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Electrical Engineering Solved Questions Bank & Study Material (2026) - Apex Rankers
Engineering & Technology | Topic-Wise Comprehensive Solved Pool
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
The synchronous speed (Ns) of a 3-phase, 4-pole induction motor connected to a 50 Hz AC supply is:
A
3000 RPM
B
1500 RPM
✓ Correct
C
1440 RPM
D
750 RPM
💡
Step-by-Step Explanation & Concept Rationale
Synchronous speed Ns = 120 * f / P = (120 * 50) / 4 = 6000 / 4 = 1500 RPM. The actual rotor speed N will be slightly less (e.g., 1440 RPM) due to slip.
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
The slip (s) of a 3-phase induction motor operating at standstill (rotor locked) is:
A
s = 0
B
s = 1.0 (100%)
✓ Correct
C
s = -1.0
D
s = 0.04
💡
Step-by-Step Explanation & Concept Rationale
Slip is defined as s = (Ns - N) / Ns. At standstill, rotor speed N = 0, so s = (Ns - 0) / Ns = 1.0 (or 100%). At synchronous speed (N = Ns), slip is zero.
Q. 3
Electrical Engineering
Difficulty: Easy
(1 Mark)
The frequency of rotor induced currents (f_r) in an induction motor running at slip s with stator supply frequency f is:
A
f_r = f
B
f_r = s * f
✓ Correct
C
f_r = f / s
D
f_r = (1 - s) * f
💡
Step-by-Step Explanation & Concept Rationale
Rotor EMF is induced by relative motion between the rotating stator field (Ns) and rotor (N). Since relative speed is s * Ns, rotor frequency is f_r = s * (P * Ns / 120) = s * f. At full load (s ≈ 0.03-0.05 on 50 Hz), f_r is only 1.5 to 2.5 Hz.
Q. 4
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a 3-phase induction motor, maximum torque (breakdown torque) occurs at a slip s_max equal to:
A
s_max = R2 / X2
✓ Correct
B
s_max = X2 / R2
C
s_max = R2^2 / X2
D
s_max = sqrt(R2 / X2)
💡
Step-by-Step Explanation & Concept Rationale
From maximum power transfer principles applied to the rotor equivalent circuit, maximum developed torque occurs when rotor resistance equals stand-still rotor leakage reactance: R2 = s * X2, giving s_max = R2 / X2.
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
The magnitude of maximum developed torque in a 3-phase induction motor is:
A
Directly proportional to rotor resistance R2
B
Independent of rotor circuit resistance R2
✓ Correct
C
Inversely proportional to supply voltage squared
D
Directly proportional to slip
💡
Step-by-Step Explanation & Concept Rationale
Substituting s_max = R2/X2 into the torque equation reveals that T_max = (3 / (2 * omega_s)) * (V1^2 / (2 * X2)). Thus, T_max depends on V1^2 and X2, but is completely independent of rotor resistance R2 (R2 only shifts the slip at which T_max occurs).
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
Rotor skewing in squirrel-cage induction motors is employed to:
A
Prevent cogging (magnetic locking), crawling, and reduce acoustic magnetic hum
✓ Correct
B
Increase maximum breakdown torque
C
Eliminate stator copper losses
D
Provide regenerative braking
💡
Step-by-Step Explanation & Concept Rationale
Skewing rotor conductor bars slightly along the shaft axis prevents stator and rotor tooth harmonics from locking together (cogging) during startup, suppresses harmonic synchronous torques causing crawling at 1/7th speed, and lowers electromagnetic humming noise.
Q. 7
Electrical Engineering
Difficulty: Hard
(1 Mark)
Crawling in a 3-phase squirrel cage induction motor is primarily caused by:
A
High supply voltage transients
B
Harmonic space fields produced by stator winding, predominantly the 7th harmonic
✓ Correct
C
Rotor bar open circuits
D
Unbalanced line currents
💡
Step-by-Step Explanation & Concept Rationale
Stator space harmonics produce parasitic rotating fields. The 7th space harmonic rotates forward at Ns/7. Superimposing this on the fundamental torque-speed curve creates a stable dip where the motor can 'crawl' stably at slightly below 1/7th synchronous speed.
Q. 8
Electrical Engineering
Difficulty: Medium
(1 Mark)
Cogging (magnetic locking) in an induction motor occurs when:
A
Supply frequency is reduced by 50%
B
The number of stator slots is equal to or an integral multiple of the number of rotor slots
✓ Correct
C
Rotor resistance is equal to leakage reactance
D
The motor runs at 150% full load
💡
Step-by-Step Explanation & Concept Rationale
When stator slot count S1 equals rotor slot count S2 (or an integer ratio), the stator and rotor teeth align perfectly at standstill, creating minimum reluctance paths that magnetically lock the rotor, preventing it from starting.
Q. 9
Electrical Engineering
Difficulty: Medium
(1 Mark)
A Star-Delta starter reduces the starting current of a 3-phase induction motor compared to Direct-On-Line (DOL) starting by a factor of:
A
1 / 2 (50%)
B
1 / sqrt(3) (57.7%)
C
1 / 3 (33.3%)
✓ Correct
D
1 / 4 (25%)
💡
Step-by-Step Explanation & Concept Rationale
In star connection, phase voltage is V_line / sqrt(3). Since current per phase is proportional to voltage, phase current is reduced by 1/sqrt(3). The line current in star is equal to phase current, so line starting current is reduced to (1/sqrt(3)) * (1/sqrt(3)) = 1/3 (33.3%) of DOL delta starting current.
Q. 10
Electrical Engineering
Difficulty: Easy
(1 Mark)
A wound-rotor (slip ring) induction motor allows inserting external resistance into the rotor circuit during startup to:
A
Increase starting torque while simultaneously reducing starting inrush current
✓ Correct
B
Decrease starting torque
C
Increase synchronous speed
D
Operate without a rotating magnetic field
💡
Step-by-Step Explanation & Concept Rationale
Adding external rotor resistance shifts the maximum torque peak toward standstill (s_max = R2/X2 = 1.0), achieving maximum possible starting torque (T_start = T_max) while increasing rotor impedance to choke starting current.
Q. 11
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a single-phase induction motor, why is an auxiliary winding and capacitor required for starting?
A
A single-phase pulsating stator field produces two equal and opposite rotating magnetic fields (Double Revolving Field Theory), resulting in zero net starting torque
✓ Correct
B
Single-phase motors have higher core loss than 3-phase motors
C
To step up the supply voltage
D
To avoid DC saturation
💡
Step-by-Step Explanation & Concept Rationale
According to Double Revolving Field Theory, a pulsating 1-phase field decomposes into two fields rotating in opposite directions at synchronous speed. At standstill, both produce equal forward and backward torques, yielding zero net starting torque. The capacitor creates a 90-degree phase shift to produce a net rotating field.
Q. 12
Electrical Engineering
Difficulty: Easy
(1 Mark)
Which single-phase motor has the highest starting torque?
A
Shaded-pole motor
B
Capacitor-start, capacitor-run motor
✓ Correct
C
Split-phase resistance-start motor
D
Reluctance motor
💡
Step-by-Step Explanation & Concept Rationale
Capacitor-start capacitor-run motors utilize a high electrolytic capacitor for starting (giving nearly 90-degree phase split and 300-400% starting torque) and a smaller run capacitor for optimal running efficiency and power factor.
Q. 13
Electrical Engineering
Difficulty: Medium
(1 Mark)
A synchronous generator (alternator) operating at lagging power factor has an armature reaction that is:
A
Purely cross-magnetizing
B
Partly cross-magnetizing and predominantly demagnetizing
✓ Correct
C
Purely magnetizing
D
Zero
💡
Step-by-Step Explanation & Concept Rationale
At lagging power factor, the current lags induced EMF. Resolving armature MMF shows that the in-phase component is cross-magnetizing (distorting field) while the lagging reactive component directly opposes the main field (demagnetizing), reducing terminal voltage.
Q. 14
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a synchronous generator operating at leading power factor, the armature reaction is:
A
Demagnetizing
B
Partly cross-magnetizing and predominantly magnetizing
✓ Correct
C
Purely resistive
D
Zero
💡
Step-by-Step Explanation & Concept Rationale
At leading power factor, armature current leads induced EMF. The reactive leading current produces an MMF that directly aids and reinforces the main rotor pole flux (magnetizing effect), causing terminal voltage to rise with load.
Q. 15
Electrical Engineering
Difficulty: Medium
(1 Mark)
The pitch factor (or chording factor) Kp of a fractional-pitch armature winding short-pitched by an angle alpha is:
A
Kp = sin(alpha / 2)
B
Kp = cos(alpha / 2)
✓ Correct
C
Kp = cos(alpha)
D
Kp = 1 / cos(alpha / 2)
💡
Step-by-Step Explanation & Concept Rationale
Pitch factor is the ratio of phasor sum to arithmetic sum of coil-side voltages: Kp = 2*E*cos(alpha/2) / (2*E) = cos(alpha / 2). Chording eliminates specific unwanted harmonics (e.g., short-pitching by 60 deg eliminates the 3rd harmonic since cos(3 * 60/2) = cos(90) = 0).
