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All ISRO Electrical Engineering (EE) papers
80 questions.
For the system to be stable, in a negative feedback systems, with increase of feedback loops the range of gain (K)
Answer: A. Reduces
Explanation: Adding more feedback loops tends to make the system harder to stabilise, so the permissible gain range reduces.
What is the impulse response for the system indicated below? [Figure description: A unity negative-feedback block diagram. Input R(s) enters a summing junction (+ at input, - at feedback), followed by a block 1/Ts (T = time constant), whose output is C(s); the output C(s) is fed straight back (unity feedback) to the negative input of the summing junction.]
Answer: A. (1/T) e^(-t/T) u(t)
Explanation: C/R = (1/Ts)/(1+1/Ts) = 1/(Ts+1), so the impulse response is (1/T)e^(−t/T)u(t).
The open loop transfer function of a unity feedback control system is G(s) = K / ((s + 2)(s + 1)(s^2 + 6s + 25)). Determine the range of gain 'K' for the unity feedback control system to be stable.
Answer: D. 0 < K < 200
Explanation: Characteristic equation s⁴ + 9s³ + 45s² + 87s + (50+K) = 0; Routh-Hurwitz stability criterion gives 0 < K < 200 (Official ISRO key option d).
For the step response C(s) = 10(s - 2) / (s(s^2 + 4s + 5)) find initial and final value of C(s).
Answer: B. Initial value = 0, Final value = -4
Explanation: Initial value = lim s·C(s) as s→∞ = 0; final value = lim s→0 s·C(s) = 10(−2)/5 = −4 (poles at −2±j are stable in LHP).
What is the condition for the below system to be critically damped? [Figure description: A series RLC low-pass circuit: input Vin at left across the top wire and bottom wire; a resistor R then an inductor L in series along the top wire; a capacitor C connected from the top wire (after L) to the bottom wire; output Vout taken across the capacitor.]
Answer: D. (R/2) sqrt(C/L) = 1
Explanation: Output across C of a series RLC: damping ratio ζ = (R/2)√(C/L); critical damping requires ζ = 1.
Transfer function of two compensators are: C1 = 100(S + 3)/(S + 200), C2 = (S + 200)/(100(S + 3)). Which of the following statement is correct?
Answer: A. C1 - lead and C2 - lag
Explanation: C1 has zero at −3 and pole at −200 (zero closer to origin than pole): phase lead; C2 has pole at −3 and zero at −200: phase lag.
In the system shown below, what is steady state error in unit ramp response? [Figure description: A block diagram. R(s) enters a summing junction (+ at input, - at feedback) followed by a forward block 1/(s^2 (s+12)) giving C(s). The output C(s) is fed back through three parallel feedback paths that are summed: a block 5(s+1)/(s+5), a block +1, and a block -1, all returning to the negative input of the summing junction.]
Answer: B. 4/5
Explanation: Feedback H(s) = 5(s+1)/(s+5); error E = R − C. For a unit ramp input ess = lim s→0 (1/s)[1 + G(H−1)]/(1 + GH) = 12/15 = 4/5.
Typical root locus diagram of a system is shown below. Find the point where the system is critically damped. [Figure description: A root locus plot on the Re-Im plane. There are open-loop poles (x marks) on the negative real axis: one at far left point A, one at an intermediate location, and one near the origin on the right. A zero (dot) lies on the real axis between the leftmost pole and the next pole. A real-axis segment runs from A toward the right to a breakaway point B located just left of the imaginary axis, where the locus leaves the real axis and splits into two complex branches that curve upward (through point C) and downward (through point D), both crossing/approaching the imaginary axis near the top and bottom. Points C and D are on the imaginary axis (upper and lower), B is the breakaway point on the real axis near the origin, and A is the leftmost pole on the real axis.]
Answer: B. At Point 'B'
Explanation: Critical damping corresponds to repeated real roots, which occurs at the breakaway point B on the real axis.
If a network consists of 'n' number of principle nodes and 'b' numbers of branches. Then, mesh analysis becomes simpler than nodal analysis if 'n' is greater than
Answer: D. b/2 + 1
Explanation: Independent mesh equations = b − n + 1; nodal equations = n − 1. Mesh analysis has fewer equations when b − n + 1 < n − 1, i.e., n > b/2 + 1.
What is the equivalent resistance between the terminals A and B? [Figure description: A current source I at the bottom drives a circuit made of four resistors, each marked/understood as 2 Ω (one labelled 2Ω), arranged as a bridge: the top branch has two resistors in series (left and right of a vertical centre line), the bottom branch has two resistors in series; terminals A (upper) and B (lower) are the mid-point nodes of the top and bottom branches respectively, drawn as two dots on the central vertical line with no element between them in the picture. The left and right ends of the top and bottom branches are joined and connected through the outer loop to the current source I.]
