9² = 0.81, a fall of about 19–20 %.
A 3-phase induction motor is driving full-load torque which is independent of speed. If line voltage drops to 90% of the rated value, percentage increase in motor copper losses
Answer: A. 23%
Explanation: Constant torque with V = 0.9 gives I ∝ 1/0.9 = 1.111; copper loss ∝ I² = 1.235, i.e. about +23 %.
The injected e.m.f in the rotor of an induction motor is of
Answer: A. The same frequency as slip frequency
Explanation: Injected rotor emf must be at slip frequency (can be in phase, anti-phase or quadrature).
If the full-load speed of a 3-phase, 50Hz induction motor is 950 r.p.m, what is its half load speed nearly equal to?
Answer: D. 975 r.p.m
Explanation: Ns = 1000 rpm; full-load slip 5 %; half-load slip ≈ 2.5 % gives about 975 rpm.
For controlling the speed of an induction motor the frequency of supply is increased by 10%. For magnetizing current to remain the same, the supply voltage must
Answer: C. Be increased by 10%
Explanation: Keeping flux (V/f) constant requires voltage to rise 10 % with a 10 % frequency rise.
The speed of an induction motor is controlled by varying the supply frequency keeping V/f constant, then
Answer: A. Breakdown torque and magnetizing current would both remain constant
Explanation: With V/f constant, flux, magnetising current and breakdown torque stay (approximately) constant.
A synchronous generator connected to an infinite bus is overexcited. Consider the only reactive power, from the point of view of the system, the machine acts as
Answer: A. a capacitor
Explanation: An overexcited generator delivers lagging reactive power, i.e. it behaves like a capacitor to the system.
The voltage stress will be maximum in an underground cable at
Answer: B. The surface of the conductor
Explanation: Electric stress in a cable is greatest at the conductor surface (E = V/(r ln(R/r))).
The dielectric strength of air at a barometric pressure of 76cm and 25 degree centigrade is
Answer: B. 21.1kv rms / cm.
Explanation: Breakdown strength of air at NTP is 30 kV/cm peak = 21.1 kV/cm rms.
The positive sequence current of a transmission line is
Answer: D. equal to negative sequence current
Explanation: Positive- and negative-sequence currents are equal in magnitude for L-L (and L-G) faults. Verify against the key.
For the fault at the terminals of a synchronous generator, the fault current is maximum for a
Answer: C. Line-to-ground fault
Explanation: At generator terminals X0 is small, so the single line-to-ground fault current can exceed the 3-phase fault current. Verify against the key.
The earth transformer is used to
Answer: B. Provide artificial neutral earthing where neutral point is not accessible
Explanation: An earthing (zig-zag) transformer provides an artificial neutral on an ungrounded system.
For differential protection of power transformer (delta-delta) the current transformers will have
Answer: C. Star-star connection
Explanation: For delta-delta power transformer differential protection, the CTs are connected in star on both sides.
For the protection of a very long extra high voltage line, the protective relay used is
Answer: D. Mho type distance relay
Explanation: Mho relays are used for long EHV lines (reactance relays suit short lines).
Resistance switching is normally employed in
Answer: D. Air-blast circuit breaker
Explanation: Resistance switching is employed in air-blast circuit breakers.
Symmetrical breaking capacity of ACB is
Answer: B. Less than asymmetrical breaking capacity
Explanation: Symmetrical breaking capacity is lower than asymmetrical breaking capacity.
By which material the fuse element is generally made
Answer: D. Silver
Explanation: Fuse elements are made of low-melting metals such as silver, tin, lead (silver key). Medium confidence.
Grounding is generally done in transmission line at
Answer: A. The supply end
Explanation: Neutral grounding is generally done at the supply end (generating station/substation).
What is the approximate value of the surge impedance loading of a 400kV , 3-phase 50Hz overhead single circuit transmission line
Answer: B. 400 MW
Explanation: SIL = V²/Zc = (400 kV)²/≈400 Ω ≈ 400 MW.
When two identical first order systems have been cascaded non-interactively the unit step response of the system will be
Answer: A. Over damped
Explanation: Two cascaded non-interacting first-order systems give a second-order system with real poles, i.e. overdamped.
Which of the following methods is most strong tool to determine the stability and the transient response of the system ?
Answer: D. Root locus.
Explanation: Root locus gives both stability and transient response (pole locations) as gain varies.
If the gain of a critically damped system is increased, it will become
Answer: A. Under damped system
Explanation: Increasing gain of a critically damped system moves poles into the complex plane: underdamped.
Phase margin of a system is used to specify
Answer: A. Relative stability
Explanation: Phase margin measures relative stability.
The rms value of an alternating current is given by steady DC current which when flowing through a given circuit for a given time produces,
Answer: A. The same heat as produced by AC when flowing through the same circuit.
Explanation: Definition of rms: same heating effect as the equivalent DC.
AC current cannot be measured directly by
Answer: C. Moving coil ammeter
Explanation: A moving-coil meter reads average value, which is zero for AC; it cannot measure AC directly.
The internal resistance of a voltmeter should be very high in order to have
Answer: C. Minimum current through the meter
Explanation: High voltmeter resistance draws minimum current from the circuit, reducing loading error.
The resistance of a thermistor
Answer: B. Decreases with the increase of temperature
Explanation: Thermistors normally have a negative temperature coefficient.