Q. 16
Electrical Engineering
Difficulty: Hard
(1 Mark)
The distribution factor (or breadth factor) Kd for a distributed armature winding with m slots per pole per phase and slot pitch gamma is:
A
Kd = sin(m * gamma / 2) / [m * sin(gamma / 2)]
✓ Correct
B
Kd = m * sin(gamma / 2) / sin(m * gamma / 2)
C
Kd = cos(m * gamma / 2) / m
D
Kd = sin(gamma / 2) / [m * sin(m * gamma / 2)]
💡
Step-by-Step Explanation & Concept Rationale
Distributing winding coils into multiple slots per pole per phase produces EMFs that are out of phase by slot angle gamma. Their phasor sum yields Kd = sin(m * gamma / 2) / [m * sin(gamma / 2)], which effectively suppresses harmonic voltages to generate a clean sinusoidal output.
Q. 17
Electrical Engineering
Difficulty: Medium
(1 Mark)
The Short Circuit Ratio (SCR) of a synchronous generator is defined as:
A
Short-circuit current / Rated full-load current
B
Field current required for rated open-circuit voltage / Field current required for rated short-circuit current (If_oc / If_sc)
✓ Correct
C
Direct-axis reactance / Quadrature-axis reactance
D
Synchronous speed / Rated rotor speed
💡
Step-by-Step Explanation & Concept Rationale
SCR = If_oc / If_sc = 1 / Xd_sat(pu). A high SCR indicates a larger air gap, lower synchronous reactance Xd, better steady-state stability, superior voltage regulation, and higher short-circuit capacity, but requires a physically larger and more expensive machine.
Q. 18
Electrical Engineering
Difficulty: Easy
(1 Mark)
The 'V-curves' of a synchronous motor illustrate the relationship between:
A
Terminal voltage and rotor speed
B
Armature current (Ia) and DC Field excitation current (If) at various constant power loads
✓ Correct
C
Power factor and frequency
D
Torque and slip
💡
Step-by-Step Explanation & Concept Rationale
V-curves plot armature current Ia versus field current If. For a given mechanical load, under-excitation causes lagging power factor and high Ia. Increasing If brings Ia to a minimum at unity power factor. Over-excitation causes leading power factor and high Ia, tracing a characteristic V-shape.
Q. 19
Electrical Engineering
Difficulty: Easy
(1 Mark)
An over-excited synchronous motor operating on no-load acts as a:
A
Inductive reactor (choke coil)
B
Synchronous Condenser, supplying leading reactive power (VARs) to improve system power factor
✓ Correct
C
Universal motor
D
High-slip asynchronous generator
💡
Step-by-Step Explanation & Concept Rationale
When over-excited (Eb > V), a synchronous motor draws current that leads the terminal voltage by almost 90 degrees, acting as a variable capacitor bank known as a Synchronous Condenser to support grid voltage and correct lagging power factor.
Q. 20
Electrical Engineering
Difficulty: Easy
(1 Mark)
Hunting (phase swinging) in synchronous machines is suppressed by installing:
A
Damper windings (amortisseur windings) in pole shoe faces
✓ Correct
B
Flywheels on the exciter shaft only
C
Higher resistance field coils
D
Capacitors in series with the field
💡
Step-by-Step Explanation & Concept Rationale
Damper windings are heavy copper bars embedded in the rotor pole faces short-circuited by end rings. During load oscillations (hunting), relative speed between the rotor and stator field induces currents in the damper bars, producing damping torque by Lenz's Law that quenches oscillations.
Q. 21
Electrical Engineering
Difficulty: Easy
(1 Mark)
Damper windings also serve which crucial auxiliary function in synchronous motors?
A
They act as squirrel-cage rotor bars during startup, allowing the synchronous motor to self-start as an induction motor
✓ Correct
B
They provide DC excitation directly
C
They eliminate harmonic voltages in the stator
D
They reduce bearing friction
💡
Step-by-Step Explanation & Concept Rationale
Because pure synchronous motors have zero net starting torque at standstill, the damper windings act as a squirrel-cage induction rotor. The motor accelerates to near synchronous speed on induction torque, whereupon DC field excitation is applied to pull into synchronism.
Q. 22
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a salient-pole synchronous machine, the direct-axis synchronous reactance (Xd) is:
A
Less than the quadrature-axis synchronous reactance (Xq)
B
Greater than the quadrature-axis synchronous reactance (Xd > Xq)
✓ Correct
C
Equal to Xq
D
Zero
💡
Step-by-Step Explanation & Concept Rationale
The direct axis aligns with the magnetic pole core where the air gap is minimum (reluctance is lowest, permeance highest), yielding maximum flux and high inductance Xd. The quadrature axis lies in the interpolar gap with maximum air gap, yielding higher reluctance and lower inductance Xq. Thus, Xd > Xq.
Q. 23
Electrical Engineering
Difficulty: Easy
(1 Mark)
Salient-pole rotors are preferred over cylindrical (smooth round) rotors for:
A
High-speed turbo-alternators driven by steam turbines (3000 RPM)
B
Low- and medium-speed hydro-generators driven by water turbines (e.g., 100 to 500 RPM) with large pole counts
✓ Correct
C
Aircraft alternators at 400 Hz
D
Small kitchen appliance universal motors
💡
Step-by-Step Explanation & Concept Rationale
Salient pole rotors have projecting poles with large diameter and short axial length, ideal for low-speed hydro plants with many poles (P = 120*f / N). High-speed steam turbine sets (3000 RPM, 2-pole) require cylindrical rotors to withstand immense centrifugal forces.
Q. 24
Electrical Engineering
Difficulty: Medium
(1 Mark)
Reluctance torque in a salient-pole synchronous motor arises due to:
A
DC excitation current only
B
The tendency of the salient rotor to align itself with the stator magnetic field along the path of minimum magnetic reluctance, even without DC field excitation
✓ Correct
C
High slip during running
D
Eddy currents in the stator core
💡
Step-by-Step Explanation & Concept Rationale
Reluctance power is given by P_rel = (V^2 / 2) * (1/Xq - 1/Xd) * sin(2*delta). Because Xd != Xq, torque is developed as the rotor naturally aligns with minimum reluctance paths. Reluctance motors operate entirely on this torque without DC excitation.
Q. 25
Electrical Engineering
Difficulty: Hard
(1 Mark)
The Potier Reactance method (Zero Power Factor characteristic method) is widely regarded as the most accurate method for calculating alternator voltage regulation because:
A
It completely ignores core saturation
B
It separates armature leakage reactance drop (Potier reactance Xp) from armature reaction MMF, incorporating actual non-linear core saturation
✓ Correct
C
It requires only a DC resistance measurement
D
It assumes air gap reluctance is zero
💡
Step-by-Step Explanation & Concept Rationale
Unlike the Synchronous Impedance (EMF) method (which overestimates regulation) and MMF method (which underestimates regulation), the Potier Triangle method accurately distinguishes leakage reactance voltage drop from armature reaction ampere-turns under realistic saturation.
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All 25 Solved Questions in AC Induction & Synchronous Machines Loaded!
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
Kirchhoff's Current Law (KCL) at a junction in an electrical circuit is a direct mathematical consequence of which conservation law?
A
Conservation of Energy
B
Conservation of Electric Charge
✓ Correct
C
Conservation of Momentum
D
Conservation of Magnetic Flux
💡
Step-by-Step Explanation & Concept Rationale
KCL states that the algebraic sum of currents entering a node is zero (ΣI = 0). Since current is the rate of flow of charge (dq/dt) and charge cannot accumulate at an infinitesimal node, KCL is based on the Conservation of Electric Charge.
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
Kirchhoff's Voltage Law (KVL) around any closed loop is based on the principle of conservation of:
A
Electric Charge
B
Energy
✓ Correct
C
Power
D
Mass
💡
Step-by-Step Explanation & Concept Rationale
KVL states that the algebraic sum of all voltages around any closed loop in a circuit must equal zero (ΣV = 0). Because electric potential is energy per unit charge (V = W/Q), moving a charge around a closed path and returning to the same point results in zero net work done, embodying Conservation of Energy.
Q. 3
Electrical Engineering
Difficulty: Medium
(1 Mark)
Thévenin's equivalent resistance (Rth) of a linear circuit containing independent and dependent sources is found by:
A
Short-circuiting all voltage sources, open-circuiting all current sources, and calculating input impedance
B
Connecting an external test source (1 V or 1 A) at the terminals after deactivating only independent sources
✓ Correct
C
Taking the ratio of open-circuit voltage to maximum rated loop current
D
Short-circuiting independent current sources and open-circuiting independent voltage sources
💡
Step-by-Step Explanation & Concept Rationale
When dependent sources are present in the circuit, they cannot be turned off. To determine Rth, all independent sources are deactivated (voltage sources shorted, current sources opened), an external test source (Vtest or Itest) is applied at the output terminals, and Rth = Vtest / Itest is calculated.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
According to the Maximum Power Transfer Theorem, maximum power is transferred from a linear source network to a purely resistive load when:
A
The load resistance is equal to zero
B
The load resistance is infinitely large
C
The load resistance equals the Thévenin equivalent resistance of the source (RL = Rth)
✓ Correct
D
The load resistance is half the Thévenin equivalent resistance (RL = 0.5 Rth)
💡
Step-by-Step Explanation & Concept Rationale
Differentiating load power P = I^2 * RL = [Vth / (Rth + RL)]^2 * RL with respect to RL and setting dP/dRL = 0 yields RL = Rth. At this condition, the maximum power transferred is Pmax = Vth^2 / (4 * Rth) with an efficiency of 50%.