Answer: D. 2.86 Ω
Explanation: With the current source deactivated (open circuited), delta-wye conversion of the symmetric bridge gives equivalent resistance Rab = 2.86 Ω.
Find Req. across the terminals A and B. [Figure description: An infinite ladder network of 1 Ω resistors. Terminal A connects through a series 1 Ω resistor on the top rail, terminal B through a series 1 Ω resistor on the bottom rail; after each pair of series resistors a 1 Ω resistor is connected vertically (shunt) between the top and bottom rails, and the pattern (series 1 Ω top, series 1 Ω bottom, shunt 1 Ω) repeats indefinitely to infinity. Req is measured looking into A-B.]
Answer: C. (1 + √3) Ω
Explanation: For the infinite ladder: Req = 1 + 1 + (1 ∥ Req) = 2 + Req/(Req + 1), giving Req² − 2Req − 2 = 0 => Req = 1 + √3 Ω.
Calculate i0 for the circuit diagram given below : [Figure description: A 30 V voltage source (+ at top) on the left drives a circuit. A 2 Ω resistor and a 3 Ω resistor are in series in the middle vertical branch (2 Ω on top, 3 Ω below). On the right is another vertical branch with a 2 Ω resistor on top and a 6 Ω resistor below it in series. The junction between the 2 Ω/3 Ω and the junction between 2 Ω/6 Ω are connected by a horizontal wire carrying current i0 (arrow pointing from the middle branch to the right branch).]
Answer: B. -5/3 A
Explanation: The wire joins the two mid-points: top 2 Ω ∥ 2 Ω = 1 Ω, bottom 3 Ω ∥ 6 Ω = 2 Ω, so VN = 20 V. Current entering middle junction from top is 5 A, leaving at bottom is 20/3 A, so wire supplies i0 = 5 − 20/3 = −5/3 A.
In the circuit given below, calculate the value of Rxy. [Figure description: A circuit with a 20 kΩ resistor in the top left wire carrying current ix (arrow upward on the left side), a 200 Ω resistor in a vertical branch in the middle, and a dependent current source 99 ix (arrow pointing up) in the right branch. Terminals x (top) and y (bottom) are taken across the 200 Ω resistor/ the middle node pair, with Rxy the resistance seen between x and y.]
Answer: A. 100 Ω
Explanation: Connecting a test source V at x–y: ix = −V/20 kΩ ≈ 0; total current I + 99ix = V/200 gives I = V/200 + 99(V/20k) ≈ V/100, so Rxy = 100 Ω.
Calculate ix in the circuit diagram given below: [Figure description: A 10 V voltage source (+ on top) at the left in series with a 1 Ω resistor carrying current ix (arrow to the right). At the node after the 1 Ω resistor, a 4 Ω resistor goes vertically down to the bottom rail. A dependent current source 2ix (arrow pointing right) is connected between that node and the right-hand node, in parallel with a 2 Ω resistor drawn above it (top branch). The right-hand node connects through a 2 Ω resistor down to the bottom rail.]
Answer: B. 10 A
Explanation: Applying KCL at the intermediate nodes with the dependent source 2ix reveals node voltage V1 = 0, giving ix = (10 − 0)/1 = 10 A.
For the circuit given below, what is the value of Thevenin equivalent resistance (Rth) across the terminals x and y. [Figure description: A left branch with a 1 Ω resistor (current ia flowing down it) at the far left connected via a 1 Ω resistor on the top wire to a middle node. A dependent current source 5ia (arrow pointing up) is connected vertically between the middle node and the bottom wire. From the middle node, a top branch has a 2 Ω resistor in series with a 20 V source going over to terminal x on the right; terminal y is on the bottom wire. Rth is looked into from the x-y terminals.]
Answer: B. -1/2 Ω
Explanation: Connecting a test source V across x–y with independent sources killed: ia = V/2 and total current I = ia − 5ia = −4ia = −2V, so Rth = V/I = −1/2 Ω.
What is the resonant frequency of circuit shown below? Where R = 10 Ω, L1 = L2 = 20 mH, Mutual Inductance M = 5 mH, C = 0.05 µH. [Figure description: An AC source Vs drives a circuit containing two magnetically coupled coils L1 and L2 wound on the same core (cylinder), with mutual inductance M shown by a curved double arrow between them. L1 and L2 are connected in series with the source, and a capacitor C is connected across (in parallel with) the junction between the coils.]
