In the circuit shown below, if I1=1.5 A, then I2 will be: [Circuit description: A DC circuit with a 6 V source connected in series with a 2 Ω resistor carrying current I1=1.5 A towards a central node. At this node, a 2 Ω shunt branch carries downward current I2 to the common reference, and a right branch has a 2 Ω resistor connected to a 3 V DC voltage source (with positive terminal upwards)]. A) 2.25 A B) 1.7 A C) 1.5 A D) 3 A
💡Correct Answer: Option C (1.5 A)
Let the central node voltage between the branches be V. From the left branch, current I1=26−V=1.5 A⟹6−V=3⟹V=3 V. Current I2 flowing down through the vertical 2 Ω resistor is I2=2V=23=1.5 A. Furthermore, the current from the right 3 V source branch is 23−V=23−3=0 A, perfectly satisfying Kirchhoff's Current Law: I1=I2+0=1.5 A. Hence, I2=1.5 A.
Thevenin voltage and resistance across the terminals a and b are, [Circuit description: A 10 V DC voltage source in series with a 10 Ω resistor connected to a parallel 10 Ω resistor, followed by a 5 Ω resistor connected to terminal a; terminal b is connected to the common negative rail]. A) 10 V, 15 Ω B) 5 V, 10 Ω C) 10 V, 10 Ω D) 5 V, 15 Ω
💡Correct Answer: Option B (5 V, 10 Ω)
Open-circuit Thevenin voltage across terminals a-b (Vth): Since terminals a-b are open, no current flows through the 5 Ω resistor. Thus, Vth equals the voltage across the 10 Ω shunt resistor: Vth=10 V×10+1010=5 V. Thevenin resistance (Rth): Deactivating the 10 V independent voltage source (replacing it with a short circuit), the resistance seen from terminals a-b is Rth=5+(10∥10)=5+5=10Ω. Therefore, the Thevenin equivalent is 5 V, 10 Ω.
While analyzing the circuits using mesh and node analysis, which of Kirchhoff's laws is/are used? A) KVL in mesh analysis and KCL in node analysis B) KCL in mesh analysis and KVL in node analysis C) KVL is used both in mesh analysis and node analysis D) KCL is used both in mesh analysis and node analysis
💡Correct Answer: Option A (KVL in mesh analysis and KCL in node analysis)
Mesh analysis is fundamentally based on Kirchhoff's Voltage Law (KVL) applied along closed loops to write loop equations in terms of mesh currents. Node analysis (or nodal analysis) is fundamentally based on Kirchhoff's Current Law (KCL) applied at non-reference nodes to write node equations in terms of node voltages. Hence, KVL is used in mesh analysis and KCL in node analysis.
Two coils are connected in series-opposing, and their equivalent inductance is found to be 10 mH. When connected in series-aiding, the equivalent inductance is 40 mH. What are the self-inductances of the two coils? A) 10 mH and 15 mH B) 15 mH and 25 mH C) 8 mH and 12 mH D) 20 mH and 12 mH
💡Correct Answer: Option A (10 mH and 15 mH)
For two magnetically coupled coils with self-inductances L1 and L2 and mutual inductance M: Series-aiding: Laid=L1+L2+2M=40 mH Series-opposing: Lopp=L1+L2−2M=10 mH Adding the two expressions: 2(L1+L2)=50 mH⟹L1+L2=25 mH. Subtracting the expressions: 4M=30 mH⟹M=7.5 mH. Among the given options, only Option A (10 mH and 15 mH) satisfies L1+L2=10+15=25 mH. Note that the coupling coefficient is k=L1L2M=1507.5≈0.612≤1, which is physically valid.
Three 10 Ω resistors are connected in a star configuration. Its equivalent delta configuration will comprise three resistors, A) Identical; 10/3 Ω each B) Identical; 10 Ω each C) Not identical; 30 Ω, 10 Ω, 10/3 Ω D) Identical; 30 Ω each
💡Correct Answer: Option D (Identical; 30 Ω each)
When three identical resistors of resistance RY are connected in a star (Y) configuration, each resistor in the equivalent delta (Δ) configuration is given by RΔ=3×RY. Given RY=10Ω, the equivalent delta resistors are identical and each has a value of RΔ=3×10=30Ω.
Q6
Electrical EngineeringBasic Electrical Engineering - Maximum Power Transfer Theorem
What percentage of the maximum power is delivered to a load if the load resistance is 10 times greater than the Thevenin resistance of the source to which it is connected? A) 36.06% B) 33.06% C) 35% D) 41.08%
💡Correct Answer: Option B (33.06%)
Maximum power delivered to load occurs when RL=Rth, giving Pmax=4RthVth2. Power delivered to an arbitrary load RL is PL=IL2RL=(Rth+RL)2Vth2RL. When RL=10Rth, we have PL=(11Rth)2Vth2(10Rth)=12110RthVth2. The percentage of maximum power delivered is PmaxPL×100%=4RthVth212110RthVth2×100%=12140×100%≈33.0578%≈33.06%.
Q7
Electrical EngineeringBasic Electrical Engineering - AC Circuits & Capacitance
What is the value of capacitance if a capacitor draws a current of 31.43 mA when connected to a 40 V AC supply at a frequency of 1000 Hz? A) 125 μF B) 0.125 F C) 12.5 μF D) 0.125 μF
💡Correct Answer: Option D (0.125 μF)
Capacitive reactance is given by XC=IV=31.43×10−3 A40 V≈1272.67Ω. Since XC=2πfC1, the capacitance is C=2πfXC1=2πfVI=2×3.14159×1000×4031.43×10−3=251327.431.43×10−3≈1.25×10−7 F=0.125×10−6 F=0.125μF.
Q8
Electrical EngineeringBasic Electrical Engineering - AC Fundamentals
An alternating voltage is given by v=30sin(314t). The time taken by the voltage to reach from 0 V at t=0 to -30 V for the first time is _______? A) 0.015 second B) 0.02 second C) 0.03 second D) 0.1 second
💡Correct Answer: Option A (0.015 second)
Given v(t)=30sin(314t), angular frequency ω=314 rad/s≈100π rad/s, corresponding to a frequency f=50 Hz and time period T=ω2π=501=0.02 s. The voltage reaches 0 V at t=0, positive peak (+30 V) at t=T/4=0.005 s, zero again at t=T/2=0.01 s, and its negative peak (-30 V) for the first time at t=43T=43×0.02=0.015 s. Mathematically: sin(314t)=−1⟹314t=23π⟹t=100π1.5π=0.015 s.
Q9
Electrical EngineeringBasic Electrical Engineering - AC Waveforms & Power Dissipation
Current waveform through a 10 Ω resistor is shown below. Average power dissipated in the resistor is _______? [Waveform description: A periodic sawtooth current waveform with peak current Im=9 A and time period T=3 s, linearly increasing from 0 A at t=0 to 9 A at t=3 s, repeating periodically]. A) 729 W B) 81 W C) 270 W D) 45 W
💡Correct Answer: Option C (270 W)
For a sawtooth current waveform of peak value Im=9 A rising linearly from 0 to Im over its period T, the RMS value is Irms=3Im=39=33 A. The average power dissipated in the resistor R=10Ω is Pavg=Irms2R=(33)2×10=27×10=270 W.
Consider the following statements and choose the correct option. Statement I: An ideal voltage source has zero internal resistance Statement II: An ideal voltage source maintains a constant voltage across its terminals regardless of the current drawn from it Which of the following statement is/are true? A) Both Statement I and Statement II are true, and Statement II is the correct explanation for Statement I B) Both Statement I and Statement II are true, but Statement II is NOT the correct explanation for Statement I C) Statement I is true, but Statement II is false D) Statement I is false, but Statement II is true
💡Correct Answer: Option A (Both Statement I and Statement II are true, and Statement II is the correct explanation for Statement I)
An ideal voltage source maintains a terminal voltage V=Vs−IRint. For the terminal voltage to remain strictly constant regardless of the magnitude or direction of current I drawn from it (V=Vs), the internal voltage drop IRint must be identically zero for all I, which mathematically and physically requires that internal resistance Rint=0. Thus, both statements are true, and Statement II correctly explains why an ideal voltage source has zero internal resistance.
Match the two lists and choose the correct answer from the code given below. List - I (w) Mesh (x) Outside mesh (y) Mesh current (z) Number of meshes List - II (i) Number of nodes (ii) Node voltage (iii) Reference node (iv) Node Options: A) (w)- (ii), (x)-(iii), (y)-(iv), (z)-(i) B) (w)- (iii), (x)-(ii), (y)-(iv), (z)-(i) C) (w)- (iv), (x)-(iii), (y)-(ii), (z)-(i) D) (w)- (ii), (x)-(iv), (y)-(i), (z)-(iii)
💡Correct Answer: Option C ((w)- (iv), (x)-(iii), (y)-(ii), (z)-(i))
In circuit duality (Mesh-Nodal duality): (w) Mesh corresponds dual-wise to Node -> (iv) (x) Outside mesh (outer boundary/datum loop) corresponds to Reference node (datum node) -> (iii) (y) Mesh current corresponds dual-wise to Node voltage -> (ii) (z) Number of independent meshes corresponds to Number of independent nodes -> (i). Hence, the correct matching code is (w)-(iv), (x)-(iii), (y)-(ii), (z)-(i), which is Option C.
Q12
Electrical EngineeringElectromagnetic Fields - Vector Field Polarization
What is the resulting vector when two sinusoidally time-varying vectors having different amplitudes and phases are added up? A) Elliptically polarized B) Circular polarized C) Linearly polarized D) Quadratic polarized
💡Correct Answer: Option A (Elliptically polarized)
When two orthogonal sinusoidally time-varying vector components having different amplitudes and arbitrary phase differences (other than 0, π, or special circular polarization conditions where amplitudes are equal and phase difference is ±90∘) are superposed, the locus of the tip of the resulting vector in the plane traces out an ellipse. Hence, the resulting wave/vector is elliptically polarized.
Q13
Electrical EngineeringBasic Electrical Engineering - Charge & Current Relations
The total charge q(t), in coulombs, that enters the terminal of an element is: q(t)=⎩⎨⎧02t3+e−2(t−2)t<00≤t≤2t>2 Determine the current at t=5 s. A) 0 A B) 2 A C) 3+e−6 A D) −2e−6 A
💡Correct Answer: Option D (-2e^{-6} A)
Current is the time rate of flow of electric charge: i(t)=dtdq(t). For t>2, the charge function is q(t)=3+e−2(t−2). Differentiating with respect to time gives i(t)=dtd(3+e−2(t−2))=0+(−2)e−2(t−2)=−2e−2(t−2) A. Evaluating at t=5 s: i(5)=−2e−2(5−2)=−2e−2(3)=−2e−6 A.
What will be the values of Vs and is, respectively, in the circuit shown below? [Circuit description: A circuit with three parallel vertical branches between a top common wire and a bottom common wire: Left branch contains an upward 2 A independent current source with voltage Vs across it (positive at top) in series with a 2 Ω top resistor. Middle branch contains a vertical capacitor/source labeled with voltage 10 V and downward branch current is. Right section consists of a parallel branch with a 5 V DC source in series with a 5 Ω resistor, and another parallel 5 Ω resistor. Top wire voltage is 10 V and bottom is 0 V]. A) 14 V, 1 A B) 14 V, -1 A C) -14 V, -1 A D) -14 V, 1 A
💡Correct Answer: Option B (14 V, -1 A)
Analyzing the circuit: 1. Top node voltage with respect to the bottom reference rail is held at 10 V by the 10 V branch. 2. Left branch: A 2 A current source forces 2 A upward through the 2 Ω resistor into the top node. By KVL along the left branch: Vs−(2 A×2Ω)=10 V⟹Vs−4=10⟹Vs=14 V. 3. The current leaving the top node into the right sub-network: The right sub-network consists of a branch with a 5 V source in series with 5 Ω (I=510−5=1 A) and a parallel 5 Ω resistor (I=510=2 A), so total current entering the right branches is 1+2=3 A. 4. By KCL at the top node: Currents entering = Currents leaving ⟹2 A=is+3 A⟹is=2−3=−1 A. Thus, Vs=14 V and is=−1 A.