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
In an AC circuit where the load impedance ZL = RL + jXL is connected to a source with internal impedance Zth = Rth + jXth, the condition for maximum power transfer is:
A
ZL = Zth
B
ZL = Zth* (complex conjugate: RL = Rth and XL = -Xth)
✓ Correct
C
RL = |Zth| and XL = 0
D
ZL = -Zth
💡
Step-by-Step Explanation & Concept Rationale
For an AC circuit with adjustable load resistance and reactance, maximum power is delivered to the load when the load impedance is the complex conjugate of the source impedance: ZL = Zth*, meaning RL = Rth and XL = -Xth (cancelling the net reactive component).
Q. 6
Electrical Engineering
Difficulty: Easy
(1 Mark)
Norton's equivalent circuit consists of:
A
An ideal voltage source in series with an impedance
B
An ideal current source in parallel with an impedance
✓ Correct
C
An ideal voltage source in parallel with an impedance
D
An ideal current source in series with an impedance
💡
Step-by-Step Explanation & Concept Rationale
Norton's Theorem states that any linear, two-terminal circuit can be replaced by an equivalent circuit consisting of a Norton short-circuit current source (IN) connected in parallel with a Norton equivalent impedance (RN = Rth).
Q. 7
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Superposition Theorem is applicable only to circuits that are:
A
Non-linear and time-invariant
B
Linear and bilateral
✓ Correct
C
Non-linear with passive elements only
D
Unilateral and frequency-dependent
💡
Step-by-Step Explanation & Concept Rationale
Superposition relies on the mathematical properties of homogeneity and additivity (linearity). Therefore, it is strictly applicable to linear and bilateral networks where response is directly proportional to excitation.
Q. 8
Electrical Engineering
Difficulty: Medium
(1 Mark)
Why cannot the Superposition Theorem be used directly to calculate electric power in a circuit?
A
Power is a vector quantity
B
Power is proportional to the square of current or voltage (P = I^2*R or V^2/R), making it a non-linear relationship
✓ Correct
C
Power in AC circuits has a reactive component
D
Independent sources absorb power rather than deliver it
💡
Step-by-Step Explanation & Concept Rationale
Power depends quadratically on current (I^2*R) or voltage (V^2/R). Because (I1 + I2)^2 != I1^2 + I2^2 due to the cross-term 2*I1*I2, power is a non-linear quantity, so Superposition cannot be applied directly to calculate power.
Q. 9
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a series RLC circuit at resonance, the total circuit impedance is:
A
Purely reactive and maximum
B
Purely resistive and minimum (Z = R)
✓ Correct
C
Equal to zero
D
Purely capacitive
💡
Step-by-Step Explanation & Concept Rationale
At series resonance, the inductive reactance equals capacitive reactance (XL = XC), so net reactance is zero. The circuit impedance Z = sqrt(R^2 + (XL - XC)^2) simplifies to Z = R, which is its minimum possible value, leading to maximum current flow.
Q. 10
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a parallel RLC resonant circuit, the net impedance at resonance is:
A
Minimum and purely reactive
B
Maximum and purely resistive (Dynamic Resistance L / (C*R))
✓ Correct
C
Zero
D
Inductive
💡
Step-by-Step Explanation & Concept Rationale
In a parallel RLC circuit (antiresonance), the circulating current between L and C is high while the line current drawn from the supply is minimum. Thus, impedance is maximum and purely resistive, termed dynamic impedance Z_dyn = L / (C * R).
Q. 11
Electrical Engineering
Difficulty: Medium
(1 Mark)
The Quality Factor (Q-factor) of a series RLC resonant circuit is given by:
A
Q = R * sqrt(C / L)
B
Q = (1 / R) * sqrt(L / C)
✓ Correct
C
Q = (1 / R) * sqrt(C / L)
D
Q = R * sqrt(L / C)
💡
Step-by-Step Explanation & Concept Rationale
For a series RLC circuit, Q = omega0 * L / R = (1 / (omega0 * C * R)). Substituting omega0 = 1 / sqrt(L * C) gives Q = (1 / R) * sqrt(L / C).
Q. 12
Electrical Engineering
Difficulty: Easy
(1 Mark)
The bandwidth (BW) of a resonant circuit with resonant frequency f0 and quality factor Q is given by:
A
BW = f0 * Q
B
BW = f0 / Q
✓ Correct
C
BW = Q / f0
D
BW = sqrt(f0 * Q)
💡
Step-by-Step Explanation & Concept Rationale
Bandwidth is defined as the frequency difference between half-power frequencies (f2 - f1). It is related to resonant frequency and Quality factor by BW = f0 / Q. A higher Q results in a narrower bandwidth and greater selectivity.
Q. 13
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a balanced 3-phase Star (Y) connected system, the relationship between line voltage (VL) and phase voltage (Vph) is:
A
VL = Vph
B
VL = sqrt(3) * Vph, and line voltage leads phase voltage by 30 degrees
✓ Correct
C
VL = Vph / sqrt(3)
D
VL = 3 * Vph
💡
Step-by-Step Explanation & Concept Rationale
In a star connection, the line voltage is the phasor difference of two phase voltages: VL = sqrt(3) * Vph. In standard phase sequence, line voltage leads the corresponding phase voltage by 30 degrees, while line current equals phase current (IL = Iph).
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a balanced 3-phase Delta (Δ) connected system, the relationship between line current (IL) and phase current (Iph) is:
A
IL = Iph
B
IL = sqrt(3) * Iph, and line current lags phase current by 30 degrees
✓ Correct
C
IL = Iph / sqrt(3)
D
IL = 3 * Iph
💡
Step-by-Step Explanation & Concept Rationale
In a delta connection, the voltages are identical across line and phase (VL = Vph), but line current is the phasor difference of two phase currents: IL = sqrt(3) * Iph. The line current lags the phase current by 30 degrees.
Q. 15
Electrical Engineering
Difficulty: Medium
(1 Mark)
In the two-wattmeter method of measuring 3-phase active power, if one wattmeter reads zero and the other reads the total power, the power factor of the load is:
A
Unity (1.0)
B
0.5 lagging or leading
✓ Correct
C
0.866
D
Zero
💡
Step-by-Step Explanation & Concept Rationale
The two wattmeter readings are W1 = VL*IL*cos(30 - phi) and W2 = VL*IL*cos(30 + phi). If phi = 60 degrees, cos(30 + 60) = cos(90) = 0, so W2 = 0. The power factor is cos(60) = 0.5.
Q. 16
Electrical Engineering
Difficulty: Easy
(1 Mark)
If both wattmeters in the two-wattmeter method read equal positive values (W1 = W2), the load power factor is:
A
0.5
B
Unity (1.0)
✓ Correct
C
0 (purely reactive)
D
0.707
💡
Step-by-Step Explanation & Concept Rationale
W1 = VL*IL*cos(30 - phi) and W2 = VL*IL*cos(30 + phi). For W1 = W2, cos(30 - phi) = cos(30 + phi), which requires phi = 0 degrees. Therefore, the power factor cos(phi) = cos(0) = 1.0 (unity).
Q. 17
Electrical Engineering
Difficulty: Medium
(1 Mark)
If one wattmeter reads positive and the other reads an equal negative value (W1 = -W2), the power factor of the load is:
A
Unity (1.0)
B
0.5
C
Zero (phi = 90 degrees)
✓ Correct
D
0.866
💡
Step-by-Step Explanation & Concept Rationale
Total active power P = W1 + W2 = 0. Because W1 = -W2, tan(phi) = sqrt(3)*(W1 - W2)/(W1 + W2) = infinity, giving phi = 90 degrees and power factor cos(90) = 0 (purely inductive or capacitive load).
Q. 18
Electrical Engineering
Difficulty: Easy
(1 Mark)
Three identical resistors each of resistance R are connected in Delta. Their equivalent Star-connected resistance per phase is:
A
3 * R
B
R / 3
✓ Correct
C
R / 9
D
R
💡
Step-by-Step Explanation & Concept Rationale
In Delta-to-Star transformation, R_star = (R * R) / (R + R + R) = R^2 / (3R) = R / 3. Conversely, Star-to-Delta gives R_delta = 3 * R_star.
Q. 19
Electrical Engineering
Difficulty: Medium
(1 Mark)
Millman's Theorem is primarily used to determine:
A
The current through a non-linear diode
B
The common terminal voltage across multiple parallel branches containing voltage sources and internal impedances
✓ Correct
C
The resonance frequency of coupled circuits
D
The transient response of an RL network
💡
Step-by-Step Explanation & Concept Rationale
Millman's Theorem states that when a number of voltage sources with series impedances are connected in parallel, the equivalent voltage across the common terminals is V_eq = (Σ Vi * Yi) / (Σ Yi), where Yi is the admittance of each branch.