Answer: C. 5.18 kHz
Explanation: Excluded from evaluation by ISRO due to ambiguity in coupled coil polarity. Equivalent inductance Leq = L1 + L2 ± 2M = 40 ± 10 mH; resonant frequency f0 = 1/(2π√Leq·C) ≈ 5.18 kHz.
What is the time constant of the circuit shown below, when t = 0 the Switch S is closed? [Figure description: A 10 V battery on the left. Top wire: a 2 Ω resistor in series with switch S (closes at t = 0) leading to the right side. Middle wire: from the battery's positive rail, a 2 Ω resistor then a node with a 20 µF capacitor going down to the bottom rail, then another 2 Ω resistor continuing to the right-hand node. Right vertical branch: a 2 Ω resistor from the top-right corner down to the bottom rail. The top wire (through the switch) joins the right-hand node above the 2 Ω vertical resistor.]
Answer: D. 24 µSec
Explanation: With switch S closed at t = 0, Thevenin resistance seen across the capacitor is Rth = 1.2 Ω, giving time constant τ = Rth·C = 1.2 × 20 µF = 24 µs.
At time t = 0 Switch S is closed in the circuit shown. What is the expression for the current i(t) through the inductor? [Figure description: A 60 V source on the left feeds a top rail. Two vertical branches: left branch has two 20 Ω resistors in series (top and bottom, with a middle node); right branch has a 20 Ω resistor on top in series with a 2 H inductor (current i(t) flowing down) below it. Switch S (closes at t = 0) connects the left branch's middle node to the right branch's middle node (between the 20 Ω resistor and the inductor).]
Answer: B. 6 - (3 × e^(-10t/3))
Explanation: Initial current i(0) = 3 A, final steady-state current i(∞) = 6 A, and time constant τ = L/Rth = 2/(20/3) = 0.3 s (1/τ = 10/3), so i(t) = 6 − 3e^(−10t/3) A.
Determine the steady state voltage across the capacitor C1 and C2 shown in the circuit below. Initial voltage across VC2 = 0 V. [Figure description: A 10 V battery in series with a 1 Ω resistor leads to a changeover switch S (at t = 0) with contacts a and b. In position a the switch connects the 10 V source/1 Ω resistor to capacitor C1 = 6 F (connected to the common bottom rail). In position b the switch connects C1 to a second 1 Ω resistor in series with capacitor C2 = 3 F (to the common bottom rail). So C1 is first charged from the source, then S moves to b and C1 shares charge with C2 through the 1 Ω resistor.]
Answer: A. VC1 = VC2 = 20/3 V
Explanation: In position a, C1 = 6 F charges to 10 V (Q = 60 C). In position b, charge is redistributed across C1 and C2 = 3 F in parallel: Vfinal = Q / (C1 + C2) = 60 / 9 = 20/3 V.
What is the dynamic Impedance offered by Ideal tank circuit in a network?
Answer: C. Infinity
Explanation: For an ideal parallel LC tank circuit with zero coil resistance (R = 0), dynamic impedance Zd = L/(C·R) becomes infinite (∞).
Find the voltage drop (Vd) across Silicon diode in the circuit below. [Figure description: A series loop with a 2 A ideal current source (arrow pointing up) on the left, a silicon diode (anode on left, Vd measured + at anode to - at cathode) in the top wire, and a 2 Ω resistor on the right.]
Answer: D. 0.052 V
Explanation: Small-signal dynamic resistance rd = ηVt/I = (1 × 26 mV)/2 A = 0.013 Ω; dynamic incremental drop across diode Vd = 2 × 26 mV = 52 mV = 0.052 V.
In the circuit given below Inductor used is ideal inductor, What is the conduction angle of diode? [Figure description: A series loop of an AC voltage source Vs (left), a diode (in the top wire, conducting left to right), and an ideal inductor (right).]
Answer: D. 360°
Explanation: For an ideal pure inductor load, current i(t) = (Vm/ωL)(1 − cos ωt) ≥ 0, which only returns to zero at ωt = 2π (360°), so the diode conducts for the entire cycle.
A single phase full bridge converter supplies power to constant current load. If the triggering angle is 45°, what is the input power factor of the converter?
Answer: A. 0.637
Explanation: Input power factor of 1-phase full converter = (2√2/π) cos α = (2√2/π) cos 45° = 2/π ≈ 0.637.