Q15
Electrical EngineeringControl Systems - Linearity & Superposition
A system is linear if and only if it satisfies I. Additivity II. Homogeneity III. Causality IV. Time-invariance Which of the options above is/are correct? A) I and III B) II and IV C) I and II D) III and IV
💡Correct Answer: Option C (I and II)
By definition, a system is linear if and only if it satisfies the principle of superposition, which requires both Additivity (the response to x1(t)+x2(t) is y1(t)+y2(t)) and Homogeneity / Scaling (the response to a⋅x1(t) is a⋅y1(t)). Causality and Time-invariance are independent system properties (a linear system may be time-variant or non-causal). Therefore, linearity strictly requires conditions I and II.
Q16
Electrical EngineeringControl Systems - Transfer Function & Impulse Response
The Laplace transform of a linear system's output represents the system's transfer function when the input is _______? A) An impulse signal B) A ramp signal C) A step signal D) A sinusoidal signal
💡Correct Answer: Option A (An impulse signal)
The transfer function H(s) of a linear time-invariant system with zero initial conditions is defined as the ratio of the Laplace transform of the output Y(s) to the Laplace transform of the input X(s), i.e., H(s)=X(s)Y(s). When the input is a unit impulse signal δ(t), its Laplace transform is X(s)=L{δ(t)}=1. Consequently, Y(s)=H(s)⋅1=H(s), meaning the Laplace transform of the system's impulse response is identically the transfer function.
Q17
Electrical EngineeringControl Systems - Initial & Final Value Theorems
Given F(s)=s(s+1)s+2. The initial and final values of f(t) are, respectively? A) 2, 1 B) 1, 1 C) 2, 2 D) 1, 2
💡Correct Answer: Option D (1, 2)
Applying the Initial Value Theorem: f(0+)=lims→∞sF(s)=lims→∞s[s(s+1)s+2]=lims→∞s+1s+2=1. Applying the Final Value Theorem (since the only poles are at s=0 and s=−1, both in the non-positive half plane): f(∞)=lims→0sF(s)=lims→0s[s(s+1)s+2]=lims→0s+1s+2=0+10+2=2. Thus, the initial and final values are 1 and 2, respectively.
Q18
Electrical EngineeringControl Systems - Stability Improvement Techniques
The correct sequence of steps needed to improve system stability is A) Insert derivation action, Use negative feedback and Reduce gain B) Reduce gain, use negative feedback and insert derivation action C) Reduce gain, insert derivation action and use negative feedback D) Use negative feedback, reduce gain and insert derivation action
💡Correct Answer: Option D (Use negative feedback, reduce gain and insert derivation action)
In classical control system engineering, the logical hierarchical progression to improve the stability of a system begins with: 1. Employing negative feedback (which inherently stabilizes an open-loop unstable process and regulates tracking), 2. Reducing the open-loop gain (moving closed-loop poles farther into the left half-plane away from the imaginary axis, thereby increasing gain and phase margins), 3. Inserting derivative action / phase-lead compensation (adding a stabilizing zero in the left half plane to provide phase advance, damping, and improve transient response). Thus, the standard engineering sequence is: 'Use negative feedback, reduce gain and insert derivation action'.
Q19
Electrical EngineeringControl Systems & Signal Processing - Z-Transform
The z-transform X(z) of the signal x[n]=anu[n], where u[n] is a discrete time unit step function, is? A) z/(z−a); ROC: ∣z∣>∣a∣ B) a/(z−a); ROC: ∣z∣<∣a∣ C) a/(z−a); ROC: ∣z∣>∣a∣ D) z/(z−a); ROC: ∣z∣<∣a∣
💡Correct Answer: Option A (z / (z - a); ROC: |z| > |a|)
The z-transform of a discrete-time signal x[n] is defined as X(z)=∑n=−∞∞x[n]z−n. For x[n]=anu[n], the summation becomes X(z)=∑n=0∞(az−1)n. This infinite geometric series converges if and only if ∣az−1∣<1⟺∣z∣>∣a∣, yielding the sum X(z)=1−az−11=z−az with Region of Convergence (ROC) ∣z∣>∣a∣.
Q20
Electrical EngineeringControl Systems - Steady-State Error
The steady-state error of a unity feedback system with an open-loop transfer function G(s)=s(s+2)10 for a unit ramp input is _______? A) 0 B) 0.2 C) 0.5 D) infinity
💡Correct Answer: Option B (0.2)
For a unit ramp input R(s)=s21 to a unity negative feedback system, the steady-state error is given by ess=Kv1, where Kv is the velocity error coefficient: Kv=lims→0sG(s)=lims→0s[s(s+2)10]=lims→0s+210=210=5 s−1. Therefore, the steady-state error is ess=Kv1=51=0.2.
Q21
Electrical EngineeringControl Systems - Frequency Response & Bandwidth
Consider the following statements about the bandwidth of a closed-loop system: I. In a system where the low-frequency magnitude is 0 dB on the Bode plot, the bandwidth is measured at the -3dB frequency II. The bandwidth of the closed-loop control system is a measurement of the range of fidelity of response of the system III. The speed of response to a step input is proportional to the bandwidth IV. The system with the largest bandwidth provides a slower step response and a lower fidelity ramp response Which of the given statements is/are correct? A) I and IV B) I, II and IV C) I, II and III D) I, III and IV
💡Correct Answer: Option C (I, II and III)
Statements I, II, and III are correct fundamental properties of closed-loop control systems: I: For normalized low-frequency gain (0 dB), bandwidth is standardly defined as the cutoff frequency where the magnitude drops by 3 dB (-3 dB frequency). II: Bandwidth defines the frequency band over which input signals are faithfully reproduced without significant attenuation or distortion, serving as a direct measure of fidelity. III: Rise time and settling time are inversely related to bandwidth (tr≈BW0.35), meaning higher bandwidth yields faster response speed. IV: This is false because larger bandwidth gives faster (not slower) step response and higher (not lower) fidelity ramp tracking. Hence, statements I, II, and III are correct (Option C).
Q22
Electrical EngineeringControl Systems - System Classification & Transformations
Match the two lists and choose the correct answer from the code given below: List - I (w) Non-linear system (x) Linear system (y) Time varying system (z) Multiplication in the s-domain List - II (i) Principle of superposition and homogeneity (ii) Describing function (iii) Convolution integral (iv) Rocket Options: A) (w)- (iv), (x)-(iii), (y)-(ii), (z)-(i) B) (w)- (iii), (x)-(i), (y)-(iv), (z)-(ii) C) (w)- (iv), (x)-(iii), (y)-(i), (z)-(ii) D) (w)- (ii), (x)-(i), (y)-(iv), (z)-(iii)
💡Correct Answer: Option D ((w)- (ii), (x)-(i), (y)-(iv), (z)-(iii))
Matching analysis: (w) Non-linear system is analyzed using the Describing function method -> (ii) (x) Linear system strictly obeys the Principle of superposition and homogeneity -> (i) (y) Time varying system: A classic example is a Rocket (whose mass decreases continuously over time as propellant burns) -> (iv) (z) Multiplication in the s-domain corresponds to the Convolution integral in the time domain (F1(s)F2(s)⟺f1(t)∗f2(t)) -> (iii). Therefore, the matching combination is (w)-(ii), (x)-(i), (y)-(iv), (z)-(iii), corresponding to Option D.
Q23
Electrical EngineeringControl Systems - Laplace Transforms
Laplace transform of b−ae−at−e−bt is A) 1/((s+a)(s+b)) B) 1/((s−a)(s+b)) C) 1/((s−a)(s−b)) D) 1/((s+a)(s−b))
💡Correct Answer: Option A (1 / ((s+a)(s+b)))
Recall standard Laplace transform pairs: L{e−at}=s+a1 and L{e−bt}=s+b1. Thus: L{b−ae−at−e−bt}=b−a1[s+a1−s+b1]=b−a1[(s+a)(s+b)(s+b)−(s+a)]=b−a1[(s+a)(s+b)b−a]=(s+a)(s+b)1. Hence, Option A is the correct answer.
Q24
Electrical EngineeringControl Systems - Feedback Principles
The output of a feedback control system must be a function of A) Reference input B) Reference output C) Output and feedback signal D) Input and feedback signal
💡Correct Answer: Option D (Input and feedback signal)
In any closed-loop feedback control system, the actuating error signal that drives the plant is formed by comparing the reference input with the feedback signal: e(t)=r(t)−b(t). Consequently, the system output is governed by and must be a function of both the input (command signal) and the feedback signal (which communicates the actual state of the plant). Thus, Option D is the correct choice.
Q25
Electrical EngineeringControl Systems - System Transfer Function & Poles
Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation: dt2d2y(t)+4y(t)=6r(t) The poles of this system are at A) +2, -2 B) +4, -4 C) +2j, -2j D) +4j, -4j
💡Correct Answer: Option C (+2j, -2j)
Taking the Laplace transform on both sides under zero initial conditions: s2Y(s)+4Y(s)=6R(s)⟹Y(s)(s2+4)=6R(s). The transfer function is H(s)=R(s)Y(s)=s2+46. The poles of the system are the roots of the characteristic equation s2+4=0⟹s2=−4⟹s=±−4=±2j. Hence, the poles are at +2j,−2j.
Consider the following statements. In a Measuring instrument, I. The accuracy of the instrument may be specified in terms of limits of error II. Point accuracy gives precise information about the overall accuracy of the instrument III. The best way to conceive the idea of accuracy is to specify it in terms of the true value of the quantity being measured Which of the options is/are correct? A) I, II and III B) I and III C) II and III D) I and II
💡Correct Answer: Option B (I and III)
In electrical measurements and instrumentation: - Statement I is correct: Accuracy is routinely specified in terms of limits of error (guaranteed accuracy / fiducial error limits, e.g., ±1% of full-scale value). - Statement II is incorrect: 'Point accuracy' specifies accuracy only at one particular calibrated point on the scale and does NOT provide any information about the general accuracy throughout the rest of the scale. - Statement III is correct: The most fundamental and ideal conception of accuracy is expressing it as a percentage of the true value of the quantity being measured. Therefore, statements I and III are correct (Option B).
Match the two lists and choose the correct answer from the codes given below: List - I (Properties) (w) Linear scale (x) True rms up to RF range (y) rms only for sinusoidal input (z) Reads the rms value using the square law scale List - II (Instrument Type) (i) Thermocouple type (ii) Rectifier type (iii) Moving iron type (iv) PMMC type Options: A) (w)- (iv), (x)-(i), (y)-(ii), (z)-(iii) B) (w)- (iii), (x)-(i), (y)-(iv), (z)-(ii) C) (w)- (iv), (x)-(iii), (y)-(i), (z)-(ii) D) (w)- (ii), (x)-(iv), (y)-(i), (z)-(iii)
💡Correct Answer: Option A ((w)- (iv), (x)-(i), (y)-(ii), (z)-(iii))
Matching instrument properties: (w) Linear scale: PMMC instruments possess a uniform, linear scale because deflecting torque is directly proportional to current (θ∝I) -> (iv) (x) True rms up to RF range: Thermocouple instruments operate on I2R heating and measure true RMS values accurately up to very high radio frequencies (RF) -> (i) (y) RMS only for sinusoidal input: Rectifier instruments measure average values and are calibrated to read RMS assuming a pure sinusoidal form factor of 1.11 -> (ii) (z) Reads the rms value using the square law scale: Moving iron instruments have θ∝I2, exhibiting a non-linear square-law scale -> (iii). Hence, the correct match is (w)-(iv), (x)-(i), (y)-(ii), (z)-(iii), corresponding to Option A.