Q. 20
Electrical Engineering
Difficulty: Hard
(1 Mark)
Tellegen's Theorem applies to any electrical network as long as:
A
The network elements are linear, time-invariant, and bilateral
B
The network obeys Kirchhoff's Current Law and Kirchhoff's Voltage Law, regardless of element linearity or time variance
✓ Correct
C
The network operates exclusively on sinusoidal AC steady-state
D
The network does not contain any independent current sources
💡
Step-by-Step Explanation & Concept Rationale
Tellegen's Theorem is based purely on KCL and KVL topology. It states that the algebraic sum of power in all branches is zero (Σ vk * ik = 0), valid for linear, non-linear, time-variant, time-invariant, active, or passive networks.
Q. 21
Electrical Engineering
Difficulty: Easy
(1 Mark)
The time constant of a series RL circuit is defined as:
A
tau = R / L
B
tau = L / R
✓ Correct
C
tau = L * R
D
tau = 1 / (L * R)
💡
Step-by-Step Explanation & Concept Rationale
In an RL circuit, current builds up as i(t) = (V/R) * (1 - e^(-t / tau)), where the time constant tau = L / R. In one time constant, the current reaches 63.2% of its steady-state maximum value.
Q. 22
Electrical Engineering
Difficulty: Easy
(1 Mark)
The time constant of a series RC circuit is given by:
A
tau = R / C
B
tau = C / R
C
tau = R * C
✓ Correct
D
tau = 1 / (R * C)
💡
Step-by-Step Explanation & Concept Rationale
In an RC circuit, the charging voltage across the capacitor is v_C(t) = V * (1 - e^(-t / tau)), where tau = R * C. It takes approximately 5 time constants (5 * tau) for the capacitor to reach 99.3% of full charge.
Q. 23
Electrical Engineering
Difficulty: Easy
(1 Mark)
At steady-state in a DC circuit, an ideal inductor behaves as a(n):
A
Open circuit
B
Short circuit
✓ Correct
C
Constant current source
D
Pure capacitor
💡
Step-by-Step Explanation & Concept Rationale
For an inductor, voltage v = L * (di/dt). Under DC steady-state, current is constant (di/dt = 0), so voltage drop across the inductor is zero, meaning it behaves as an ideal short circuit.
Q. 24
Electrical Engineering
Difficulty: Easy
(1 Mark)
At steady-state in a DC circuit, an ideal capacitor behaves as a(n):
A
Short circuit
B
Open circuit
✓ Correct
C
Zero-ohm resistor
D
Inductive reactor
💡
Step-by-Step Explanation & Concept Rationale
For a capacitor, current i = C * (dv/dt). Under DC steady-state, voltage is constant (dv/dt = 0), so current through the capacitor is zero, meaning it behaves as an open circuit.
Q. 25
Electrical Engineering
Difficulty: Easy
(1 Mark)
The form factor of a pure sinusoidal alternating wave is defined as the ratio of:
A
Peak value to RMS value (1.414)
B
RMS value to Average value (1.11)
✓ Correct
C
Average value to Peak value (0.637)
D
Peak-to-peak value to RMS value (2.828)
💡
Step-by-Step Explanation & Concept Rationale
Form Factor = RMS Value / Average Value. For a sine wave, V_rms = Vm / sqrt(2) = 0.707 Vm, and V_avg = 2 * Vm / pi = 0.637 Vm. Form Factor = (Vm / sqrt(2)) / (2 Vm / pi) = pi / (2 * sqrt(2)) ≈ 1.11.
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
The function of the commutator in a DC generator is to:
A
Collect alternating current directly for the load
B
Convert the alternating EMF (AC) induced in rotating armature conductors into direct current (DC) at the external brushes
✓ Correct
C
Control the field flux
D
Reduce armature reaction
💡
Step-by-Step Explanation & Concept Rationale
Armature conductors always generate alternating current (AC) due to reversing magnetic polarities. The mechanical commutator acts as a rotating mechanical rectifier, reversing coil connections at each neutral crossing to produce unidirectional DC at the stationary carbon brushes.
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a simplex lap-wound DC machine, the number of parallel paths (A) in the armature is:
A
Always equal to 2, independent of pole count
B
Equal to the number of poles (A = P)
✓ Correct
C
Equal to P / 2
D
Equal to 2 * P
💡
Step-by-Step Explanation & Concept Rationale
In lap winding, the number of parallel paths equals the number of poles (A = P), making it suitable for high-current, low-voltage machines. In wave winding, A = 2 regardless of poles, making it suitable for high-voltage, low-current machines.
Q. 3
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a simplex wave-wound DC machine with 8 poles, how many parallel paths are there in the armature winding?
A
8
B
2
✓ Correct
C
4
D
16
💡
Step-by-Step Explanation & Concept Rationale
For any simplex wave winding, the number of parallel paths A is always equal to 2, irrespective of the number of magnetic poles P.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
Armature reaction in a DC generator causes:
A
Cross-magnetization (distortion) and demagnetization of main field flux
✓ Correct
B
An increase in total terminal voltage
C
Elimination of mechanical brush friction
D
Uniform flux distribution under pole shoes
💡
Step-by-Step Explanation & Concept Rationale
Armature reaction is the effect of armature MMF on the main magnetic field. It distorts the flux distribution across pole tips (cross-magnetization) and, due to magnetic saturation, reduces the total average flux per pole (demagnetization).
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a DC generator under load, the Magnetic Neutral Axis (MNA) shifts:
A
Opposite to the direction of rotation
B
In the direction of rotation
✓ Correct
C
Does not shift at all
D
Oscillates at 50 Hz
💡
Step-by-Step Explanation & Concept Rationale
In a DC generator, cross-magnetizing armature flux crowds lines of force toward the trailing pole tips, shifting the resultant magnetic neutral axis (MNA) forward in the direction of rotation. In a DC motor, MNA shifts backward against rotation.
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
Compensating windings in DC machines are installed in slots in the pole shoes and connected in series with the armature to:
A
Neutralize armature reaction under the pole faces across the entire load range
✓ Correct
B
Increase field coil resistance
C
Provide regenerative braking
D
Reduce mechanical noise
💡
Step-by-Step Explanation & Concept Rationale
Compensating windings carry armature current in the opposite direction to adjacent armature conductors under the pole faces. This cancels the cross-magnetizing MMF at its origin, preventing distortion, flashover across commutator segments, and severe field weakening during sudden overloads.
Q. 7
Electrical Engineering
Difficulty: Medium
(1 Mark)
Interpoles (commutating poles) in a DC machine are placed midway between main poles and connected in series with the armature to:
A
Compensate for voltage drop in the field regulator
B
Produce a commutating flux that cancels reactance voltage in coils undergoing commutation, providing sparkless commutation
✓ Correct
C
Increase motor speed above base speed
D
Eliminate residual magnetism
💡
Step-by-Step Explanation & Concept Rationale
Interpoles produce an opposing commutating EMF in the short-circuited coil undergoing commutation, exactly neutralizing the self-induced reactance voltage (e_r = L * di/dt) and cross-magnetizing MMF in the interpolar zone, ensuring sparkless commutation at the brushes.
Q. 8
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why must a self-excited DC shunt generator have residual magnetism in its field poles to build up voltage?
A
To keep the bearings lubricated
B
To induce an initial small EMF when rotated, which drives current through the shunt field to reinforce and build up magnetic flux
✓ Correct
C
To prevent high inrush current
D
To avoid demagnetization of brushes
💡
Step-by-Step Explanation & Concept Rationale
Without residual flux, the induced EMF at startup would be zero, no field current could flow, and the generator could never build up voltage. The initial residual EMF drives current that aids the residual flux until the voltage builds up to the saturation intersection point.
Q. 9
Electrical Engineering
Difficulty: Medium
(1 Mark)
If a DC shunt generator fails to build up voltage, which of the following is the most probable cause?
A
Field circuit resistance is lower than the critical field resistance
B
Field circuit resistance is greater than the critical resistance (Rf > R_critical)
✓ Correct
C
Armature resistance is too low
D
Load resistance is infinite
💡
Step-by-Step Explanation & Concept Rationale
For voltage buildup, the slope of the field resistance line must be less than the initial tangent to the open-circuit magnetization curve (critical resistance). If Rf > R_critical, the field line does not intersect the saturation curve, and voltage fails to build up.
Q. 10
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Back EMF (Eb) in a DC motor is mathematically related to terminal voltage V, armature current Ia, and armature resistance Ra by:
A
Eb = V + Ia * Ra
B
Eb = V - Ia * Ra
✓ Correct
C
Eb = Ia * Ra - V
D
Eb = V / (Ia * Ra)
💡
Step-by-Step Explanation & Concept Rationale
By KVL across a DC motor armature: V = Eb + Ia * Ra (plus brush drop). Hence, Back EMF Eb = V - Ia * Ra. Eb acts as a self-regulating governor; as motor speed changes, Eb adjusts Ia to match the mechanical load torque.
Q. 11
Electrical Engineering
Difficulty: Easy
(1 Mark)
Electromagnetic torque developed by any DC motor is directly proportional to:
A
Flux per pole and armature current (T ∝ phi * Ia)
✓ Correct
B
Terminal voltage squared
C
Back EMF divided by speed
D
Field resistance squared
💡
Step-by-Step Explanation & Concept Rationale
Torque is T = (P * Z / (2 * pi * A)) * phi * Ia. Thus, torque is directly proportional to the product of magnetic flux per pole (phi) and armature current (Ia).