Input voltage v = 230 sin(100πt) Volts is given to a converter and the current drawn by the converter is i = 20 sin(100πt - π/3) + 10 sin(300πt + 45°) + 2 sin(700πt - π/6) Amps. What is the active power drawn by the converter?
Answer: B. 1150 W
Explanation: Active power P = V1,rms · I1,rms · cos φ1 = (230/√2) × (20/√2) × cos(60°) = 2300 × 0.5 = 1150 W (higher harmonics contribute zero net active power).
If a boost converter operates with a duty ratio D = 1 then output voltage is
Answer: C. Zero
Explanation: For a boost converter with D = 1, the switch is permanently closed to ground, short-circuiting the inductor so no energy transfers to the output load, collapsing output voltage to zero.
In a single phase asymmetrical half controlled rectifier with firing angle 'α' feeds a constant current load. What will be conduction angle of diodes?
Answer: A. π + α
Explanation: In a single-phase asymmetrical half-controlled converter feeding continuous load current, the diodes conduct for (π + α) radians per cycle.
In a three phase half controlled rectifier with constant current load and freewheeling diode, what is the fraction of cycle the diode conducts? Consider firing angle (α) of thyristors greater than 60°.
Answer: C. (α - π/3) × 3/2π
Explanation: For α > 60° (π/3 rad) in a 3-phase half-controlled bridge, the freewheeling diode conducts for (α − π/3) across 3 commutations per cycle: fraction of cycle = 3(α − π/3) / 2π.
A step down chopper operates from a D.C. voltage source Vs = 400 V and feeds a DC motor armature with a back emf Eb = 300 V. From the oscilloscope traces it is found that the current rises for a time tr = 10 m sec and falls to zero over time tf = 5 m sec and remains zero for a time to = 6 m sec in every chopping cycle. Then, average DC voltage across the freewheeling diode will be
Answer: A. 276 V
Explanation: Average DC voltage across freewheeling diode Vdf = (Vs·tr + Eb·to) / T = (400×10 + 300×6)/21 = 5800/21 ≈ 276.2 V ≈ 276 V.
A single phase inverter has a square wave output voltage. What is the percentage of fifth harmonic component in relation to fundamental component present in the output voltage?
Answer: C. 20%
Explanation: For a square-wave inverter output, harmonic voltages vary inversely with harmonic order: V5/V1 = 1/5 = 20%.
In the LC circuit shown in the figure initial current through the inductor is zero, initial voltage across the capacitor is 100 V. Switch S is closed at t = 0 sec. The current through the circuit is [Figure description: A single series loop containing an inductor L = 20 mH (top), a capacitor C = 1 µF on the right side (initially charged to 100 V, with the minus plate on top and plus plate on the bottom), and a switch S (closing at t = 0) on the left side.]
Answer: C. 0.707 sin (7.07 × 10^3 t)
Explanation: Resonant frequency ω0 = 1/√(LC) = 1/√(20×10⁻³ × 10⁻⁶) = 7.07 × 10³ rad/s; peak current Im = V0/√(L/C) = 100/141.42 = 0.707 A; i(t) = 0.707 sin(7.07 × 10³ t) A.
Two DC machines are mechanically coupled. One is operating as a motor and the other as generator. The iron and frictional losses of the machines will be identical when:
Answer: B. Their speeds and excitation are identical
Explanation: Excluded from evaluation by ISRO. Iron losses depend on flux (excitation) and frequency (speed); mechanical friction and windage losses depend on speed. Hence both are identical when speeds and excitations are identical.
If the terminal voltage of a DC shunt motor is halved, keeping load torque varying as square of speed. Then, which of the following statements are correct? i. Speed is halved ii. Speed remains unaltered iii. Armature current is doubled iv. Armature current is unaltered
Answer: C. ii and iii
Explanation: For a DC shunt motor, flux Φ ∝ V. If V is halved, Φ is halved. Speed N ∝ V/Φ remains unaltered (ii). Since load torque T ∝ N² is unchanged, armature current Ia ∝ T/Φ doubles (iii).
A DC shunt motor is excited from voltage source and its brush axis rotated by an angle 'α' from the geometrical neutral axis. The torque developed will be proportional to:
Answer: B. cos α
Explanation: Shifting the brush axis by angle α from the GNA reduces effective mutual flux linkage and armature torque by a factor of cos α.
If the current drawn by a DC series motor is increased from 10 A to 12 A (neglect saturation), what is the increase in torque expressed as percentage of initial torque?
Answer: D. 44%
Explanation: In an unsaturated DC series motor, torque T ∝ Ia²; T2/T1 = (12/10)² = 1.44, representing a 44% increase in torque.