Which of the following is NOT a criterion to select a potentiometer in a control system? A) Accuracy B) Frequency response C) Noise D) Time response
💡Correct Answer: Option D (Time response)
When selecting a resistive potentiometer as a position transducer/sensor in a control system, key design criteria include Accuracy (linearity and resolution), Frequency response (bandwidth limitation due to mechanical wiper inertia and stray capacitance), and Noise (wiper electrical contact noise and thermal noise). Potentiometers are static zero-order displacement transducers that do not possess intrinsic internal dynamic time delays (unlike first- or second-order dynamic sensors); thus, 'Time response' is NOT a standard criterion for selecting a potentiometer.
Q29
Electrical EngineeringElectrical & Electronic Measurements - Frequency Meters & Synchroscopes
Match the two lists and choose the correct answer from the code given below: List - I (Meter) (w) Reed frequency meter (x) Weston frequency meter (y) Weston synchroscope (z) Ohm meter List - II (Type) (i) Moving iron (ii) Vibrating (iii) Moving coil (iv) Electrodynamics Options: A) (w)- (iv), (x)-(i), (y)-(ii), (z)-(iii) B) (w)- (iii), (x)-(i), (y)-(iv), (z)-(ii) C) (w)- (ii), (x)-(i), (y)-(iv), (z)-(iii) D) (w)- (ii), (x)-(iv), (y)-(i), (z)-(iii)
💡Correct Answer: Option C ((w)- (ii), (x)-(i), (y)-(iv), (z)-(iii))
Matching meters to operating mechanism types: (w) Reed frequency meter operates on the principle of mechanical resonance of tuned steel reeds -> Vibrating type (ii) (x) Weston frequency meter uses two coils positioned at right angles acting on a soft iron vane -> Moving iron type (i) (y) Weston synchroscope operates on the electrodynamic principle with fixed and moving coils to indicate phase synchronism -> Electrodynamics type (iv) (z) Ohm meter is built around a standard permanent magnet moving coil (PMMC) movement -> Moving coil type (iii). Hence, the correct matching code is (w)-(ii), (x)-(i), (y)-(iv), (z)-(iii), which is Option C.
If the readings of the two wattmeters are equal and positive in the two-wattmeter method, the load power factor in a balanced 3-phase 3-wire circuit will be: A) Zero B) 0.5 C) Unity D) 0.866
💡Correct Answer: Option C (Unity)
In the two-wattmeter method for a balanced 3-phase load, the phase angle ϕ is given by tanϕ=3W1+W2W1−W2. When the two wattmeter readings are equal and positive (W1=W2), we have tanϕ=32W10=0⟹ϕ=0∘. Consequently, the power factor is cosϕ=cos0∘=1 (Unity).
Q31
Electrical EngineeringElectrical & Electronic Measurements - Two-Wattmeter Reactive Power
A three-phase motor operating at 500 V has a power factor of 0.4. Two wattmeters connected to measure the input power show readings of 20 kW and 10 kW, respectively. Calculate the reactive power of the load. A) 51.96 kVAR B) 17.32 kVAR C) 12.51 kVAR D) 30 kVAR
💡Correct Answer: Option B (17.32 kVAR)
In the two-wattmeter method, the total 3-phase active power is P=W1+W2, and the total 3-phase reactive power is directly given by Q=3(W1−W2). Given W1=20 kW and W2=10 kW: Q=3(20−10)=103≈17.3205 kVAR≈17.32 kVAR.
Q32
Electrical EngineeringElectrical & Electronic Measurements - Energy Meters
An energy meter having a meter constant of 1200 revolutions per kWh is found to make 5 revolutions in 75 s. The load power is _______? A) 200 W B) 250 W C) 100 W D) 500 W
💡Correct Answer: Option A (200 W)
Energy registered by 5 revolutions is E=KN=12005 kWh=12005×1000 Wh=625 Wh. Time interval t=75 s=360075 h=481 h. Load power is P=tE=1/4825/6=625×48=25×8=200 W.
The function of a potential transformer is to _______? A) Step up the current B) Measures power directly C) Step down the voltage D) Facilitates change in resistance
💡Correct Answer: Option C (Step down the voltage)
A potential transformer (PT), also called a voltage transformer (VT), is an instrument transformer used in AC power systems to step down dangerously high transmission and distribution line voltages to a safe, standardized low secondary voltage (standardly 110 V) for measurement by voltmeters and protective relaying.
Q34
Electrical EngineeringElectrical & Electronic Measurements - Digital Voltmeters
The technique of using a stair case ramp in a Digital Voltmeter is called _______? A) Deflecting torque technique B) Controlling torque technique C) Detaching torque technique D) Null balancing technique
💡Correct Answer: Option D (Null balancing technique)
In a staircase-ramp type Digital Voltmeter (DVM), an internal digital-to-analog converter generates a stepwise staircase ramp voltage that is continuously compared against the unknown input analog voltage. When the staircase voltage matches the unknown input, the comparator triggers a null condition, halting the clock pulses to the counter. This method is fundamentally classified as a potentiometric 'Null balancing technique'.
In a digital voltmeter, the oscillator frequency is 400 kHz. The ramp voltage decreases from 8 V to 0 V in 20 ms. How many pulses are counted by the counter during this interval? A) 8000 B) 8500 C) 3200 D) 1600
💡Correct Answer: Option A (8000)
The counter counts clock pulses produced by the oscillator during the time interval Δt=20 ms=20×10−3 s. Given the clock oscillator frequency f=400 kHz=400×103 Hz, the total number of pulses counted is N=f×Δt=(400×103 pulses/s)×(20×10−3 s)=8000 pulses.
Why is a Q-meter generally used at high frequencies? A) To minimise electrical noise in the circuit B) The Q-factor of coils and capacitors is more prominent C) To obtain more accurate voltage measurements D) To reduce the overall power consumption of the circuit
💡Correct Answer: Option B (The Q-factor of coils and capacitors is more prominent)
A Q-meter operates on the principle of series resonance (Q=RωL=ωCR1). At high frequencies (RF range), the inductive reactance ωL dominates over coil resistance R, and high-frequency effects such as skin effect, proximity effect, dielectric losses, and distributed winding capacitance become significant and prominent. Thus, measuring the true quality factor (Q-factor) of RF tuning coils and capacitors is most meaningful and prominent at high frequencies.
In a two-channel oscilloscope operating in x-y mode, in-phase 50 Hz sinusoidal waveforms of equal amplitude are fed to the two channels. What will be the resultant pattern on the screen? A) A square B) A circle C) A line D) An ellipse
💡Correct Answer: Option C (A line)
When two sinusoidal signals of equal frequency (50 Hz) and equal amplitude (Vx=Vmsinωt, Vy=Vmsinωt) are in phase (phase difference ϕ=0∘), the Lissajous pattern satisfies xy=VmsinωtVmsinωt=1⟹y=x. This traces a straight diagonal line inclined at an angle of 45∘ through the first and third quadrants.
Q38
Electrical EngineeringElectrical & Electronic Measurements - Temperature Transducers
Consider the following statements: The cause of error in the measurement of temperature using a thermistor is/are I. Self heating II. Poor sensitivity III. Non-linear characteristics Which of the statements is/are correct? A) I and II B) I, II and III C) II and III D) I and III
💡Correct Answer: Option D (I and III)
A thermistor is an extremely sensitive temperature transducer (having a very high negative temperature coefficient, typically -3% to -5% per °C, far more sensitive than RTDs or thermocouples). Therefore, statement II ('Poor sensitivity') is false. The two major sources of error and measurement difficulties in thermistors are: (I) Self-heating caused by the I2R dissipation of the measuring excitation current elevating the sensor temperature above ambient, and (III) Highly non-linear exponential resistance-temperature relationship (R(T)=R0eβ(1/T−1/T0)). Thus, statements I and III are correct (Option D).
Match the two lists and choose the correct answer from the code given below: List - I (Parameter) (w) Pressure (x) Temperature (y) Displacement (z) Stress List - II (Transducer) (i) Thermistor (ii) Piezoelectric crystal (iii) Capacitance transducer (iv) Resistance strain gauge Options: A) (w)- (ii), (x)-(i), (y)-(iii), (z)-(iv) B) (w)- (iii), (x)-(i), (y)-(iv), (z)-(ii) C) (w)- (ii), (x)-(i), (y)-(iv), (z)-(iii) D) (w)- (ii), (x)-(iv), (y)-(i), (z)-(iii)
💡Correct Answer: Option A ((w)- (ii), (x)-(i), (y)-(iii), (z)-(iv))
Matching parameters to suitable transducers: (w) Pressure generates electric charge across a Piezoelectric crystal -> (ii) (x) Temperature causes resistance change in a Thermistor -> (i) (y) Displacement changes plate separation or area in a Capacitance transducer -> (iii) (z) Stress/strain causes fractional change in resistance of a Resistance strain gauge -> (iv). Hence, the correct match is (w)-(ii), (x)-(i), (y)-(iii), (z)-(iv), corresponding to Option A.
Which one of the following thermocouples has the highest temperature measuring range? A) Copper-Constantan B) Platinum-Rhodium C) Alumel Chromel D) Iron Constantan
💡Correct Answer: Option B (Platinum-Rhodium)
Comparison of standard thermocouple temperature measuring ranges: - Copper-Constantan (Type T): -200 °C to +350 °C - Iron-Constantan (Type J): -40 °C to +750 °C - Chromel-Alumel (Type K): -200 °C to +1250 °C - Platinum-Rhodium (Type R / Type S / Type B): Noble metal thermocouples capable of continuous measurement up to 1600 °C - 1700 °C (and short-term up to 1800 °C). Therefore, Platinum-Rhodium has the highest temperature measuring range.
A pn junction diode's dynamic conductance is directly proportional to: A) Applied voltage B) Thermal voltage C) Current D) Resistance
💡Correct Answer: Option C (Current)
The dynamic (or AC) resistance of a forward-biased p-n junction diode is given by rd=dIdV=IηVT, where VT is the thermal voltage and I is the diode forward current. Consequently, the dynamic conductance gd=rd1=ηVTI. Thus, dynamic conductance is directly proportional to the diode current I.
Q42
Electrical EngineeringElectronics - Diode Small-Signal Model
Which of the following statements is correct? Under small signal operation of a diode _______? A) Its bulk resistance increases B) It behaves as a clipper C) It acts like a closed switch D) Its junction resistance predominates
💡Correct Answer: Option D (Its junction resistance predominates)
In small-signal AC operation of a semiconductor diode, the diode is modeled as an incremental dynamic junction resistance (rj=IDQηVT) in series with the semiconductor ohmic bulk resistance (rb). At typical small-signal operational bias currents, the incremental non-linear barrier/junction resistance rj is significantly larger than the minute ohmic bulk resistance (rb≈1 to 2Ω), so the junction resistance predominates.
At 25°C, the collector-emitter voltage drop of a silicon transistor at saturation is approximately _______? A) 0.3 V B) 1.7 V C) 0.7 V D) 0.5 V
💡Correct Answer: Option A (0.3 V)
For a standard silicon bipolar junction transistor (BJT) operating in the saturation region at room temperature (25°C), the base-emitter saturation voltage VBE(sat) is approximately 0.7 V to 0.8 V, while the collector-emitter saturation voltage drop VCE(sat) is typically around 0.2 V to 0.3 V (standardly taken as 0.3 V in textbook problems).
Q44
Electrical EngineeringElectronics - Field Effect Transistors (FET)
Consider the following statements related to the Field effect transistor. I. Its operation depends upon the flow of the majority carriers only II. It has a high input resistance III. It is suitable for high frequency IV. Its operation depends upon the flow of both majority and minority carriers The correct statements is/are _______? A) I and II only B) I, II and III only C) II and IV only D) III and IV only
💡Correct Answer: Option B (I, II and III only)
Field Effect Transistors (FETs): - Statement I is correct: A FET is a unipolar device whose conduction relies exclusively on the flow of majority charge carriers (electrons in n-channel, holes in p-channel). - Statement II is correct: Due to reverse-biased gate junction (JFET) or insulated oxide gate dielectric (MOSFET), the input resistance is exceedingly high (108Ω to 1014Ω). - Statement III is correct: Because there is no minority carrier storage time during switching, FETs exhibit fast switching speeds and are well suited for high-frequency applications. - Statement IV is false (BJTs are bipolar, whereas FETs are strictly unipolar). Thus, statements I, II, and III are correct (Option B).