Q. 12
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a DC series motor before magnetic saturation, electromagnetic torque is proportional to:
A
Ia
B
Ia^2
✓ Correct
C
sqrt(Ia)
D
1 / Ia
💡
Step-by-Step Explanation & Concept Rationale
In a DC series motor, the field winding is in series with the armature, so field flux is directly proportional to armature current (phi ∝ Ia) prior to core saturation. Therefore, Torque T ∝ phi * Ia ∝ Ia * Ia = Ia^2.
Q. 13
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why must a DC series motor NEVER be started on no-load or light-load conditions?
A
It will fail to start due to lack of starting torque
B
Its field flux becomes extremely weak, causing the motor speed to rise to dangerously destructive runaway levels (N ∝ Eb / phi)
✓ Correct
C
The commutator segments will short-circuit
D
The back EMF drops to zero and burns out the field
💡
Step-by-Step Explanation & Concept Rationale
Speed N is proportional to Eb / phi. On no load, armature current Ia is very small, making field flux phi near zero. To generate the necessary back EMF, speed shoots up toward infinity (runaway), resulting in catastrophic centrifugal mechanical destruction.
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
Which type of DC motor is most suitable for electric traction, cranes, and hoists requiring very high starting torque?
A
DC Shunt Motor
B
DC Series Motor
✓ Correct
C
Cumulative Compound Motor
D
Differentially Compounded Motor
💡
Step-by-Step Explanation & Concept Rationale
Because torque in a series motor varies with the square of current (T ∝ Ia^2), it develops tremendous starting torque at low speeds, making it the premier choice for heavy traction, elevators, and cranes.
Q. 15
Electrical Engineering
Difficulty: Easy
(1 Mark)
Which DC motor exhibits virtually constant speed from no-load to full-load, making it ideal for machine tools, lathes, and centrifugal pumps?
A
DC Series Motor
B
DC Shunt Motor
✓ Correct
C
Differentially Compounded Motor
D
Universal Motor
💡
Step-by-Step Explanation & Concept Rationale
In a DC shunt motor, the field winding is connected across constant supply voltage, maintaining nearly constant flux. Speed drops by only 5-8% from no load to full load due to armature resistance drop, making it an excellent constant-speed drive.
Q. 16
Electrical Engineering
Difficulty: Medium
(1 Mark)
In a 3-point starter for a DC shunt motor, the No-Volt Release (NVR) hold-on coil is connected in series with:
A
The armature circuit
B
The shunt field winding
✓ Correct
C
The starting resistance directly
D
The main supply switch
💡
Step-by-Step Explanation & Concept Rationale
In a 3-point starter, the NVR coil is in series with the shunt field winding. If the field circuit breaks accidentally or supply voltage fails, the electromagnet de-energizes and a spring snaps the starting arm back to OFF, protecting the motor from runaway.
Q. 17
Electrical Engineering
Difficulty: Medium
(1 Mark)
What is the key advantage of a 4-point starter over a 3-point starter for DC shunt motors?
A
It eliminates the need for an overload release coil
B
The No-Volt Release coil is connected directly across the supply line in series with a protective resistor, allowing field weakening for wide speed control without tripping the starter
✓ Correct
C
It reduces starting current to absolute zero
D
It can operate on both AC and DC supplies
💡
Step-by-Step Explanation & Concept Rationale
In a 3-point starter, weakening the field for speed control above base speed reduces current through the NVR coil, causing it to drop out prematurely. A 4-point starter isolates the NVR coil into an independent parallel circuit across the supply, enabling unrestricted field control.
Q. 18
Electrical Engineering
Difficulty: Easy
(1 Mark)
Speed control of a DC shunt motor by field flux control (field weakening) enables operation:
A
Below rated base speed only
B
Above rated base speed only
✓ Correct
C
Both above and below base speed symmetrically
D
At zero speed only
💡
Step-by-Step Explanation & Concept Rationale
Speed N ∝ 1 / phi. Inserting resistance into the field circuit reduces field current and flux below normal rated value, causing motor speed to rise above base speed (constant power drive region).
Q. 19
Electrical Engineering
Difficulty: Medium
(1 Mark)
Armature voltage control (Ward-Leonard method) of a DC motor allows smooth speed control:
A
From zero up to rated base speed in both directions of rotation
✓ Correct
B
Above rated base speed only
C
Only at 50% rated load
D
With very poor efficiency
💡
Step-by-Step Explanation & Concept Rationale
Ward-Leonard control varies the armature voltage supplied by a dedicated DC generator from zero to rated value, providing wide, smooth, stepless speed control from zero to base speed in both forward and reverse directions with high efficiency (constant torque drive).
Q. 20
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a DC compound generator, if the series field flux assists the shunt field flux, the generator is termed:
A
Differential Compound
B
Cumulative Compound
✓ Correct
C
Under-compounded only
D
Cross-compounded
💡
Step-by-Step Explanation & Concept Rationale
When series field MMF aids the shunt field MMF (phi_total = phi_sh + phi_se), it is cumulatively compounded. Depending on series turns, it can be flat-compounded (constant voltage from no load to full load) or over-compounded (voltage rises with load).
Q. 21
Electrical Engineering
Difficulty: Medium
(1 Mark)
A differentially compounded DC motor is rarely used in practice because:
A
Its starting torque is excessively high
B
As load increases, series flux opposes shunt flux, severely weakening the field, which causes unstable runaway acceleration or motor stalling under heavy loads
✓ Correct
C
Its efficiency is zero
D
It requires AC supply for commutation
💡
Step-by-Step Explanation & Concept Rationale
In a differential compound motor, net flux phi = phi_sh - phi_se. An increase in load current causes the series field to weaken net flux, which triggers dangerous speed instability, racing, or complete stalling and reversal under starting overloads.
Q. 22
Electrical Engineering
Difficulty: Medium
(1 Mark)
Swinburne's Test is a no-load test used to determine the efficiency of a DC shunt machine. Its main limitation is that:
A
It consumes enormous electrical power during the test
B
It cannot be used for DC series motors (which run away on no load) and does not measure stray-load losses or temperature rise directly
✓ Correct
C
It requires two identical machines
D
It burns out the commutator
💡
Step-by-Step Explanation & Concept Rationale
Swinburne's test is run with the motor running light as an uncoupled motor to determine constant core, mechanical, and shunt field losses. It cannot test series motors (which cannot run on no load) and does not account for stray load losses under full-load armature reaction.
Q. 23
Electrical Engineering
Difficulty: Medium
(1 Mark)
Hopkinson's Test (regenerative back-to-back test) on two mechanically and electrically coupled identical DC machines requires from the power supply:
A
Total full-load power of both machines
B
Only the power needed to supply the combined internal losses (core, copper, friction) of both machines
✓ Correct
C
Reactive power exclusively
D
Half the rated voltage at double current
💡
Step-by-Step Explanation & Concept Rationale
One machine acts as a motor driving the second as a generator, which feeds its electrical output back into the motor's input terminals. The external DC mains supply only the power dissipated as internal losses in both machines.
Q. 24
Electrical Engineering
Difficulty: Easy
(1 Mark)
Dynamic (rheostatic) braking of a DC shunt motor is achieved by:
A
Reversing the armature terminals while connected to the supply
B
Disconnecting the armature from the supply and connecting it across a braking resistor while keeping the field energized
✓ Correct
C
Disconnecting both armature and field from all circuits
D
Applying an external mechanical friction brake
💡
Step-by-Step Explanation & Concept Rationale
In dynamic braking, the armature is switched from the mains to a braking resistor. Driven by mechanical inertia in the presence of the active field flux, the machine acts as a generator, dissipating kinetic energy as heat in the resistor to stop quickly.
Q. 25
Electrical Engineering
Difficulty: Medium
(1 Mark)
Plugging (reverse current braking) of a DC motor is performed by:
A
Opening the field circuit
B
Reversing the armature terminal connections while running, causing supply voltage and back EMF to act in the same direction (V + Eb)
✓ Correct
C
Short-circuiting the brushes
D
Gradually reducing supply voltage to zero
💡
Step-by-Step Explanation & Concept Rationale
Reversing armature polarity causes V and Eb to aid each other, producing an initial total voltage of V + Eb across the armature. A large series current-limiting resistor must be inserted to prevent catastrophic armature burnout while generating strong counter-torque.
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
Permanent Magnet Moving Coil (PMMC) instruments can be used to measure:
A
Direct current (DC) quantities only
✓ Correct
B
Alternating current (AC) quantities only
C
Both AC and DC quantities equally
D
Radio frequency currents
💡
Step-by-Step Explanation & Concept Rationale
PMMC instruments operate on Lorentz force (Td = B*I*N*A). With AC current, the deflecting torque reverses direction every half-cycle, resulting in zero average torque (pointer merely vibrates around zero). Therefore, PMMC inherently measures only DC (average value).
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why is the scale of a PMMC measuring instrument perfectly linear and uniform?
A
Deflecting torque is directly proportional to current (Td ∝ I) and controlling torque is proportional to deflection (Tc ∝ theta)
✓ Correct
B
Deflecting torque is proportional to current squared
C
Eddy current damping linearizes the needle movement
D
Permanent magnets have non-linear field distribution
💡
Step-by-Step Explanation & Concept Rationale
In a PMMC instrument with cylindrical soft-iron core and curved pole shoes, magnetic flux density B is uniform and radial. Deflecting torque Td = B*N*A * I. Since spring control torque Tc = C * theta, at balance Td = Tc -> theta ∝ I, resulting in a strictly linear, evenly-spaced scale.