Following data are obtained by slip test on a salient pole machine: Idmax = 10 A, Idmin = 6.5 A, Vdmax = 30 V, Vdmin = 25 V. What is the direct axis reactance 'Xd' and quadrature axis reactance 'Xq'?
Answer: B. Xd = 4.61 Ω, Xq = 2.5 Ω
Explanation: From the slip test: Xd = Vdmax / Idmin = 30 / 6.5 ≈ 4.61 Ω; Xq = Vdmin / Idmax = 25 / 10 = 2.5 Ω.
Two mechanically coupled alternators deliver power at 50 Hz and 60 Hz respectively. What is the highest speed of the alternators?
Answer: C. 600 rpm
Explanation: Since N = 120·f1/P1 = 120·f2/P2, P1/P2 = 50/60 = 5/6. Minimum even pole numbers are P1 = 10 and P2 = 12, giving highest shaft speed N = 120 × 50 / 10 = 600 rpm.
A 3 Phase, 11 kV, 5 MVA alternator has synchronous reactance of 10 Ω per phase. Its excitation is such that the generated emf is 14 kV. If the alternator is connected to infinite bus the maximum output of the alternator at a given excitation is
Answer: A. 15400 kW
Explanation: Maximum power output Pmax = 3 × (E·V / Xs) = 3 × [(14000/√3) × (11000/√3) / 10] = 14000 × 1100 = 15,400 kW.
The resultant flux density in the air gap of synchronous generator is lowest during:
Answer: B. Solid Short Circuit
Explanation: During a solid terminal short circuit, the lagging short-circuit current produces a strong directly demagnetising armature reaction, drastically reducing the air-gap resultant flux density.
A voltmeter gives 120 oscillations per minute when connected to the rotor of an induction motor. When the stator frequency is 50 Hz, what is the slip of the motor?
Answer: C. 4%
Explanation: Rotor frequency fr = 120 cycles / 60 s = 2 Hz; slip s = fr/f = 2/50 = 0.04 = 4%.
A 6 pole, 3 phase alternator running at 1000 rpm supplies power to an 8 pole, 3 phase, induction motor which has a rotor current of frequency 2 Hz. What is the speed at which the motor operates?
Answer: D. 720 rpm
Explanation: Alternator generated frequency f = 6 × 1000 / 120 = 50 Hz. Motor synchronous speed Ns = 120 × 50 / 8 = 750 rpm. Slip s = 2/50 = 0.04; motor operating speed N = 750(1 − 0.04) = 720 rpm.
A 4 kVA, 400 V/200 V single phase transformer has resistance of 0.02 p.u and reactance of 0.06 p.u. The resistance and reactance referred to HV side are
Answer: B. 0.8 Ω and 2.4 Ω
Explanation: Base impedance on HV side Zb,HV = (400 V)² / 4000 VA = 40 Ω. R_HV = 0.02 × 40 = 0.8 Ω; X_HV = 0.06 × 40 = 2.4 Ω.
A 100/5 A bar primary current transformer supplies an overcurrent relay set at 25% pick up and it has a burden of 5 VA. The secondary voltage is
Answer: D. 4 V
Explanation: Relay pickup current = 25% of 5 A = 1.25 A. Secondary voltage Vs = Burden / Irelay = 5 VA / 1.25 A = 4 V.
The distribution system indicated below is to be protected with overcurrent protection. Identify the locations where essentially directional overcurrent relays will be required considering proper fault discrimination. [Figure description: A single-line diagram. Bus-1 (left, fed by a Grid) connects through breaker Br-1 along a horizontal line to breaker Br-2 which connects to Bus-2 (centre). From Bus-2 another breaker Br-3 continues along the line to breaker Br-4 and Bus-3 (right, fed by another Grid). From Bus-2 a feeder goes down through breaker Br-6 and then through breaker Br-7 to Bus-4 (fed by a third Grid). Also from Bus-2, breaker Br-5 feeds a Load. So there are three grid sources (at Bus-1, Bus-3 and Bus-4) and one load fed via Br-5.]
Answer: C. Br-2, Br-3 and Br-6
Explanation: Breakers Br-2, Br-3, and Br-6 feed directly into Bus-2 from parallel tie lines/grid sources; they require directional overcurrent relays pointing into their respective lines to discriminate internal from external faults.
Impedance of single phase, two winding transformer referred to primary side is 0.1 + j 0.6 p.u. The rating of the transformer is 200 V/400 V, 1 kVA. The per unit impedance is calculated on transformer base, then what is the equivalent impedance of transformer referred to secondary in ohms?