Q45
Electrical EngineeringElectronics - MOSFET Body Effect
Body effect in MOSFETs primarily results in A) Increase in the value of transconductance B) Decrease in the value of transconductance C) Change in the value of the threshold voltage D) Increase in the value of the output resistance
💡Correct Answer: Option C (Change in the value of the threshold voltage)
In MOSFETs, when a reverse bias voltage is applied between the source and substrate/body (VSB>0), the width of the depletion layer under the channel widens, requiring a larger gate-to-source voltage to achieve inversion. This substrate bias effect (body effect) alters the threshold voltage according to Vth=Vth0+γ(2ϕF+VSB−2ϕF). Thus, the primary direct impact is a change (increase for NMOS) in the threshold voltage.
Which of the following is NOT true for direct coupled amplifiers? A) Low cost B) Operating point shifts with temperature variations C) Less stable due to bias drift D) Can amplify high-frequency signals
💡Correct Answer: Option D (Can amplify high-frequency signals)
Direct-coupled amplifiers connect stages directly without coupling capacitors or transformers, making them low cost and uniquely capable of amplifying extremely low frequency and DC signals (zero frequency). However, at high frequencies, the parasitic capacitances of the transistors bypass the signal, making them unsuitable for high frequencies compared to tuned or RF amplifiers. Therefore, 'Can amplify high-frequency signals' is NOT true.
Consider the following circuits: I. Oscillator II. Emitter follower III. Power amplifier Which of the above circuit(s) employ feedback? A) I only B) II and III C) II only D) I and II
💡Correct Answer: Option D (I and II)
- Circuit I (Oscillator): Employs regenerative (positive) feedback to sustain oscillations without an external input signal. - Circuit II (Emitter follower): An unbypassed emitter resistor provides 100% negative voltage-series feedback, resulting in unity voltage gain, high input impedance, and low output impedance. - Circuit III (Power amplifier): Conventional basic power amplifiers (like Class A, B, or C stages) do not intrinsically rely on feedback for their basic amplification definition (unless feedback is intentionally added externally). Thus, circuits that fundamentally employ feedback are I and II (Option D).
Consider the following statements: I. Astable multivibrator can be used for generating a square wave II. Bistable multivibrator can be used for storing binary information Which of these statements is/are correct? A) I and II B) II only C) I only D) Neither I nor II
💡Correct Answer: Option A (I and II)
Both statements are correct: - Statement I: An astable multivibrator has no stable states and continuously alternates between two quasi-stable states, functioning as a free-running relaxation oscillator that produces a square or rectangular wave output. - Statement II: A bistable multivibrator (flip-flop / latch) has two stable states and can remain indefinitely in either state until triggered, making it the fundamental memory element for storing one bit of binary data.
An Op-Amp has a slew rate of 2 V/μsec. If the peak output is 12 V, what will be the power bandwidth? A) 36.5 KHz B) 22.5 KHz C) 26.5 KHz D) 12.5 KHz
💡Correct Answer: Option C (26.5 KHz)
The slew rate limited maximum frequency (power bandwidth) for an undistorted sinusoidal output of peak amplitude Vm is given by fmax=2πVmSR. Given SR=2 V/\mus=2×106 V/s and Vm=12 V: fmax=2×π×122×106=12π106=37.6991000000≈26525.8 Hz≈26.5 kHz.
Q50
Electrical EngineeringElectronics - Operational Amplifiers & Active Loads
Active load is primarily used in the collector of the differential amplifier of an OPAMP to _______? A) Increase the output resistance B) Increase the differential gain Ad C) Handle large signals D) Provide symmetry
💡Correct Answer: Option B (Increase the differential gain A_d)
In an operational amplifier differential input stage, using active loads (such as a current mirror active load) replaces passive load resistors with the high dynamic output impedance of a transistor. Because voltage gain is proportional to load resistance (Ad=−gmro), active loading provides an extremely large effective load resistance without requiring impractical silicon chip area or high DC supply voltages, thereby vastly increasing the open-loop differential gain Ad.
Q51
Electrical EngineeringDigital Electronics - Number Systems
The decimal equivalent of the hexadecimal number (BAD)16 is _______? A) 4739 B) 5929 C) 3416 D) 2989
💡Correct Answer: Option D (2989)
Converting the hexadecimal number (BAD)16 to decimal: In hexadecimal: B=11, A=10, D=13. Decimal value =(11×162)+(10×161)+(13×160) =(11×256)+(10×16)+(13×1) =2816+160+13=2989. Hence, (BAD)16=(2989)10.
In Boolean algebra, if F=(A+B)(A′+C), then A) F=AC+A′B B) F=AC+BC C) F=A′B+BC D) F=AA′+A′B+BC
💡Correct Answer: Option A (F = AC + A'B)
Expanding the product of sums: F=(A+B)(A′+C)=AA′+AC+BA′+BC. Since AA′=0, this becomes F=AC+A′B+BC. By the Boolean consensus theorem (or resolving consensus on A and A′), the term BC is redundant because BC=BC(A+A′)=ABC+A′BC. Thus, AC+ABC=AC and A′B+A′BC=A′B. Therefore, F=AC+A′B.
The output of the logic gate in the figure is: [Circuit description: A 2-input XOR gate with an inversion bubble at the output, representing an XNOR gate, where input 1 is variable A and input 2 is permanently connected to ground (0)]. A) 0 B) 1 C) A' D) A
💡Correct Answer: Option C (A')
The diagram displays a 2-input XNOR gate (an Exclusive-OR gate with an inversion bubble at the output). The truth function of a 2-input XNOR gate is F=A⊕B=AB+A′B′. Here, the lower input is tied to Ground, so B=0. Substituting B=0: F=A(0)+A′(1)=0+A′=A′. Thus, the output is A′.
The AND function can be realized by using only 'N' NOR gates. What is 'N' equal to? A) 2 B) 3 C) 4 D) 5
💡Correct Answer: Option B (3)
To implement a 2-input AND function (Y=AB) using only NOR gates: By De Morgan's laws: AB=A+B=NOR(A,B). 1. NOR gate 1 (with inputs tied together) produces A. 2. NOR gate 2 (with inputs tied together) produces B. 3. NOR gate 3 takes A and B as inputs, producing A+B=AB. Thus, exactly N=3 NOR gates are required to realize the AND function.
Q55
Electrical EngineeringDigital Electronics - De Morgan's Theorems
According to De Morgan's second theorem: A) A NAND gate is always complementary to an AND gate B) A NAND gate equivalent to a bubbled NAND gate C) A NAND gate is equivalent to a bubbled AND gate D) A NAND gate is equivalent to a bubbled OR gate
💡Correct Answer: Option D (A NAND gate is equivalent to a bubbled OR gate)
De Morgan's laws state: 1. A+B=A⋅B (A NOR gate is equivalent to a bubbled AND gate). 2. A⋅B=A+B (A NAND gate is equivalent to an OR gate with inverted/bubbled inputs, i.e., a bubbled OR gate). Therefore, according to De Morgan's second theorem, a NAND gate is equivalent to a bubbled OR gate.
The Boolean expression for the output Y in the logic circuit is: [Circuit description: Input A passes into a NOT gate whose output along with another branch of A feeds an XOR gate; the output of this XOR gate enters a 3-input NAND gate along with inputs B and C. Note: A is also fed directly into the 3-input gate at top rail]. A) AB'C B) A'BC C) ABC D) A'B'C'
💡Correct Answer: Option B (A'BC)
In this standard textbook logic puzzle: 1. Input A is inverted to produce A'. The XOR gate receives A and A', producing A⊕A′=1. 2. The second stage gate is a 3-input NAND gate with an active-low inverter configuration or an AND-gate arrangement with complemented output. Tracing the final output gives Y=A′BC.
The digital circuit shown below represents which of the following? [Circuit description: A D flip-flop with its inverted output Q' fed back to an XOR gate with external input X; the output of the XOR gate is connected to data input D: D=X⊕Q (or X⊕Q′), forming a toggle flip-flop]. A) T flip-flop B) JK flip-flop C) Clocked RS flip-flop D) Ring-counter
💡Correct Answer: Option A (T flip-flop)
Connecting an XOR gate to the D input of a D flip-flop with one input of the XOR gate fed by the output Q and the other by input T yields D=T⊕Q=TQ+TQ. For a D flip-flop, the next state is Qnext=D=T⊕Q. When T=0, Qnext=Q (no change); when T=1, Qnext=Q (toggle). This is the exact characteristic equation of a T (Toggle) flip-flop.
Consider the following statements: I. SRAM is made up of flip-flops II. SRAM stores a bit by maintaining a stable on/off state of its transistors III. DRAM has high speed and low density IV. DRAM is cheaper than SRAM Which of the above statements is/are correct? A) I, II and III B) I, II and IV C) II, III and IV D) I, III and IV
💡Correct Answer: Option B (I, II and IV)
- Statement I is correct: Static RAM (SRAM) memory cells are cross-coupled bistable flip-flops (typically 6 transistors per cell). - Statement II is correct: SRAM retains data by holding a stable on/off transistor state as long as power is applied without needing refresh cycles. - Statement III is incorrect: Dynamic RAM (DRAM) is slower than SRAM and has HIGH density (one transistor and one capacitor per bit cell), not low density. - Statement IV is correct: Because each DRAM cell requires only a single transistor and capacitor compared to 6 transistors for SRAM, DRAM is much cheaper per bit. Thus, statements I, II, and IV are correct (Option B).
Which one of the following is the software interrupt of an 8085 microprocessor? A) RST 7.5 B) INTR C) TRAP D) RST 7
💡Correct Answer: Option D (RST 7)
The 8085 microprocessor features five hardware interrupts: TRAP (non-maskable), RST 7.5, RST 6.5, RST 5.5, and INTR. In contrast, its software interrupts are the eight Restart instructions: RST 0, RST 1, RST 2, RST 3, RST 4, RST 5, RST 6, and RST 7 (vectored to addresses n×8, e.g., RST 7 vectors to 0038H). Thus, RST 7 is a software interrupt.
In the 8085 microprocessor, during the PUSH PSW operation, the stack pointer is: A) Decremented by one B) Incremented by one C) Decremented by two D) Incremented by two
💡Correct Answer: Option C (Decremented by two)
In the 8085 microprocessor, the stack grows downward in memory. The instruction PUSH PSW pushes the contents of the Program Status Word (the Accumulator and the Flag register) onto the stack. Since PSW is a 16-bit quantity (two 8-bit bytes: first Accumulator at SP−1, then Flags at SP−2), the 16-bit Stack Pointer (SP) register is decremented by two (SP←SP−2).
Q61
Electrical EngineeringPower Electronics - Power Semiconductor Devices
Match the two lists and choose the correct answer from the codes given below. List-I (w) MOSFET (x) GTO (y) UJT (z) TRIAC List-II (i) Turn off by a negative gate pulse (ii) Bi-directional switching (iii) High-speed switching (iv) Triggering circuit Options: A) (w)- (iii), (x)-(i), (y)-(iv), (z)-(ii) B) (w)- (ii), (x)-(i), (y)-(iii), (z)-(iv) C) (w)- (iv), (x)-(iii), (y)-(ii), (z)-(i) D) (w)- (iii), (x)-(i), (y)-(ii), (z)-(iv)
💡Correct Answer: Option A ((w)- (iii), (x)-(i), (y)-(iv), (z)-(ii))
Matching power semiconductor devices to their key characteristics: (w) Power MOSFET is a majority carrier unipolar device with negligible storage delay, providing High-speed switching -> (iii) (x) Gate Turn-Off Thyristor (GTO) can be turned off by applying a negative gate current pulse -> (i) (y) Uni-Junction Transistor (UJT) exhibits negative resistance and is standardly used in relaxation oscillator Triggering circuits for thyristors -> (iv) (z) TRIAC conducts in both directions across MT1 and MT2, providing Bi-directional switching -> (ii). Hence, the correct matching code is (w)-(iii), (x)-(i), (y)-(iv), (z)-(ii), which corresponds to Option A.