The early effect in a bipolar junction transistor is caused by
Answer: C. Large collector-base reverse bias
Explanation: Large collector-base reverse bias widens the collector depletion region, reducing base width (Early effect).
Fermi level for an n-type semiconductor lies
Answer: B. Near conduction band
Explanation: In n-type material the Fermi level lies near the conduction band.
For a forward biased pn-junction diode diffusion capacitance varies
Answer: A. Linearly with current
Explanation: Diffusion capacitance C_D = τ I/V_T, proportional to current.
In a multi-stage R-C coupled amplifier, the coupling capacitor
Answer: A. Limits the low frequency response
Explanation: Coupling capacitors (with input resistance) form a high-pass filter, limiting the low-frequency response.
An operation amplifier should preferably have
Answer: A. Low out put impedance
Explanation: An op-amp should have low output impedance (and high input impedance).
The output voltage of an operational amplifier is ?
Answer: B. 180 degree out of phase from the input.
Explanation: Taken as the inverting configuration: output is 180° out of phase. Verify against the key.
A class-A transformer coupled, transistor power amplifier is required to deliver a power output of 10 Watts. The maximum power rating of the transistor should be less than
Answer: C. 20 W
Explanation: For class-A transformer-coupled, efficiency is at most 50 %, so the transistor must dissipate 2×Po = 20 W.
Which of the following Boolean algebra rules is correct?
Answer: C. A+ĀB = A+B
Explanation: A + ĀB = A + B (absorption/redundancy law).
In an all NOR gate realization of a combinational circuit all EVEN and ODD level gates behave like
Answer: B. AND and OR
Explanation: In NOR-NOR realization of combinational logic, odd level gates act as OR and even level gates act as AND, or equivalently alternate levels act as AND and OR logic.
Use of a reverse conducting thyristor in place of antiparallel combination of thyristor and feedback diode in an inverter:
Answer: C. Minimizes the effects of load inductance on the commutation performance
Explanation: Use of a reverse conducting thyristor (RCT) integrates the anti-parallel diode into a single wafer, minimizing lead inductance and its effects on commutation performance.
In a resonance pulse inverter:
Answer: D. DC saturation of transformer core is minimized
Explanation: In a resonant pulse inverter, DC saturation of the transformer core is minimized.
The vectors x1 = (1,2,4), x2 = (2,-1,3), x3 = (0,1,2), x4 = (-3,7,2) are
Answer: B. Linearly dependent
Explanation: Any set of 4 vectors in 3-dimensional space R³ is linearly dependent since the dimension of R³ is 3.
Characteristic roots of matrix A and A^T will be
Answer: B. Same
Explanation: det(A − λI) = det((A − λI)ᵀ) = det(Aᵀ − λI), so matrix A and its transpose Aᵀ have the exact same characteristic roots.
The minimum point of the function (x^3/3) – x is at
Answer: A. x = 1
Explanation: f'(x) = x² − 1 = 0 gives critical points x = ±1. Since f''(x) = 2x, f''(1) = 2 > 0 indicates a local minimum at x = 1.
The area bounded by the curves y^2 = 9x, x – y + 2 = 0 is given by
Answer: B. 1/2
Explanation: Intersection points of y² = 9x and x = y − 2 give y = 3 and y = 6. Area = ∫₃⁶ [(y − 2) − y²/9] dy = [y²/2 − 2y − y³/27]₃⁶ = (−2) − (−2.5) = 1/2.
The integrating factor of equation sec^2 y dy/dx + x tan y = x^3 is
Answer: A. [e]^(x^2/2)
Explanation: Substitute u = tan y, then du/dx + x·u = x³, which is a linear differential equation. The integrating factor is e^(∫ x dx) = e^(x²/2).
An urn contains 5 black and 5 white balls. The probability of drawing two balls of the same colour
Answer: B. 4/9
Explanation: P(same colour) = P(2 Black) + P(2 White) = (5/10 × 4/9) + (5/10 × 4/9) = 20/90 + 20/90 = 40/90 = 4/9.
Ten percent of screws produced in a certain factory turn out to be defective. Find the probability that in a sample of 10 screws chosen at random, exactly two will be defective.
Answer: A. 0.2
Explanation: Using binomial distribution: P(X = 2) = ¹⁰C₂ × (0.1)² × (0.9)⁸ = 45 × 0.01 × 0.4305 ≈ 0.1937 ≈ 0.2.
The equation x^3 – x^2 + 4x – 4 = 0 is to be solved using the Newton-Raphson method. If x =2 is taken as the initial approximation of the solution, then the next approximation using the method will be
Answer: B. 4/3
Explanation: f(2) = 2³ − 2² + 4(2) − 4 = 8; f'(x) = 3x² − 2x + 4, so f'(2) = 12. Then x₁ = x₀ − f(x₀)/f'(x₀) = 2 − 8/12 = 4/3.
The unique polynomial P(x) of degree 2 such that: P(1) = 1, P(3) = 27, P(4) = 64 is
Answer: A. 8x^2 – 19x + 12
Explanation: For P(x) = 8x² − 19x + 12: P(1) = 8 − 19 + 12 = 1, P(3) = 72 − 57 + 12 = 27, and P(4) = 128 − 76 + 12 = 64. Matches all conditions.