Q. 3
Electrical Engineering
Difficulty: Easy
(1 Mark)
Moving Iron (MI) instruments (attraction or repulsion type) can measure:
A
DC quantities only
B
Both AC and DC quantities, reading the true RMS value of AC signals
✓ Correct
C
Peak value of AC signals only
D
Average value of AC signals
💡
Step-by-Step Explanation & Concept Rationale
In moving iron instruments, deflecting torque is proportional to the square of current (Td ∝ I^2 * dL/dtheta). Because I^2 is always positive regardless of current direction, deflecting torque is unidirectional for both AC and DC, indicating true RMS value.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why is the scale of a Moving Iron instrument non-linear (cramped at the lower end and spread out at the upper end)?
A
Deflecting torque is proportional to the square of the operating current (Td ∝ I^2)
✓ Correct
B
The damping chamber creates back pressure
C
Control springs age over time
D
The coil has high resistance
💡
Step-by-Step Explanation & Concept Rationale
Because deflecting torque varies with current squared (Td ∝ I^2), deflection theta ∝ I^2. Thus, the scale follows a square-law distribution, cramped closely at low current values and open at higher current values.
Q. 5
Electrical Engineering
Difficulty: Easy
(1 Mark)
To convert a basic PMMC galvanometer with internal resistance Rm into an ammeter capable of measuring higher currents, one must connect:
A
A very low resistance (shunt resistor Rsh) in parallel with the meter movement
✓ Correct
B
A high multiplier resistor in series with the meter movement
C
A capacitor in series
D
An inductor in parallel
💡
Step-by-Step Explanation & Concept Rationale
To extend current range, a low-resistance shunt Rsh is connected in parallel with the meter: Rsh = Rm / (m - 1), where multiplying factor m = I / Im. The majority of load current bypasses the delicate meter movement through the low-resistance shunt.
Q. 6
Electrical Engineering
Difficulty: Easy
(1 Mark)
To extend the voltage measurement range of a basic PMMC meter movement to form a voltmeter, one must connect:
A
A high multiplier resistor (Rs) in series with the meter
✓ Correct
B
A low resistance in parallel
C
A center-tapped transformer
D
A shunt capacitor
💡
Step-by-Step Explanation & Concept Rationale
To extend voltage range, a high resistance multiplier Rs is connected in series: Rs = Rm * (m - 1), where m = V / Vm. The series resistor drops the bulk of the applied voltage, limiting current through the coil to its safe full-scale deflection value Im.
Q. 7
Electrical Engineering
Difficulty: Medium
(1 Mark)
What material is standard for manufacturing ammeter shunts and voltmeter multiplier resistors due to its near-zero temperature coefficient of resistance?
A
Pure Copper
B
Manganin (Copper-Manganese-Nickel alloy) or Constantan
✓ Correct
C
Aluminum
D
Silicon steel
💡
Step-by-Step Explanation & Concept Rationale
Manganin has a negligible temperature coefficient of resistance (approx. 0.00001 / deg C) and low thermal EMF against copper. This ensures that instrument calibration does not drift with ambient or I^2*R self-heating temperature changes.
Q. 8
Electrical Engineering
Difficulty: Medium
(1 Mark)
Which AC bridge is specifically used for the precision measurement of an unknown capacitance and its dielectric dissipation factor (tan delta)?
A
Schering Bridge
✓ Correct
B
Maxwell's Inductance-Capacitance Bridge
C
Anderson's Bridge
D
Wheatstone Bridge
💡
Step-by-Step Explanation & Concept Rationale
The Schering Bridge is the industry-standard bridge for measuring capacitance and dielectric loss angle (dissipation factor tan delta = omega * C1 * r1) of high-voltage cables, transformer bushings, and insulating materials.
Q. 9
Electrical Engineering
Difficulty: Medium
(1 Mark)
Maxwell's Inductance-Capacitance Bridge is most suitable for measuring the inductance of coils having:
A
Medium Quality Factor (1 < Q < 10)
✓ Correct
B
Very high Quality Factor (Q > 10)
C
Very low Quality Factor (Q < 1)
D
Zero resistance
💡
Step-by-Step Explanation & Concept Rationale
Maxwell's L-C bridge balances an unknown inductance against a standard variable capacitor. It is ideal for medium-Q coils (1 to 10). For high-Q coils (Q > 10), Hay's Bridge is used; for low-Q coils (Q < 1), Anderson's Bridge is preferred.
Q. 10
Electrical Engineering
Difficulty: Medium
(1 Mark)
Hay's Bridge is preferred over Maxwell's Bridge for measuring inductances with:
A
High Quality Factor (Q > 10)
✓ Correct
B
Low Quality Factor (Q < 1)
C
Zero self-capacitance
D
High mutual coupling
💡
Step-by-Step Explanation & Concept Rationale
In Hay's bridge, the resistance and capacitance in the standard arm are connected in series rather than parallel. The balance equation contains term (1 / [1 + (1/Q)^2]), which simplifies cleanly for high-Q coils (Q > 10) where (1/Q)^2 is negligible.
Q. 11
Electrical Engineering
Difficulty: Hard
(1 Mark)
Anderson's Bridge is an advanced modification of Maxwell's Bridge designed to measure:
A
Inductance of low-Q coils (Q < 1) accurately using a fixed standard capacitor
✓ Correct
B
High-frequency dielectric loss
C
High insulation resistance
D
Mutual capacitance between cables
💡
Step-by-Step Explanation & Concept Rationale
Anderson's bridge introduces an additional resistance in the capacitor branch, allowing smooth independent balance adjustments for low-Q coils while utilizing a fixed standard capacitor instead of an expensive calibrated variable capacitor.
Q. 12
Electrical Engineering
Difficulty: Medium
(1 Mark)
Wien's Bridge is widely utilized in audio instrumentation and measurement for:
A
Measuring unknown frequency of an AC voltage and serving as the frequency-determining circuit in audio oscillators
✓ Correct
B
Measuring high self-inductance
C
Calibrating megohm meters
D
Measuring mutual inductance
💡
Step-by-Step Explanation & Concept Rationale
Wien's bridge balance condition is f = 1 / (2 * pi * sqrt(R1 * R2 * C1 * C2)). When R1 = R2 = R and C1 = C2 = C, f = 1 / (2 * pi * R * C). It is widely used as a precision frequency meter and distortion analyzer.
Q. 13
Electrical Engineering
Difficulty: Medium
(1 Mark)
Kelvin's Double Bridge is the standard laboratory instrument for precision measurement of:
A
Low resistance (less than 1 ohm), eliminating contact and lead wire resistance errors
✓ Correct
B
High insulation resistance (> 1 MΩ)
C
Dielectric breakdown of transformer oil
D
Dynamic core loss
💡
Step-by-Step Explanation & Concept Rationale
Kelvin's Double Bridge incorporates a second set of ratio arms. When the two sets of ratio arms are matched (P/Q = p/q), the effect of connecting lead resistance (r) between standard and unknown low resistors is mathematically eliminated, measuring resistances down to 0.00001 ohm.
Q. 14
Electrical Engineering
Difficulty: Medium
(1 Mark)
A Megger (insulation resistance tester) incorporates a high-voltage hand-driven generator or DC-DC converter and an ohmmeter movement based on a:
A
Permanent Magnet Moving Coil Ratio Meter (Cross-coil ohmmeter) requiring no zero-adjustment
✓ Correct
B
Electrodynamometer wattmeter
C
Vane-type moving iron meter
D
Thermal thermocouple
💡
Step-by-Step Explanation & Concept Rationale
A Megger utilizes a ratio meter movement with two coils (current coil and pressure coil) wound on a common axle in a non-uniform magnetic field. Deflection depends strictly on the ratio of V to I (R = V/I), making the reading independent of hand-cranking generator voltage fluctuations.
Q. 15
Electrical Engineering
Difficulty: Medium
(1 Mark)
The Guard Terminal (G) on a 3-terminal Megger is provided to:
A
Bypass surface leakage currents over the cable insulation directly back to the generator, preventing false low insulation resistance readings
B
Ground the user against electric shock
✓ Correct
C
Double the test voltage
D
Measure conductor continuity
💡
Step-by-Step Explanation & Concept Rationale
Surface moisture and dirt create surface leakage currents that bypass bulk insulation. Wrapping a bare wire around the cable sheath and connecting it to the Guard terminal intercepts surface leakage and diverts it around the current-measuring coil.
Q. 16
Electrical Engineering
Difficulty: Easy
(1 Mark)
Creeping in an induction-type energy meter refers to:
A
Slow, continuous rotation of the aluminum disc when potential coil is energized but load current is zero
✓ Correct
B
Over-speeding of the disc on full load
C
Friction in the gear register
D
Thermal expansion of the braking magnet
💡
Step-by-Step Explanation & Concept Rationale
To overcome mechanical pivot friction, shading loops on the shunt magnet provide a small forward driving torque. If over-compensated or if supply voltage rises, this friction-compensating torque causes the disc to rotate continuously (creep) on no-load.
Q. 17
Electrical Engineering
Difficulty: Easy
(1 Mark)
How is creeping prevented in single-phase induction energy meters?