Which one of the following is the most suitable device for a DC-DC converter? A) BJT B) MOSFET C) GTO D) Thyristor
💡Correct Answer: Option B (MOSFET)
DC-DC converters (switch-mode choppers like Buck, Boost, and Buck-Boost converters) operate at high switching frequencies (typically tens to hundreds of kHz) to minimize the physical size, weight, and cost of inductors and filter capacitors. Power MOSFETs are voltage-controlled majority carrier devices that have very low switching losses and extremely fast turn-on/turn-off times, making them the most suitable and dominant switching device for low-to-medium voltage DC-DC converters.
An SCR is rated for 650 V PIV. What is the voltage (rms) for which the device can be operated if the voltage safety factor is 2? A) 240 V rms B) 325 V rms C) 650 V rms D) 230 V rms
💡Correct Answer: Option D (230 V rms)
The voltage safety factor (VSF) for a thyristor is defined as VSF=2×VrmsPIV=Peak operating supply voltagePeak repetitive reverse voltage. Given PIV=650 V and VSF=2: 2=2×Vrms650⟹Vrms=22650=1.4142325≈229.81 V≈230 V rms. Hence, the device can be safely operated on a 230 V rms AC supply.
Consider the following statements: Phase-controlled converters at small values of output voltage have I. Large harmonics in the utility system II. Poor power factor III. High efficiency IV. Notches in the line voltage waveform Which of the above statements is/are correct? A) I and II B) II, III and IV C) I, II and IV D) I and IV
💡Correct Answer: Option C (I, II and IV)
When phase-controlled converters operate at large firing angles α (corresponding to small output DC voltage Vdc=Vdocosα): - Statement I is correct: Current distortion increases, introducing substantial low-order harmonic currents into the utility AC source. - Statement II is correct: Displacement power factor is cosα, which approaches zero as α→90∘; combined with distortion factor, total power factor is extremely poor. - Statement III is incorrect: Efficiency is low due to large reactive power and harmonic heating losses. - Statement IV is correct: Commutation overlap (di/dt during thyristor commutation through source inductance) creates deep voltage notches in the supply line voltage waveform. Thus, statements I, II, and IV are correct (Option C).
The average output of a semi-converter connected to a 120 V, 50 Hz supply and firing angle of π/2 is: A) 54.02 V B) 56.02 V C) 108.04 V D) Zero
💡Correct Answer: Option A (54.02 V)
The average output voltage of a single-phase semi-converter (half-controlled bridge) is given by V0=πVm(1+cosα). Given supply voltage Vrms=120 V, peak voltage Vm=1202≈169.7056 V. For firing angle α=π/2=90∘, we have cos(90∘)=0. Therefore, V0=π1202(1+0)=3.14159169.7056≈54.018 V≈54.02 V.
The circuit shown in the figure below will work as which one of the following? [Circuit description: A DC-DC converter circuit consisting of input voltage V1, a controlled series switch T, an inductor L in the shunt branch connected to ground, a diode with cathode connected to the switch-inductor node and anode towards the parallel RC filter stage and load R, output voltage V2 with reversed polarity]. A) Buck-Boost converter B) Buck converter C) Boost converter D) Dual converter
💡Correct Answer: Option A (Buck-Boost converter)
In the shown topology, a controlled switch T is in series with the input, the inductor L is connected in the shunt (parallel) branch, and the diode points in reverse direction to feed the output capacitor and load. When switch T is ON, energy is stored in inductor L. When switch T is turned OFF, the inductor voltage reverses and discharges its stored energy through the diode into the load, producing an inverting output voltage V2=−1−DDV1. This is the standard inverting Buck-Boost converter topology.
PWM switching is preferred in voltage source inverters for the purpose of _______? A) Controlling output voltage B) Output harmonics C) Reducing filter size D) All of the above
💡Correct Answer: Option D (All of the above)
Pulse Width Modulation (PWM) techniques in Voltage Source Inverters (VSI) provide multiple critical engineering advantages: 1. Controlling output voltage by varying the modulation index (ma) without requiring a variable DC link voltage. 2. Minimizing lower-order output harmonics by shifting dominant harmonic energy to higher frequencies around the carrier frequency. 3. Reducing filter size because high-frequency ripple is easily attenuated using significantly smaller, lighter, and less costly L-C filter components. Therefore, 'All of the above' is correct.
In a three-phase VSI, with 120° conduction mode, how many switches conduct at a time? A) 1 B) 2 C) 3 D) 6
💡Correct Answer: Option B (2)
In a three-phase bridge inverter consisting of 6 switches: - In 180° conduction mode, each switch conducts for 180°, and at any instant exactly 3 switches are conducting simultaneously. - In 120° conduction mode, each switch conducts for 120°, followed by a 60° non-conduction interval for that phase leg. Consequently, at any given instant, exactly 2 switches (one from the positive group and one from the negative group of different legs) conduct simultaneously.
For a 1-phase full-bridge inverter fed from 48 V DC and connected to a load resistance of 2.4 ohms, the rms value of the fundamental component of the output voltage is _______? A) 20 V B) 21.6 V C) 43.2 V D) 34.4 V
💡Correct Answer: Option C (43.2 V)
For a single-phase full-bridge square-wave inverter fed from DC input Vs=48 V, the peak amplitude of the fundamental harmonic of output voltage is Vo1,max=π4Vs. The RMS value of this fundamental component is Vo1,rms=2Vo1,max=π24Vs=π22Vs≈0.9003Vs. Substituting Vs=48 V: Vo1,rms=π22×48=3.1415996×1.4142=3.14159135.76≈43.21 V≈43.2 V.
Q70
Electrical EngineeringElectrical Drives - Four-Quadrant DC Drives
For a low-speed, high-power reversible operation, the most suitable drive is: A) Current source inverter-fed AC drives B) Voltage source inverter-fed AC drives C) Dual converter-fed DC drives D) None of the above
💡Correct Answer: Option C (Dual converter-fed DC drives)
For heavy industrial applications requiring high power, low-speed operation, high starting torque, and seamless four-quadrant reversible operation (such as steel rolling mills, mine hoists, and cranes), Dual Converter-fed DC drives are the classic and most suitable drive. A dual converter comprises two back-to-back full converters that inherently provide bidirectional voltage and bidirectional current flow without mechanical contactors.
Q71
Electrical EngineeringElectrical Machines - Transformer Losses & Maximum Efficiency
A 2 kVA transformer has an iron loss of 150 watts and a full-load copper loss of 250 watts. The maximum efficiency of the transformer would occur when the total loss is _______? A) 500 W B) 275 W C) 450 W D) 300 W
💡Correct Answer: Option D (300 W)
In a transformer, maximum efficiency occurs at that fraction of load where the variable copper loss equals the constant iron loss: Pcu=Pi. Given iron loss Pi=150 W, the copper loss at the condition of maximum efficiency must also be 150 W. Consequently, the total losses at maximum efficiency are Ptotal=Pi+Pcu=150 W+150 W=300 W.
Q72
Electrical EngineeringElectrical Machines - Types of Transformers
Match the two lists and choose the correct answer from the codes given below. List-I (Transformer) (w) Power transformer (x) Auto transformer (y) Welding transformer (z) Isolation transformer List-II (Voltage ratio) (i) 230 V / 230 V (ii) 220 V / 240 V (iii) 400 V / 100 V (iv) 132 kV / 11 kV Options: A) (w)- (iv), (x)-(ii), (y)-(iii), (z)-(i) B) (w)- (ii), (x)-(i), (y)-(iii), (z)-(iv) C) (w)- (iv), (x)-(iii), (y)-(ii), (z)-(i) D) (w)- (iv), (x)-(ii), (y)-(i), (z)-(iii)
💡Correct Answer: Option A ((w)- (iv), (x)-(ii), (y)-(iii), (z)-(i))
Matching transformer types to representative voltage ratings: (w) Power transformer operates at high transmission voltage levels, such as 132 kV / 11 kV -> (iv) (x) Auto transformer is economically used where transformation ratio is close to unity, such as 220 V / 240 V -> (ii) (y) Welding transformer steps down voltage to low voltage and high current (drooping characteristic), such as 400 V / 100 V -> (iii) (z) Isolation transformer provides electrical safety isolation with a 1:1 transformation ratio, i.e., 230 V / 230 V -> (i). Hence, the correct match is (w)-(iv), (x)-(ii), (y)-(iii), (z)-(i), which is Option A.
Q73
Electrical EngineeringElectrical Machines - Transformer Equivalent Circuit & OC Test
Under what condition of the transformer will the equivalent circuit appear as shown below? [Circuit description: An AC voltage source V1 connected across only the parallel exciting branch consisting of core-loss resistance R01 in parallel with magnetizing reactance X01, with the series winding resistance and leakage reactance omitted/negligible and secondary open]. A) Short circuit B) Open circuit C) Under rated load D) Under load and no load
💡Correct Answer: Option B (Open circuit)
During the Open-Circuit (No-Load) test of a transformer, the secondary winding is left completely open-circuited (I2=0). The current drawn by the primary is only the small no-load exciting current I0 (2% to 5% of rated current). Because I0 is so small, the voltage drop across the primary series winding resistance and leakage reactance (R1 and X1) is negligible and ignored, leaving only the parallel magnetizing branch (R01∥X01) across the applied voltage V1.
Q74
Electrical EngineeringElectrical Machines - Transformer Maximum Efficiency
If P1 and P2 be the iron and copper losses of a transformer at full load, and the maximum efficiency of the transformer is at 75% of the full load, then what is the ratio of P1 and P2? A) 10/16 B) 9/16 C) 13/16 D) 3/16
💡Correct Answer: Option B (9/16)
Let x be the fraction of full load at which maximum efficiency occurs. Here x=75%=10075=43. The condition for maximum efficiency is: Iron Loss = Copper Loss at that load fraction P1=x2P2=(43)2P2=169P2. Therefore, the ratio of P1 to P2 is P2P1=169.
Q75
Electrical EngineeringElectrical Machines - Transformer Voltage Regulation
A transformer can exhibit negative voltage regulation under which load power factor condition? A) Only at zero power factor lagging B) At unity power factor C) At a lagging power factor (but not zero power factor) D) At the leading power factor for certain values
💡Correct Answer: Option D (At the leading power factor for certain values)
The approximate per-unit voltage regulation of a transformer is given by VR=Rpucosϕ±Xpusinϕ, where the negative sign applies to a leading power factor. Under leading power factor conditions (capacitive load), if Xpusinϕ>Rpucosϕ, the voltage regulation becomes negative (the secondary terminal voltage rises on load above the no-load value). Thus, negative voltage regulation occurs at leading power factor for certain values (specifically when tanϕ>XpuRpu).
Q76
Electrical EngineeringElectrical Machines - Parallel Operation of DC Generators
A pair of similar DC shunt generators operates in parallel and supplies to a common load. It is required to switch off Machine No. 1 and allow Machine No. 2 to supply the entire load. The following operations are to be used to achieve this: I. Switch off the main switch of machine No. 1 II. Reduce the field current of machine No. 1 III. Increase the field current of machine No. 2 IV. Ensure that machine No. 1 just floats The correct sequence of these operations is: A) II, IV, III, I B) II, III, IV, I C) IV, III, II, I D) III, II, IV, I
💡Correct Answer: Option B (II, III, IV, I)
To smoothly transfer load from Machine No. 1 to Machine No. 2 operating in parallel and then disconnect Machine 1 without sparking or disturbance to busbar voltage: 1. Gradually reduce the field current of Machine 1 (II) and simultaneously increase the field current of Machine 2 (III) to maintain constant busbar voltage while shifting load current to Machine 2. 2. Continue this adjustment until the ammeter of Machine 1 reads zero, i.e., ensuring Machine 1 just floats on the busbars (IV). 3. Open the circuit breaker / main switch of Machine 1 (I). Hence, the correct operational sequence is II, III, IV, I (Option B).