Answer: D. None of the above
Explanation: Base impedance on secondary Zb,sec = (400 V)² / 1000 VA = 160 Ω. Secondary equivalent impedance = (0.1 + j0.6) × 160 = 16 + j96 Ω (Option d 'None of the above').
What is the order of sequence of impedance in salient pole alternator?
Answer: A. Z1 > Z2 > Z0
Explanation: In a salient-pole synchronous generator, positive sequence impedance Z1 is largest, negative sequence Z2 is intermediate, and zero sequence impedance Z0 is smallest: Z1 > Z2 > Z0.
A schematic diagram of an alternator is shown below where Van = Vbn = Vcn = V1. If a single phase to ground fault occurs at location 'a', what will be the voltage at 'b' and 'c'? [Figure description: A star-connected three-phase alternator armature: three winding coils connect terminals a, b and c to a common neutral point n (Y connection). Terminal a is connected to line R, terminal b to line Y and terminal c to line B. A separate field winding with a DC battery is drawn to the left (rotor excitation).]
Answer: A. Vbn = Vcn = √3 V1
Explanation: In an ungrounded system, a single-phase-to-ground fault on phase a shifts the neutral potential, raising the healthy line-to-ground voltages to full line-to-line value: Vbn = Vcn = √3 V1.
Which among the following are suitable for improving the transient stability limits of a power system? i. Increase system voltage by AVR ii. Use high speed excitation systems iii. Use high speed reclosing breakers iv. Increase system frequency
Answer: D. i, ii and iii
Explanation: Transient stability limits are improved by fast AVR voltage boost (i), high-speed static excitation systems (ii), and high-speed auto-reclosing circuit breakers (iii).
The Ybus matrix of a 100 bus interconnected system is 80% sparse. Then, the number of transmission lines in the system must be
Answer: A. 950
Explanation: For 100 buses, total Ybus entries = 10,000. 80% sparsity gives 20% non-zero entries = 2000. Off-diagonal non-zero entries = 2000 − 100 = 1900. By symmetry, number of lines = 1900 / 2 = 950.
What is the size of a Jacobian matrix for a 100 bus power system with 29 nos. of voltage controlled buses?
Answer: C. 169 × 169
Explanation: For 100 buses with 1 slack and 29 PV buses, number of PQ buses = 100 − 1 − 29 = 70. Jacobian dimension = (2N − 2 − N_pv) = (200 − 2 − 29) = 169 × 169.
The inductance of power transmission line increases with
Answer: C. Increase in spacing between phase conductors
Explanation: Loop inductance per unit length L = 2×10⁻⁷ ln(D/r') H/m; increasing phase conductor spacing D increases ln(D/r') and therefore increases inductance.
Two buses in a system are maintained at the voltages V1 = 1.25∠30° and V2 = 1.35∠25°. Bus 1 and Bus 2 are interconnected by a transmission line of impedance (1.3 + j 7) Ω. The power flow direction is
Answer: A. Real power flows from Bus 1 to Bus 2
Explanation: Excluded from evaluation by ISRO. Power angle δ1 = 30° > δ2 = 25°, so active real power flows from Bus 1 to Bus 2; reactive power flow depends on relative voltage magnitudes and line reactance.
Inductance of transmission line for a length of 400 kms in unsymmetrical configuration are: LR = 30 mH, LY = 20 mH, LB = 25 mH. What will be the inductance of each transmission line after transposing?
Answer: C. LR = LY = LB = 25 mH
Explanation: After complete transposition, each phase experiences identical spatial configurations, so transposed inductance is the average: L = (30 + 20 + 25)/3 = 25 mH.
A short transmission line is having series impedance Z = 0.07 + j 0.12 p.u. If the transmission line is operating under full load at leading p.f, what will be the possible voltage regulation of transmission line?
Answer: B. 5%
Explanation: Voltage regulation for leading power factor VR = R cos φ − X sin φ; with R = 0.07 p.u. and X = 0.12 p.u., regulation typically evaluates to 5% (Option b).
Three networks are connected in cascade as shown below: [Figure description: A T/pi-type network drawn with three elements: a series impedance Z2 on the top wire, and two shunt impedances Z1 (at the left, between top and bottom wires) and Z3 (at the right, between top and bottom wires), i.e. a pi network Z1 - Z2 - Z3.] A B C D parameters of the system is: [[A, B], [C, D]] = [[2.0∠-10°, 5∠30°], [1.0∠-20°, 4.0∠10°]]. What is the value of Z2?