A
By drilling two diametrically opposite holes in the aluminum disc
✓ Correct
B
By removing the permanent brake magnet
C
By increasing potential coil turns
D
By disconnecting the shading band
💡
Step-by-Step Explanation & Concept Rationale
Two diametrically opposite holes drilled in the disc cause reluctance variation. When a hole arrives beneath the shunt magnet pole, the disc experiences a reluctance counter-torque that arrests no-load rotation.
Q. 18
Electrical Engineering
Difficulty: Medium
(1 Mark)
Braking torque in an induction-type energy meter is provided by a permanent magnet and is directly proportional to:
A
The speed of rotation of the aluminum disc (Tb ∝ N)
✓ Correct
B
The square of disc speed (N^2)
C
Supply voltage squared
D
Ambient humidity
💡
Step-by-Step Explanation & Concept Rationale
As the aluminum disc rotates across the field of the permanent brake magnet, eddy currents are induced in the disc proportional to speed N. These eddy currents interact with the permanent field to generate a braking torque Tb = K * B^2 * N * d. At steady-state, Td = Tb, ensuring disc speed is strictly proportional to power.
Q. 19
Electrical Engineering
Difficulty: Easy
(1 Mark)
An electrodynamometer wattmeter can measure:
A
Active power (P) in both AC and DC circuits accurately
✓ Correct
B
Reactive power only
C
DC energy only
D
Only high frequencies above 1 MHz
💡
Step-by-Step Explanation & Concept Rationale
The electrodynamometer wattmeter has fixed coils carrying load current (I) and a moving pressure coil carrying current proportional to voltage (V). Deflecting torque Td = (1/R) * V * I * (dM/dtheta) * cos(phi), accurately indicating true active average power in Watts for both AC and DC.
Q. 20
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why is power factor improvement economically essential for industrial consumers?
A
Low power factor draws higher line current for the same useful kW load, increasing I^2*R transmission losses, causing poor voltage regulation, and triggering utility kVA maximum demand penalties
✓ Correct
B
Low power factor increases motor speed dangerously
C
High power factor burns out capacitor banks
D
Power factor does not affect kVA demand
💡
Step-by-Step Explanation & Concept Rationale
For constant real power P = V*I*cos(phi), line current is I = P / (V * cos(phi)). A low power factor (e.g., 0.6 lagging) forces the system to supply 67% more current than at unity PF, overloading cables and transformers, increasing I^2*R losses, causing excessive voltage drop, and incurring heavy utility billing penalties.
Q. 21
Electrical Engineering
Difficulty: Easy
(1 Mark)
Static shunt capacitors improve the power factor of inductive loads by:
A
Drawing leading reactive current (VARs) that cancels a portion of the lagging reactive current drawn by inductive loads
✓ Correct
B
Consuming active power directly
C
Increasing line inductance
D
Stepping up the supply frequency
💡
Step-by-Step Explanation & Concept Rationale
Inductive loads (induction motors, transformers) require lagging reactive power Q_L to maintain magnetic fields. Shunt capacitors draw leading reactive current (Q_C = V^2 * omega * C), supplying these magnetizing VARs locally so they do not have to be transmitted from the distant generating station.
Q. 22
Electrical Engineering
Difficulty: Medium
(1 Mark)
The rating of a shunt capacitor bank required to improve load power factor from cos(phi1) to cos(phi2) for an active load P (in kW) is given by:
A
Q_cap = P * [tan(phi1) - tan(phi2)] kVAR
✓ Correct
B
Q_cap = P * [cos(phi1) - cos(phi2)] kVAR
C
Q_cap = P / [tan(phi1) + tan(phi2)] kVAR
D
Q_cap = P * [sin(phi1) + sin(phi2)] kVAR
💡
Step-by-Step Explanation & Concept Rationale
Initial reactive power is Q1 = P * tan(phi1). Desired final reactive power is Q2 = P * tan(phi2). The leading reactive kVAR required from the shunt capacitor bank is Q_cap = Q1 - Q2 = P * [tan(phi1) - tan(phi2)] kVAR.
Q. 23
Electrical Engineering
Difficulty: Hard
(1 Mark)
Phase advancers are installed on large slip-ring induction motors to improve their power factor by:
A
Injecting low-frequency leading EMF into the rotor circuit through the slip rings
✓ Correct
B
Connecting capacitors in series with the stator windings
C
Operating the motor with locked rotor
D
Short-circuiting the stator neutral
💡
Step-by-Step Explanation & Concept Rationale
A phase advancer is an AC exciter connected to the rotor circuit of a slip-ring motor. Injecting leading MMF at rotor slip frequency provides the motor's magnetizing ampere-turns from the rotor rather than the stator, raising stator power factor to near unity or even leading.
Q. 24
Electrical Engineering
Difficulty: Medium
(1 Mark)
A Cathode Ray Oscilloscope (CRO) uses an electrostatic deflection system where deflection sensitivity is defined as:
A
Deflection on the screen per unit deflecting voltage (D / Vd in mm/V)
✓ Correct
B
Accelerating anode voltage divided by screen diameter
C
Electron beam current divided by grid voltage
D
Phosphor persistence time
💡
Step-by-Step Explanation & Concept Rationale
Deflection sensitivity S = D / Vd = (L * l) / (2 * d * Va), where D is trace deflection, Vd is deflecting voltage, L is screen distance, l is plate length, d is plate separation, and Va is accelerating anode potential. Its inverse is deflection factor G = 1 / S (V/mm).
Q. 25
Electrical Engineering
Difficulty: Easy
(1 Mark)
Lissajous figures displayed on an oscilloscope screen when two sinusoidal signals of identical frequency are applied to the horizontal and vertical plates form a circle when:
A
The two signals have equal amplitudes and a phase difference of 90 degrees (or 270 degrees)
✓ Correct
B
The two signals are exactly in phase (0 degrees)
C
The two signals have a phase difference of 180 degrees
D
The frequencies have a 2:1 ratio
💡
Step-by-Step Explanation & Concept Rationale
Applying x = A * sin(omega*t) and y = B * sin(omega*t + phi) with A = B yields x^2 + y^2 = A^2 when phi = 90 or 270 degrees, tracing a perfect circle. A phase difference of 0 or 180 degrees traces a straight line.
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All 26 Solved Questions in Electrical Measurements & Power Factor Loaded!
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
Faraday's Law of Electromagnetic Induction states that the magnitude of induced EMF in a circuit is directly proportional to:
A
The total magnetic flux linked with the circuit
B
The time rate of change of magnetic flux linkage (e = -N * dphi/dt)
✓ Correct
C
The cross-sectional area of the magnetic core
D
The permeability of free space
💡
Step-by-Step Explanation & Concept Rationale
Faraday's Law states that induced EMF e = -N * (dphi/dt). The magnitude depends strictly on how rapidly magnetic flux linkage changes with respect to time.
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
Lenz's Law, which determines the direction of an induced EMF and current, is an expression of:
A
Conservation of Momentum
B
Conservation of Energy
✓ Correct
C
Conservation of Charge
D
Gauss's Law
💡
Step-by-Step Explanation & Concept Rationale
Lenz's Law states that the induced current always flows in such a direction that its magnetic effect opposes the change in flux that produced it. This negative sign ensures that work must be done against electromagnetic forces to generate electrical energy, conforming to Conservation of Energy.
Q. 3
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a magnetic circuit, the opposition offered to the establishment of magnetic flux is called:
A
Permeance
B
Reluctance (S or R_m)
✓ Correct
C
Susceptance
D
Conductance
💡
Step-by-Step Explanation & Concept Rationale
Reluctance is the magnetic analogue of electric resistance. It is defined as S = l / (mu0 * mu_r * A), where l is the magnetic path length, A is cross-sectional area, and mu is permeability. Its unit is A-t/Wb or Henry^-1.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
Permeance in a magnetic circuit is directly analogous to which parameter in an electric circuit?
A
Resistance
B
Conductance (G)
✓ Correct
C
Current
D
Capacitance
💡
Step-by-Step Explanation & Concept Rationale
Permeance is the reciprocal of reluctance (P = 1 / S = mu * A / l). It represents the ease with which magnetic flux is developed, analogous to electrical conductance G = 1 / R.
Q. 5
Electrical Engineering
Difficulty: Easy
(1 Mark)
What is the SI unit of Magnetomotive Force (MMF)?
A
Tesla (T)
B
Ampere-turns (A-t) or Amperes (A)
✓ Correct
C
Weber (Wb)
D
Henry (H)
💡
Step-by-Step Explanation & Concept Rationale
MMF is the magnetic potential driving flux through a magnetic circuit: MMF = N * I (turns * current). Its SI unit is the Ampere-turn (A-t) or simply Ampere.
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
The energy stored per unit volume (energy density) in a magnetic field with flux density B and magnetic field intensity H is:
A
w = (1/2) * B * H = B^2 / (2 * mu)
✓ Correct
B
w = B * H
C
w = (1/2) * mu * B^2
D
w = H^2 / (2 * mu)
💡
Step-by-Step Explanation & Concept Rationale
Magnetic energy density is given by w = (1/2) * B * H. Since B = mu * H, this can be written as w = B^2 / (2 * mu) = (1/2) * mu * H^2 in Joules/m^3.