Q77
Electrical EngineeringElectrical Machines - DC Motor Armature Reaction
In a DC motor, what is the effect of armature reaction under load conditions? A) Strengthens the main field flux B) Reduces armature current C) Shifts the neutral plane in the direction of rotation D) Has no effect on commutation
💡Correct Answer: Option C (Shifts the neutral plane in the direction of rotation)
Note on standard DC machine theory: In a DC generator, armature reaction shifts the Magnetic Neutral Axis (MNA) forward in the direction of rotation. In a DC motor, the armature current direction is reversed with respect to rotation, so armature reaction shifts the MNA backward (opposite to rotation). However, among the four provided choices (A: Strengthens main flux [false], B: Reduces armature current [false], D: Has no effect [false]), C is the intended official answer based on the classic general neutral plane shift question phrasing in examination keys.
Q78
Electrical EngineeringElectrical Machines - DC Generator Losses & Efficiency
The maximum efficiency occurs in a separately excited DC generator when the terminal voltage is 220 V and the induced emf is 240 V. If the armature resistance is 0.2 Ω, then stray losses will be? A) 1000 W B) 2000 W C) 1500 W D) 4000 W
💡Correct Answer: Option B (2000 W)
In a DC generator: Eg=V+IaRa⟹240 V=220 V+Ia(0.2Ω)⟹Ia(0.2)=20 V⟹Ia=0.220=100 A. The armature copper loss at this condition is Pcu=Ia2Ra=(100)2×0.2=10000×0.2=2000 W. The condition for maximum efficiency in a separately excited generator is that variable armature copper loss equals constant/stray losses: Pstray=Pcu=2000 W. Thus, stray losses are 2000 W.
Q79
Electrical EngineeringElectrical Machines - Magnetic Axes in DC Machines
In DC machines, the field-flux axis and armature-mmf axis are, respectively, along A) Direct axis and indirect axis B) Direct axis and inter-polar axis C) Quadrature axis and direct axis D) Quadrature axis and inter-polar axis
💡Correct Answer: Option B (Direct axis and inter-polar axis)
In DC machines, the main field poles produce a magnetic flux aligned along the centerline of the pole cores, which is designated as the Direct axis (d-axis). The cross-magnetizing armature magnetomotive force (mmf) acts along the axis electrically perpendicular to the direct axis (at 90 electrical degrees), which coincides with the geometric interpolar space and interpolar poles, known as the inter-polar axis (or Quadrature axis). Thus, they are along the Direct axis and inter-polar axis, respectively.
Q80
Electrical EngineeringElectrical Machines - DC Machine EMF Equation
A DC machine operating at 750 rpm produces an induced emf of 220 V. If the machine runs at 700 rpm and the induced emf increases to 250 V, what is the percentage increase in the field flux? A) 17% B) 11.25% C) 42.4% D) 21.7%
💡Correct Answer: Option D (21.7%)
The induced emf in a DC machine is proportional to flux and rotational speed: E=kϕN⟹ϕ∝NE. Initial state: ϕ1∝750220=7522≈0.29333. Second state: ϕ2∝700250=7025=145≈0.35714. The ratio of fluxes is ϕ1ϕ2=220/750250/700=700×220250×750=70×2225×75=15401875≈1.21753. The percentage increase in field flux is (ϕ1ϕ2−1)×100%=(1.21753−1)×100%≈21.75%≈21.7%.
Q81
Electrical EngineeringElectrical Machines - Synchronous Motor V-Curves
The V-curve of a synchronous motor shows the variation of: A) Armature current and DC excitation at constant load B) Supply voltage and field current at constant excitation C) Power factor and supply voltage during shunting D) Supply voltage and excitation current at constant load
💡Correct Answer: Option A (Armature current and DC excitation at constant load)
The V-curves of a synchronous motor are plots of armature current (Ia) versus field/excitation current (If) at constant real power loads. At under-excitation, the motor operates at a lagging power factor and draws high armature current; as excitation increases, power factor reaches unity where armature current is minimized; with over-excitation, power factor becomes leading and current increases again, creating a characteristic V-shaped curve.
Q82
Electrical EngineeringElectrical Machines - Alternator Voltage Regulation
Which of the following methods gives a higher-than-actual value of voltage regulation in an alternator? A) ZPF method B) MMF method C) EMF method D) ASA method
💡Correct Answer: Option C (EMF method)
The EMF method (Synchronous Impedance method) assumes that magnetic saturation is absent and uses unsaturated synchronous reactance Xs, which treats armature reaction entirely as an equivalent leakage reactance voltage drop. Because the actual core saturates under loaded conditions, the calculated open-circuit voltage E0 and corresponding voltage regulation are significantly higher than the actual value. Hence, the EMF method is also known as the 'Pessimistic method'.
Match the two lists and choose the correct answer from the code given below: List-I (Machine) (w) DC motor (x) DC Generator (y) Alternator (z) Induction motor List-II (Graph) (i) Circle Diagram (ii) V-Curve (iii) Open Circuit Characteristics (iv) Speed-torque Characteristics Options: A) (w)- (iv), (x)-(ii), (y)-(iii), (z)-(i) B) (w)- (ii), (x)-(iv), (y)-(iii), (z)-(i) C) (w)- (iv), (x)-(iii), (y)-(ii), (z)-(i) D) (w)- (iv), (x)-(iii), (y)-(i), (z)-(ii)
💡Correct Answer: Option C ((w)- (iv), (x)-(iii), (y)-(ii), (z)-(i))
Matching machines with their characteristic operational diagrams: (w) DC motor is characterized by its Speed-torque Characteristics -> (iv) (x) DC Generator magnetization is represented by its Open Circuit Characteristics (OCC) -> (iii) (y) Alternator / Synchronous machine operation is depicted by V-Curves (armature current vs field current at constant power) -> (ii) (z) Induction motor performance across all operating regimes is determined using the Circle Diagram -> (i). Hence, the correct matching code is (w)-(iv), (x)-(iii), (y)-(ii), (z)-(i), which is Option C.
Q84
Electrical EngineeringElectrical Machines - Synchronous Motor Power & Torque
A 440 V, three-phase, 10-pole and 50 Hz synchronous motor delivering a torque of 50/π Nm delivers a power of: A) 50 W B) 500 W C) 1500 W D) 1000 W
💡Correct Answer: Option B (500 W)
Synchronous speed in rpm is Ns=P120f=10120×50=600 rpm. Angular velocity is ωs=602πNs=602π×600=20π rad/s. Mechanical power developed is P=T×ωs=(π50 Nm)×(20π rad/s)=50×20=1000 W. Hence, Option D (1000 W) is the correct answer.
A permanent magnet stepper motor with 8 poles in the stator and 6 poles in the rotor will have a step angle of? A) 14 degree B) 15 degree C) 30 degree D) 45 degree
💡Correct Answer: Option B (15 degree)
The step angle β of a stepper motor with stator poles Ns and rotor teeth/poles Nr is given by β=Ns×Nr∣Ns−Nr∣×360∘. Given Ns=8 and Nr=6: β=8×6∣8−6∣×360∘=482×360∘=241×360∘=15∘.
Q86
Electrical EngineeringElectrical Machines - Two-Phase AC Servomotor
The torque speed characteristic of a two phase induction motor is largely affected by A) Reactance-to-resistance ratio B) Speed C) Voltage D) Supply voltage frequency
💡Correct Answer: Option A (Reactance-to-resistance ratio)
In a two-phase AC servomotor (two-phase induction motor used in control systems), a linear drooping torque-speed characteristic with negative slope throughout the operational speed range is essential to prevent single-phasing and ensure positive damping. This desired characteristic is achieved by designing the rotor with a very high resistance, resulting in a low reactance-to-resistance ratio (X/R). Thus, the ratio of reactance to resistance (X/R) is the primary governing design parameter.
Q87
Electrical EngineeringElectrical Machines - Induction Motor Torque Equation
When the supply voltage for an induction motor is reduced by 10%, the maximum running torque will nearly _______? A) Decrease by 10% B) Increase by 20% C) Increase by 10% D) Decrease by 20%
💡Correct Answer: Option D (Decrease by 20%)
The maximum developed torque (breakdown torque Tmax) of an induction motor is directly proportional to the square of the supply voltage: Tmax∝V2. When the supply voltage is reduced by 10%, the new voltage is V′=0.9V. The new maximum torque becomes Tmax′∝(0.9V)2=0.81V2=0.81Tmax. The reduction in maximum running torque is (1−0.81)×100%=19%≈20%. Thus, it will nearly decrease by 20%.
Q88
Electrical EngineeringElectrical Machines - Induction Motor Torque Relations
The rated slip of an induction motor at full load is 5% while the ratio of starting current to full load current is 4. The ratio of starting torque to full load torque would be A) 0.6 B) 2.0 C) 0.8 D) 3.1
💡Correct Answer: Option C (0.8)
The relationship between starting torque and full-load torque in an induction motor is given by TflTst=(IflIst)2×sfl. Given IflIst=4 and full load slip sfl=5%=0.05: TflTst=(4)2×0.05=16×0.05=0.8.
For a given applied voltage and current, the speed of the universal motor will be A) Higher in AC excitation than in DC excitation B) Higher in DC excitation than in AC excitation C) Same in both DC and AC excitation D) Dangerously high in DC excitation
💡Correct Answer: Option B (Higher in DC excitation than in AC excitation)
A universal motor is a series wound motor designed to operate on either DC or single-phase AC. When operated on AC, the inductive reactance of the field and armature windings produces a significant reactive voltage drop (IX), which reduces the net back EMF available (Eb=V2−(IX)2−IR), leading to a lower operating speed. Under DC excitation, there is zero inductive reactance drop, so back EMF is higher (Eb=V−IR), resulting in a higher speed for the same terminal voltage and current.
Q90
Electrical EngineeringElectrical Machines - Speed Control of Induction Motors
Which of the following method(s) are suitable for the speed control of squirrel cage induction motors? I. Voltage control II. Rotor resistance control III. Frequency control IV. Pole changing method Select the correct answer using the codes given below: A) I, III and IV B) II, III and IV C) I, II and III D) II and IV
💡Correct Answer: Option A (I, III and IV)
In a squirrel cage induction motor, the rotor conductors are permanently short-circuited by end rings, meaning external resistance cannot be added to the rotor circuit (eliminating method II, Rotor resistance control, which is applicable only to slip-ring / wound rotor induction motors). Therefore, speed control of squirrel cage motors can only be implemented from the stator side via: (I) Stator voltage control, (III) Frequency / V/f control, and (IV) Pole changing method. Hence, I, III, and IV are suitable (Option A).
Q91
Electrical EngineeringElectrical Machines - Induction Motor Stability & Crawling
As shown in the figure below, the required load torque line intersects the resultant torque-speed characteristic of a 3-phase squirrel cage induction motor at points P, Q and R. Which is/are the stable operating point(s)? [Diagram description: A standard induction motor torque-speed curve with an initial dip showing cogging/crawling harmonics intersecting a horizontal required load torque line at point P (on positive slope region), point Q (on negative slope dip), and point R (on standard negative slope stable operating region near synchronous speed)]. A) P and Q B) Q and R C) Only R D) P and R
💡Correct Answer: Option D (P and R)
According to the criterion of steady-state electric drive stability, an operating point is stable if dωmdTL−dωmdTe>0. For a constant load torque (TL=const⟹dωmdTL=0), stability requires dωmdTe<0 (the motor torque must decrease with increasing speed, i.e., a negative slope with respect to speed). At point Q, the motor torque is increasing with speed (positive slope), making Q unstable. At points P and R, the characteristic has a negative slope with respect to speed (motor torque decreases as speed increases), satisfying the stability criterion. Hence, P and R are both stable operating points (point P corresponds to stable crawling, and point R to normal rated running).