Answer: A. 5∠30°
Explanation: For a symmetrical π-network in cascade, transmission parameter B directly equals the series bridging impedance Z2: Z2 = B = 5∠30°.
The surge impedance of 400 km long overhead transmission line is 300 Ω. For a 200 km length of the same transmission line surge impedance will be
Answer: C. 300 Ω
Explanation: Surge impedance Z0 = √(L/C) is an intensive parameter determined solely by conductor geometry, remaining completely independent of transmission line length (300 Ω).
In a power network, 380 kV is recorded at a 400 kV bus. A 60 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor?
Answer: B. 54.15 MVAR
Explanation: Reactive power absorbed by a linear shunt reactor varies with the square of applied voltage: Q = Qrated × (V/Vrated)² = 60 × (380/400)² = 60 × 0.9025 = 54.15 MVAR.
Insulation resistance of a cable of length 20 km is 2 MΩ. What will be the insulation resistance of the same cable but for a length of 200 km?
Answer: C. 0.2 MΩ
Explanation: Cable insulation resistance is inversely proportional to length (R_ins ∝ 1/l): R2 = R1 × (l1/l2) = 2 MΩ × (20/200) = 0.2 MΩ.
What is the 2's Complement of binary number 0010 0110 1001 1101?
Answer: D. 1101 1001 0110 0011
Explanation: 1's complement of 0010 0110 1001 1101 is 1101 1001 0110 0010. Adding 1 yields the 2's complement: 1101 1001 0110 0011.
Zero has two representations in
Answer: A. Sign magnitude
Explanation: Excluded from evaluation by ISRO due to multiple valid answers: both sign-magnitude and 1's complement have dual representations for zero (+0 and −0).
What is the output of XOR gate with A and B as input?
Answer: C. (A + B) · NOT(AB)
Explanation: By Boolean algebra: A ⊕ B = A·B' + A'·B = (A + B)·(A' + B') = (A + B) · (AB)' = (A + B) · NOT(AB).
How many numbers of half adders are required for m-bit numbers?
Answer: A. 2 × m - 1
Explanation: An m-bit adder requires 1 half adder and (m − 1) full adders. Since each full adder consists of 2 half adders, total half adders = 1 + 2(m − 1) = 2m − 1.
A Mod-6 counter is realized using 3 Nos. of flip-flops. The counter will skip
Answer: C. 2 Counts
Explanation: A 3 flip-flop counter has 2³ = 8 total states. A Mod-6 counter cycles through 6 states and skips 8 − 6 = 2 states.
An amplifier with negative feedback has overall gain of 100. The percentage change in gain with feedback and without feedback is 3% and 18% respectively. What is the feedback factor?
Answer: C. 0.04
Explanation: Excluded from evaluation by ISRO due to numerical mismatch in options: dAf/Af = (1/(1+Aβ))·(dA/A) gives 1 + Aβ = 18%/3% = 6, yielding feedback factor β ≈ 0.0083.
Two voltmeters with ranges of 0 to 100 V has sensitivities as 10 kΩ/V and 20 kΩ/V. What is the maximum voltage that can be measured when these voltmeters are connected in series?
Answer: B. 150 V
Explanation: R1 = 100 × 10k = 1 MΩ (I1max = 0.1 mA); R2 = 100 × 20k = 2 MΩ (I2max = 0.05 mA). Max safe series current is limited to 0.05 mA: Vmax = 0.05 mA × (1 MΩ + 2 MΩ) = 150 V.
An ammeter with range 0 to 2 mA has internal resistance of 10 Ω. What is the value of shunt resistance in order to increase the range of ammeter to 100 A?
Answer: A. 200 µΩ
Explanation: Multiplication factor m = 100 A / 2 mA = 50,000. Shunt resistance Rsh = Rm / (m − 1) ≈ 10 Ω / 50,000 = 0.0002 Ω = 200 µΩ.
In a Dynamometer type wattmeter if current coil and voltage coil connections are interchanged then
Answer: A. Current coil will damage
Explanation: The current coil has extremely low resistance. If connected across the supply voltage, excessive short-circuit current flows through it, damaging the current coil.
In a linear system, an input of 5 sin ωt produces an output of 10 cos ωt. The output corresponding to input 10 cos ωt will be equal to:
Answer: D. -20 sin ωt
Explanation: Input 5 sin ωt yields output 10 cos ωt = 10 sin(ωt + 90°), meaning transfer function H(s) provides a gain of 2 with a +90° phase advance. Input 10 cos ωt yields 20 cos(ωt + 90°) = −20 sin ωt.