Q92
Electrical EngineeringElectrical Machines - Induction Motor Power Flow
A 3-phase induction motor operating at a slip of 5% develops 20 kW rotor power output. What is the corresponding rotor copper loss in the operating condition? A) 750 W B) 850 W C) 1052 W D) 1200 W
💡Correct Answer: Option C (1052 W)
In an induction motor, the power flow ratio in the rotor is given by: Rotor input power (Pg) : Rotor copper loss (Pcu) : Mechanical power developed (Pm) = 1:s:(1−s). Given rotor power developed Pm=20 kW=20000 W and slip s=5%=0.05: PmPcu=1−ss⟹Pcu=1−ssPm=1−0.050.05×20000=0.950.05×20000=191×20000≈1052.63 W≈1052 W.
Consider the following statements: If a 3-phase squirrel cage induction machine operates at a slip of minus 0.05, then the machine will: I. Draw electrical power from the mains II. Draw mechanical power through the shaft III. Deliver electrical power to the mains Which of the above statements is/are correct? A) I, II and III B) I and II C) II and III D) I and III
💡Correct Answer: Option C (II and III)
When an induction machine is driven by a prime mover above synchronous speed (N>Ns), its slip becomes negative (s=−0.05). In this generating mode (Induction Generator): - Mechanical power is supplied into the rotor through the shaft by the prime mover (Statement II is correct). - The developed torque reverses, and the machine delivers active electrical power back into the AC supply mains (Statement III is correct). (Note: While it absorbs reactive power for excitation, active power is delivered to the mains, making statements II and III the standard description of generator operation). Thus, II and III are correct (Option C).
A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in a clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 ohms, then the effective rotor resistance in the backward branch of the equivalent circuit is _______? A) 2.0 ohm B) 4.0 ohm C) 78 ohm D) 156 ohm
💡Correct Answer: Option A (2.0 ohm)
Synchronous speed Ns=4120×50=1500 rpm. Forward slip sf=15001500−1425=150075=0.05. Backward slip sb=2−sf=2−0.05=1.95. In the double revolving field theory equivalent circuit, the rotor branch for backward field has effective resistance sb0.5r2 (or total branch resistance sbr2). With standstill rotor resistance r2=7.8Ω: sbr2=1.957.8=4.0Ω, and per revolving half-field it is sb0.5r2=1.950.5×7.8=2.0Ω. Therefore, the effective rotor resistance in the backward branch is 2.0Ω.
Q95
Electrical EngineeringElectrical Machines - Two-Phase AC Servomotor
The ratio of rotor reactance to the rotor resistance for a two-phase servo motor _______. A) Is equal to that of a normal induction motor B) Is greater than that of a normal induction motor C) May be less or greater than that of a normal induction motor D) Is less than that of a normal induction motor
💡Correct Answer: Option D (Is less than that of a normal induction motor)
In a two-phase AC servomotor, the rotor is intentionally constructed with high resistance (using thin drag cups or high-resistivity brass/aluminum conductors) to ensure that the maximum torque occurs at a high slip (sm>1). This design provides a strictly linear and negative torque-speed slope over the entire operating range and prevents single-phasing. Consequently, the rotor resistance R2 is much higher and the ratio of rotor reactance to resistance (X2/R2) is substantially lower than that of a conventional induction motor.
Q96
Electrical EngineeringPower Systems - Nuclear Power Generation
Match the two lists and choose the correct answer from the code given below. List-I (Material) (w) Deuterium (x) Heavy water (y) Uranium-235 (z) Thorium-232 List-II (Use) (i) A fissile fuel for a nuclear reactor (ii) A fertile fuel for a nuclear reactor (iii) Used as a fuel in a fusion reactor (iv) Used as a moderator in a nuclear reactor Options: A) (w)- (iv), (x)-(ii), (y)-(iii), (z)-(i) B) (w)- (ii), (x)-(iv), (y)-(iii), (z)-(i) C) (w)- (iii), (x)-(iv), (y)-(i), (z)-(ii) D) (w)- (iv), (x)-(iii), (y)-(i), (z)-(ii)
💡Correct Answer: Option C ((w)- (iii), (x)-(iv), (y)-(i), (z)-(ii))
Matching nuclear materials to applications: (w) Deuterium (heavy hydrogen isotope) is the primary fusion fuel in thermonuclear reactors (D−T or D−D fusion) -> (iii) (x) Heavy water (D2O) slows down fast neutrons and serves as a Moderator in PHWR/CANDU reactors -> (iv) (y) Uranium-235 undergoes thermal neutron fission directly and is a Fissile fuel -> (i) (z) Thorium-232 cannot undergo direct thermal fission but breeds fissile U-233 upon neutron capture, acting as a Fertile fuel -> (ii). Hence, the correct matching code is (w)-(iii), (x)-(iv), (y)-(i), (z)-(ii), corresponding to Option C.
Q97
Electrical EngineeringPower Systems - Reactive Power Compensation
In a 400 kV power system, the voltage measured at a 400 kV bus is 360 kV. Determine the reactive power absorbed by a shunt reactor rated at 50 MVAR, 400 kV, connected to this bus. A) 61.7 MVAR B) 40.5 MVAR C) 55.5 MVAR D) 46.5 MVAR
💡Correct Answer: Option B (40.5 MVAR)
The reactive power absorbed by a constant-impedance linear shunt reactor is directly proportional to the square of the operating bus voltage: Q∝V2⟹Qactual=Qrated×(VratedVactual)2. Given Qrated=50 MVAR, Vrated=400 kV, and operating voltage Vactual=360 kV: Qactual=50×(400360)2=50×(0.9)2=50×0.81=40.5 MVAR.
Q98
Electrical EngineeringPower Systems - Nuclear Reactor Control
In a nuclear reactor, chain reaction is controlled by introducing _______? A) Cadmium rods B) Iron rods C) Graphite rods D) Brass rods
💡Correct Answer: Option A (Cadmium rods)
In a nuclear reactor, control rods are inserted into or withdrawn from the reactor core to absorb excess neutrons and precisely regulate the fission chain reaction multiplication factor (k). Cadmium and Boron have exceptionally high neutron capture cross-sections and are standardly fabricated into control rods. (Graphite is used as a moderator to slow down neutrons, not as a control absorber rod). Thus, Cadmium rods are used.
Q99
Electrical EngineeringPower Systems - Hydroelectric & Pumped Storage Power Plants
Consider the following statements regarding the pumped storage plants: I. A pumped storage plant is a peak load plant II. The starting time of a pumped storage plant is very long III. Reversible turbines and pumps are very suitable for pumped storage plants IV. Pumped storage plants can be used for load frequency control Which of the above statements is/are correct? A) I and III B) I and II C) III and IV D) I, III and IV
💡Correct Answer: Option D (I, III and IV)
Analyzing pumped storage hydro plants: - Statement I is correct: Pumped storage plants operate as peak-load generating stations, discharging water during high-tariff peak demand hours and pumping it back during off-peak hours. - Statement II is incorrect: Hydroelectric and pumped storage units have very quick start-up times (typically 1 to 3 minutes), not long start-up times. - Statement III is correct: Reversible Francis pump-turbines coupled to motor-generators are standard, efficient, and cost-effective. - Statement IV is correct: Due to rapid governor responsiveness and dispatchability, pumped storage plants provide excellent spinning reserve and automatic load-frequency control (LFC). Thus, statements I, III, and IV are correct (Option D).
Q100
Electrical EngineeringPower Systems - Nuclear Power Plant Instrumentation & Safety
A microprocessor-based safety control system installed in a nuclear power plant must be stress tested under which of the following conditions? I. Ageing due to radiation II. Thermal stresses III. Seismic vibration Which of the above statements is/are correct? A) I and II only B) II and III only C) I and III only D) All of the above
💡Correct Answer: Option D (All of the above)
Safety-grade Instrumentation and Control (I&C) systems in nuclear facilities must undergo comprehensive Environmental Qualification (EQ) testing (per IEEE 323 and IEEE 344 standards). This qualification mandates rigorous testing for: (I) Radiation-induced ageing and total ionizing dose effects, (II) Harsh environmental thermal stresses and temperature/humidity cycling, and (III) Dynamic seismic shock and vibration shaking table tests to ensure fail-safe shutdown during earthquakes. Therefore, 'All of the above' is correct.
Q101
Electrical EngineeringPower Systems - Hydropower Engineering & Hydrology
With reference to a hydropower station, the graphical representation of the discharge as a function of time is known as: A) Monograph B) Load duration curve C) Hydrograph D) Waterfilling curve
💡Correct Answer: Option C (Hydrograph)
A Hydrograph is the fundamental graphical plot showing stream discharge / river water flow rate (Q in m3/s) as a function of chronological time (hours, days, or months). In contrast, a Flow Duration Curve sorts discharges in descending order, and a Load Duration Curve plots electrical power demand against time.
Q102
Electrical EngineeringPower Systems - Hydroelectric Generation
Assuming 100% efficiency and taking the density of water as 1000 kg/m³, what is the power output of a hydroelectric generator unit operating with a head of 1.0 m and a discharge rate of 1.0 cubic meters per second? A) 2.91 kW B) 9.81 kW C) 19.50 kW D) 6.95 kW
💡Correct Answer: Option B (9.81 kW)
Theoretical water power output is given by P=ρ⋅g⋅Q⋅H⋅η. Given water density ρ=1000 kg/m3, gravitational acceleration g=9.81 m/s2, discharge Q=1.0 m3/s, head H=1.0 m, and efficiency η=1.0 (100%): P=1000×9.81×1.0×1.0×1.0=9810 W=9.81 kW.
Q103
Electrical EngineeringPower Systems - Insulation Coordination & EHV Lines
The insulation of modern EHV lines is designed based on A) Switching voltage B) Corona C) Radio interference D) Lightning
💡Correct Answer: Option A (Switching voltage)
For transmission systems operating at extra-high voltages (EHV and UHV, typically 400 kV and above), insulation levels and tower clearances are dictated by transient switching surges (internal switching overvoltages) rather than external lightning overvoltages. Lightning overvoltages are roughly independent of system voltage and dominate insulation coordination only up to 220 kV. Beyond 300 kV-400 kV, switching overvoltages become the decisive design criterion.
Q104
Electrical EngineeringPower Systems - Transmission Line Impedance & Parallel Lines
Consider two parallel short transmission lines of impedances ZA and ZB, respectively, as shown in the figure below. Currents IA and IB both are lagging, and the sending-end voltage is Vs. If the reactance to resistance ratio of both the impedances ZA and ZB are equal, then the total current I will be _______? [Circuit description: Parallel lines between sending end Vs and receiving end VR, branch A carrying IA through ZA and branch B carrying IB through ZB, total current I=IA+IB]. A) Lag both IA and IB B) Lead both IA and IB C) Lag one of IA and IB D) Both IA and IB in phase
💡Correct Answer: Option D (Both I_A and I_B in phase)
Since both parallel lines share the identical voltage drop ΔV=Vs−VR, the branch currents are IA=ZAΔV and IB=ZBΔV. The impedance angles are given by θA=tan−1(RAXA) and θB=tan−1(RBXB). Given that the reactance-to-resistance ratios are equal (RAXA=RBXB), the impedance phase angles are identical (θA=θB). Consequently, IA and IB lag the voltage drop ΔV by the exact same angle θ, meaning IA and IB are strictly in phase with each other, and total current I=IA+IB is in phase with both IA and IB.
Q105
Electrical EngineeringPower Systems - Mechanical Design of Transmission Lines
Galloping in transmission line conductors arises generally due to A) Adoption of horizontal conductor configurations B) Heavyweight of the line conductors C) Asymmetrical layers of ice formation D) None of the above
💡Correct Answer: Option C (Asymmetrical layers of ice formation)
Conductor galloping is a high-amplitude (several meters), low-frequency (0.1 to 1 Hz) wind-induced aerodynamic oscillation of overhead transmission conductors. It occurs when ice coats the windward side of the round conductor asymmetrically, creating an aerodynamic airfoil shape. Strong crosswinds acting on this non-circular, iced aerodynamic profile generate unstable aerodynamic lift forces that trigger large vertical galloping oscillations.
Q106
Electrical EngineeringPower Systems & EM Theory - Travelling Waves & Reflection Coefficient
The reflection coefficient for the transmission line at point P, shown in the figure below, is? [Diagram description: A transmission line with surge impedance Z0=400Ω terminating at point P with a shunt-connected grounding resistive load of RL=400Ω]. A) 1 B) 0 C) 1.5 D) 0.5
💡Correct Answer: Option B (0)
The voltage reflection coefficient at a load termination is given by Γ=ZL+Z0ZL−Z0. Here, the transmission line has surge impedance Z0=400Ω and terminates into a matched resistive load ZL=400Ω. Substituting these values gives Γ=400+400400−400=8000=0. Thus, there is zero reflection (perfect impedance match).
Q107
Electrical EngineeringPower Systems & EM Theory - Transmission Line Electrical Length
A 100 MHz signal is transmitted over a line with propagation speed v=2×108 m/s. Find the electrical length of a 1.5-meter line (in degrees)? A) 270 degree B) 298 degree C) 265 degree D) 115 degree
💡Correct Answer: Option A (270 degree)
Signal wavelength is λ=fv=100×106 Hz2×108 m/s=2.0 m. The phase propagation constant is β=λ2π=22π=π rad/m=180∘/m. The electrical length for a line of physical length l=1.5 m is θ=β⋅l=(180∘/m)×1.5 m=270∘.
Q108
Electrical EngineeringPower Systems & EM Theory - Smith Chart & VSWR
A point located on the outer edge of a Smith chart represents: A) Infinity VSWR B) Infinite Gain C) Zero VSWR D) Frequency outside the desired band
💡Correct Answer: Option A (Infinity VSWR)
The boundary (outer circular perimeter) of a standard Smith chart corresponds to the circle of reflection coefficient magnitude ∣Γ∣=1. The Voltage Standing Wave Ratio is defined as S=1−∣Γ∣1+∣Γ∣. When ∣Γ∣=1, the denominator vanishes, resulting in S=∞ (Infinity VSWR). This boundary represents purely reactive terminations (pure inductors, capacitors, open circuits, or short circuits) where total reflection occurs.
Q109
Electrical EngineeringPower Systems - Travelling Waves & Refraction
A travelling wave due to lightning, with an incident voltage V, travels along an overhead line with a surge impedance of 400 Ω and reaches a cable with a surge impedance of 40 Ω. What is the voltage that enters the cable at the junction? A) 1/11 V B) 4/11 V C) 1 V D) 2/11 V
💡Correct Answer: Option D (2/11 V)
When an incident voltage wave V travelling along a line of surge impedance Z1=400Ω strikes a junction connected to a cable of surge impedance Z2=40Ω, the transmitted (refracted) voltage wave V′′ entering the cable is given by the transmission coefficient: V′′=Z1+Z22Z2V=400+402×40V=44080V=112V.
Q110
Electrical EngineeringPower Systems - Surge Impedance Loading (SIL)
What is the surge impedance loading of a lossless 400 kV, 3-phase, 50 Hz overhead transmission line with an average surge impedance of 400 ohms? A) 400 MW B) 4003 MW C) 400 kW D) 4003 kW
💡Correct Answer: Option A (400 MW)
Surge Impedance Loading (SIL) of a three-phase transmission line is defined as SIL=ZcVL2, where VL is the rated line-to-line operating voltage in kV and Zc is the surge impedance in ohms. Given VL=400 kV and Zc=400Ω: SIL=400(400)2=400 MW.
Q111
Electrical EngineeringPower Systems - Transmission Line Performance & Tuned Lines
When the sending end voltage and current are numerically equal to the receiving end voltage and current, respectively, then the line is called _______? A) A transposed line B) A tuned line C) A long line D) A short line
💡Correct Answer: Option B (A tuned line)
When the sending end voltage and current are numerically equal to the receiving end voltage and current, respectively, then the line is called _______? A transmission line whose distributed inductance and capacitance parameters (or auxiliary terminal reactive networks) are chosen such that ∣Vs∣=∣Vr∣ and ∣Is∣=∣Ir∣ regardless of the line length or load is termed a 'Tuned power line'. In a tuned line, the natural propagation angle corresponds to an integer multiple of wavelengths (or quarter/half wave tuning), ensuring that voltage and current profiles at the receiving end are identically reproduced at the sending end.
Q112
Electrical EngineeringPower Systems - Distribution Systems
Why is a ring main distribution system preferred over a radial system? I. Voltage drop in the feeder is less II. Power factor is higher III. Supply is more reliable Select the correct answer using the code given below: A) I and III B) II and III C) I and II D) All of the above
💡Correct Answer: Option A (I and III)
Comparing distribution system topologies: - Statement I is correct: In a ring main system, consumers are fed from two parallel paths around the closed loop, which halves the effective feeder impedance and significantly reduces the total voltage drop. - Statement II is incorrect: Power factor is entirely determined by consumer load impedances and is not increased by changing the distribution feeder topology. - Statement III is correct: In the event of a fault in any feeder section, isolators isolate the faulty segment while consumers continue to be fed from the alternate path around the ring, providing much greater supply reliability. Thus, statements I and III are correct (Option A).
Q113
Electrical EngineeringPower Systems - Overhead Line Insulators & String Efficiency
The unequal voltage distribution across the units in a suspension-type insulator string is caused by: A) Unequal self-capacitance of the units B) Non-uniform distance of separation of the units from the tower body C) Non-uniform distance between the cross-arm and the units D) The existence of stray capacitance between the metallic junction of units and the tower body
💡Correct Answer: Option D (The existence of stray capacitance between the metallic junction of units and the tower body)
In a suspension insulator string, all disc insulator units possess identical mutual/self-capacitance (C). However, each metallic pin/cap linking adjacent discs forms a shunt/stray capacitance (C1=kC) to the earthed grounded steel tower body and cross-arm. Because charging currents leak through these shunt capacitances to earth, the current flowing down through successive discs is not uniform, causing the disc closest to the energized line conductor to carry the largest current and experience the greatest voltage stress.
Q114
Electrical EngineeringPower Systems - Automatic Load Frequency Control (ALFC)
In a load-frequency control system with free governor action, an increase in load demand under steady-state conditions is accommodated by: A) Only by increasing the generator excitation to compensate for the load demand B) Only by decreasing the load demand due to a drop in system frequency C) Partly by increased generation and partly by a decrease in load demand D) Partly by increased generation and partly by increased generator excitation
💡Correct Answer: Option C (Partly by increased generation and partly by a decrease in load demand)
Under free governor operation (uncontrolled primary speed droop control), when an incremental load step ΔPL occurs, the system frequency drops by Δf. This drop produces two balancing steady-state actions: 1. The turbine speed governors sense the speed decline and open steam/water valves, increasing turbine generation by ΔPG=−RΔf. 2. Frequency-dependent consumer loads (motors, pumps) naturally reduce their power absorption by ΔPD=DΔf. Thus, the load increase is accommodated partly by increased generation and partly by a decrease in load demand (due to frequency sensitivity of loads).
Q115
Electrical EngineeringPower Systems - Fault Analysis & Symmetrical Components
At the point of fault, the positive sequence voltage component becomes zero during which type of fault? A) L-L-G fault B) L-L-L fault C) L-L fault D) L-G fault
💡Correct Answer: Option B (L-L-L fault)
During a symmetrical three-phase short-circuit fault (L-L-L or L-L-L-G fault) at the fault point, all three phase voltages collapse identically to zero (Va=Vb=Vc=0). By symmetrical component transformation, the positive sequence voltage is Va1=31(Va+αVb+α2Vc)=0. For all unsymmetrical faults (L-G, L-L, L-L-G), the positive sequence voltage at the fault point remains non-zero. Thus, Va1=0 strictly occurs during a three-phase symmetrical (L-L-L) fault.
Which of the following is a key advantage of using SF6 gas in circuit breakers? A) Low cost B) Non-toxic C) Low pressure requirement D) Excellent insulating and arc-quenching properties
💡Correct Answer: Option D (Excellent insulating and arc-quenching properties)
Sulfur hexafluoride (SF6) is an electronegative gas that possesses extraordinary dielectric strength (approx. 2.5 to 3 times that of air at atmospheric pressure) and phenomenal arc-extinguishing capability (approx. 100 times better than air). Its rapid ability to capture free electrons forming heavy negative ions enables near-instantaneous post-arc dielectric recovery at current zero. Hence, 'Excellent insulating and arc-quenching properties' is the primary advantage.
Which type of protection is provided on a generator to protect against stator insulation failure? A) Differential protection B) Thermocouple actuated alarm C) Reverse power relay D) Overcurrent relay
💡Correct Answer: Option A (Differential protection)
Stator insulation failure leads to phase-to-phase and phase-to-earth short-circuit winding faults within the generator stator. Merz-Price circulating current differential protection (biased percentage differential relaying) is the universal, standard primary protection employed for alternator stator windings to instantaneously isolate internal insulation breakdowns while remaining completely immune to external through-faults.
Q118
Electrical EngineeringPower Systems - Earthing & Electrical Safety
Standard wire gauge used for earthing lead should not be thinner than _______? A) 12 SWG wire B) 16 SWG wire C) 8 SWG wire D) 14 SWG wire
💡Correct Answer: Option C (8 SWG wire)
According to Indian Electricity (IE) Rules and IS 3043 (Code of Practice for Earthing), the cross-sectional area of an earth continuity conductor and main earth lead must be sufficiently thick to carry maximum prospective fault currents without fusing or excessive heating. Under standard electrical wiring rules, the earth continuity lead should not be smaller/thinner than 8 SWG (Standard Wire Gauge) galvanized iron/copper wire.
Q119
Electrical EngineeringUtilization of Electrical Energy - Illumination
A 250 V lamp has 20 lm/W and draws a current of 0.6 A. Its total flux in lumens is _______? A) 150 B) 3000 C) 3500 D) 37.5
💡Correct Answer: Option B (3000)
The electrical power consumed by the lamp is P=V×I=250 V×0.6 A=150 W. Given luminous efficacy of the lamp ηlum=20 lumens/Watt: Total luminous flux emitted is Φ=P×ηlum=150 W×20 lm/W=3000 lumens.
Q120
Electrical EngineeringUtilization of Electrical Energy - Electric Heating
Recognise the heating type shown in the given figure? [Figure description: Electrodes immersed into a container where high-resistive powder is placed over the charge, connected to DC or AC supply, heating the charge through resistance of the powder]. A) Direct induction heating B) Indirect resistance heating C) Indirect arc heating D) Direct resistance heating
💡Correct Answer: Option B (Indirect resistance heating)
In indirect resistance heating, the current passes through a separate high-resistance heating element or high-resistive powder placed above or around the charge. The heat generated in this high-resistive layer is transferred to the charge beneath it via conduction and radiation. Because the electric current does not pass directly through the charge material itself (as it would in direct resistance heating), this process is classified as Indirect Resistance Heating.
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About JKSSB Junior Engineer (Electrical) 2025 Previous Year Paper
This page provides the full solved question paper for the JKSSB Junior Engineer (Electrical) examination conducted in 2025. Every MCQ is presented with verified answer keys and detailed bilingual explanations to support concept building.
Key subjects covered in this paper include Electrical Engineering. Practicing authentic previous year questions is the proven way to master question patterns and improve accuracy for upcoming exams across Jammu & Kashmir.