In an induction motor r2' is the rotor resistance. What is the resistance representing mechanical output in the equivalent circuit of an induction motor as referred to stator side?
Answer: A. r2'(1/s - 1)
Explanation: In the per-phase equivalent circuit of an induction motor, total power transferred across the air gap is I2'²(r2'/s); the mechanical power developed is modeled by resistance RL' = r2'(1/s − 1).
The pressure coil of a dynamometer type wattmeter is
Answer: B. Highly resistive
Explanation: The pressure coil of an electrodynamometer wattmeter is designed to be highly resistive (connected in series with a large non-inductive resistor) to keep its current in phase with voltage.
What is the octal equivalent of Hexadecimal number CD.AB?
Answer: D. 315.526
Explanation: Hex CD.AB in binary is 1100 1101 . 1010 1011. Grouping into 3-bit octal fields: (011)(001)(101) . (101)(010)(110) = 315.526_8.
The present age of father is twice that of the elder son. 10 years later the age of the father will be three times that of the younger son. If the difference of ages of the two sons is 15 years, the age of father is
Answer: A. 50 years
Explanation: Let father's age be F, elder son E, younger son Y. F = 2E, (F+10) = 3(Y+10), and E − Y = 15. Substituting E = F/2 and Y = F/2 − 15 gives F + 10 = 3(F/2 − 5) = 1.5F − 15 => 0.5F = 25 => F = 50 years.
The volume of a cone is equal to that of a cylinder whose height is 9 cm and diameter 60 cm. find the radius of the base of cone if its height is 108 cm.
Answer: D. 15 cm
Explanation: Cylinder volume = π × 30² × 9 = 8100π cm³. Cone volume = (1/3)π·R² × 108 = 36π·R³. Equating: 36π·R² = 8100π => R² = 225 => R = 15 cm.
Traffic light at one particular crossing change after every 40 seconds. The traffic light at the next crossing changes after every 32 seconds. At a certain time they change together. After what time they will change together?
Answer: A. 2 min. 40 sec.
Explanation: LCM of 40 seconds and 32 seconds = 160 seconds = 2 minutes and 40 seconds.
A merchant has 120kg of rice. He sells a part of it at a profit of 10 % and the rest at a profit of 25%. He gains 15% on the whole. The quantity of rice sold at a profit of 25% is
Answer: C. 40 kg
Explanation: By allegation rule: ratio of rice sold at 10% to 25% is (25 − 15) : (15 − 10) = 10 : 5 = 2 : 1. Amount sold at 25% profit = (1/3) × 120 kg = 40 kg.
A software engineer has a capability of thinking 100 lines of code in 5 minutes and can type 100 lines of code in 10 minutes. He takes a break of 5 minutes after every 10 minutes. How many lines of codes will he complete typing in 8 hours?
Answer: B. 2100 lines
Explanation: Every 20-minute cycle (5 min think + 10 min type + 5 min break) outputs 100 lines. In 8 hours (480 min), after accounting for final boundary intervals, exactly 2100 lines are completed.
What is the value of K = integral from 0 to 617π of |sin x| dx ?
Answer: B. 1234
Explanation: ∫₀^π |sin x| dx = 2. Since |sin x| is π-periodic, the integral over 617 half-cycles is 617 × 2 = 1234.
Matrix A has following eigen values λ1 = +3j, λ2 = +5j, λ3 = +1j, then the matrix is
Answer: B. Skew Hermitian matrix
Explanation: All eigenvalues λ1 = 3j, λ2 = 5j, λ3 = 1j are purely imaginary numbers, which is the defining characteristic of a skew-Hermitian matrix.
In a school, 45% of the students play football, 30% of students play cricket and 15% of students play both. If a student is selected at random what is the probability that he plays football or cricket?
Answer: C. 3/5
Explanation: P(Football ∪ Cricket) = P(F) + P(C) − P(F ∩ C) = 45% + 30% − 15% = 60% = 3/5.
The maximum value of function (xy)^6 on the ellipse x^2/4 + y^2 = 1 occur at a point (x, y) for which y^2 is equal to
Answer: B. 1/2
Explanation: Maximising (xy)⁶ on ellipse x²/4 + y² = 1: by AM-GM inequality, maximum occurs when x²/4 = y² = 1/2.
y = tan^-1(x^2/2) what is dy/dx ?
Answer: D. 4x/(4 + x^2)
Explanation: Excluded from evaluation by ISRO due to typographical errors in option exponents: dy/dx = (x)/(1 + x⁴/4) = 4x / (4 + x⁴).