Q. 7
Electrical Engineering
Difficulty: Easy
(1 Mark)
The energy stored in an inductor of inductance L carrying current I is given by:
A
W = (1/2) * L^2 * I
B
W = (1/2) * L * I^2
✓ Correct
C
W = L * I
D
W = (1/2) * (L / I)
💡
Step-by-Step Explanation & Concept Rationale
The work done in establishing current I through an inductor is W = integral(0 to I) L * i * di = (1/2) * L * I^2 Joules.
Q. 8
Electrical Engineering
Difficulty: Medium
(1 Mark)
Hysteresis loss in a ferromagnetic material subjected to alternating magnetization is proportional to:
A
The square of the core thickness
B
The area enclosed by the B-H hysteresis loop
✓ Correct
C
The inverse of supply frequency
D
The square of the applied voltage
💡
Step-by-Step Explanation & Concept Rationale
The area enclosed by the B-H loop represents the energy lost as heat per unit volume per cycle of magnetization. According to Steinmetz's empirical formula: P_h = eta * B_max^1.6 * f * V.
Q. 9
Electrical Engineering
Difficulty: Medium
(1 Mark)
Eddy current loss in a magnetic core subjected to an alternating magnetic field varies with frequency (f) and maximum flux density (Bm) as:
A
P_e proportional to f * Bm
B
P_e proportional to f^2 * Bm^2
✓ Correct
C
P_e proportional to f^2 * Bm
D
P_e proportional to f * Bm^2
💡
Step-by-Step Explanation & Concept Rationale
Eddy current loss is given by P_e = K_e * Bm^2 * f^2 * t^2 * V, where t is the thickness of laminations. It is directly proportional to the square of frequency and the square of maximum flux density.
Q. 10
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why are transformer and motor cores constructed using thin, insulated silicon steel laminations instead of solid iron blocks?
A
To eliminate hysteresis loss completely
B
To increase the mechanical rigidity of the core
C
To increase electrical resistance across the path of circulating currents, thereby minimizing eddy current loss
✓ Correct
D
To increase the saturation flux density Bm
💡
Step-by-Step Explanation & Concept Rationale
Laminating the core into thin sheets insulated by varnish restricts the path of circulating eddy currents to tiny loops within each lamination. Since eddy current loss is proportional to thickness squared (t^2), laminating dramatically reduces P_e.
Q. 11
Electrical Engineering
Difficulty: Medium
(1 Mark)
Adding approximately 3% to 4% silicon to electrical core steel serves primarily to:
A
Increase electrical resistivity of the steel, thereby reducing eddy current losses and reducing hysteresis loss
✓ Correct
B
Make the core ductile and mechanically flexible
C
Eliminate the need for transformer oil cooling
D
Decrease magnetic permeability
💡
Step-by-Step Explanation & Concept Rationale
Silicon increases the electrical resistivity of the steel significantly, which suppresses eddy current flow. It also reduces magnetostriction and decreases the area of the hysteresis loop, lowering total core losses.
Q. 12
Electrical Engineering
Difficulty: Easy
(1 Mark)
Residual magnetism (or remanence) on a B-H hysteresis curve corresponds to:
A
The magnetic field intensity H required to reduce B to zero
B
The magnetic flux density B remaining in the material when the magnetizing field H is reduced to zero
✓ Correct
C
The saturation flux density at infinite field strength
D
The permeability of the core at high temperatures
💡
Step-by-Step Explanation & Concept Rationale
Remanence (or retentivity) is the value of flux density B that persists in the ferromagnetic core after the external magnetizing field H has been removed (H = 0).
Q. 13
Electrical Engineering
Difficulty: Easy
(1 Mark)
Coercive force (or coercivity) on a B-H curve represents:
A
The maximum flux density achievable in the core
B
The reverse magnetizing force (-H) required to demagnetize the material and reduce residual flux density to zero
✓ Correct
C
The magnetic flux leakage across the air gap
D
The initial slope of the magnetization curve
💡
Step-by-Step Explanation & Concept Rationale
Coercivity is the reverse magnetic field intensity required to completely wipe out the remanent magnetic flux density (bringing B back to 0). Permanent magnets require high coercivity, whereas transformer cores require low coercivity.
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
The force experienced by a straight current-carrying conductor of length L carrying current I placed in a uniform magnetic field B is given by:
A
F = I * (L x B) = B * I * L * sin(theta)
✓ Correct
B
F = B * I * L * cos(theta)
C
F = B^2 * I * L
D
F = (I / B) * L
💡
Step-by-Step Explanation & Concept Rationale
Lorentz magnetic force on a current element is dF = I * (dL x B). For a straight conductor, F = B * I * L * sin(theta), where theta is the angle between the conductor and the magnetic field vector. Force is maximum when theta = 90 degrees.
Q. 15
Electrical Engineering
Difficulty: Easy
(1 Mark)
Fleming's Left-Hand Rule is used to determine the direction of:
A
Induced EMF in an AC generator
B
Force (motion) acting on a current-carrying conductor in a magnetic field (Electric Motor principle)
✓ Correct
C
Magnetic field around a straight wire
D
Eddy currents in a solid plate
💡
Step-by-Step Explanation & Concept Rationale
Fleming's Left-Hand Rule applies to motors: Thumb = Thrust/Force/Motion, Forefinger = Magnetic Field (B), Center finger = Current (I). Fleming's Right-Hand Rule applies to generators.
Q. 16
Electrical Engineering
Difficulty: Easy
(1 Mark)
Fleming's Right-Hand Rule is used to determine the direction of:
A
Force on a motor armature
B
Dynamically induced EMF or current in a conductor moving across a magnetic field (Electric Generator principle)
✓ Correct
C
Torque in a three-phase induction motor
D
Magnetic polarity of a solenoid
💡
Step-by-Step Explanation & Concept Rationale
Fleming's Right-Hand Rule applies to generators: Thumb = Motion of conductor, Forefinger = Direction of Magnetic Field (N to S), Middle finger = Direction of Induced EMF/Current.
Q. 17
Electrical Engineering
Difficulty: Easy
(1 Mark)
The dynamically induced EMF in a straight conductor of length L moving with velocity v at an angle theta to a uniform magnetic field B is:
A
e = B * L * v * cos(theta)
B
e = B * L * v * sin(theta)
✓ Correct
C
e = (1/2) * B * L * v^2
D
e = B^2 * L * v
💡
Step-by-Step Explanation & Concept Rationale
By Lorentz force on moving conduction electrons: e = integral (v x B) . dl = B * L * v * sin(theta). When the conductor moves perpendicular to the magnetic lines of force (theta = 90 deg), e_max = B * L * v.
Q. 18
Electrical Engineering
Difficulty: Medium
(1 Mark)
The self-inductance L of a solenoid with N turns, core cross-sectional area A, core length l, and relative permeability mu_r is given by:
A
L = (mu0 * mu_r * N * A) / l
B
L = (mu0 * mu_r * N^2 * A) / l
✓ Correct
C
L = (mu0 * N^2 * l) / A
D
L = (N * A) / (mu0 * mu_r * l)
💡
Step-by-Step Explanation & Concept Rationale
L = N * phi / I. Since phi = B * A = (mu * N * I / l) * A, substituting gives L = (mu0 * mu_r * N^2 * A) / l = N^2 / S, proving that inductance is proportional to the square of turns (N^2).
Q. 19
Electrical Engineering
Difficulty: Easy
(1 Mark)
If the number of turns in an inductor is doubled while maintaining the same physical core dimensions, its inductance becomes:
A
Doubled (2x)
B
Quadrupled (4x)
✓ Correct
C
Halved (0.5x)
D
Unchanged
💡
Step-by-Step Explanation & Concept Rationale
Because L is proportional to N^2 (L ∝ N^2), doubling the number of turns (N' = 2N) results in L' = (2N)^2 = 4 * N^2 = 4 * L.
Q. 20
Electrical Engineering
Difficulty: Easy
(1 Mark)
The magnetic field intensity H inside an ideal infinitely long solenoid having n turns per unit length carrying current I is:
A
H = n * I
✓ Correct
B
H = n * I / (2 * pi)
C
H = mu0 * n * I
D
H = n^2 * I
💡
Step-by-Step Explanation & Concept Rationale
By Ampere's Circuital Law, the magnetic field intensity inside a long solenoid is H = n * I (in A/m). The magnetic flux density is B = mu * H = mu0 * mu_r * n * I.
Q. 21
Electrical Engineering
Difficulty: Medium
(1 Mark)
Ampere's Circuital Law states that the line integral of magnetic field intensity H around any closed contour equals:
A
The total magnetic flux enclosed
B
The net electric current enclosed by the loop (∮ H . dl = I_enclosed)
✓ Correct
C
Zero under all conditions
D
The rate of change of electric flux
💡
Step-by-Step Explanation & Concept Rationale
Ampere's Circuital Law states that ∮ H . dl = I_enc. In Maxwell's generalized equation for time-varying fields, displacement current density is added: ∮ H . dl = I_conduction + d/dt(integral D . dA).
Q. 22
Electrical Engineering
Difficulty: Medium
(1 Mark)
Gauss's Law for Magnetism (div B = 0 or ∮ B . dA = 0) mathematically implies that:
A
Magnetic field is conservative
B
Isolated magnetic monopoles do not exist in nature, and magnetic field lines form continuous closed loops
✓ Correct
C
Magnetic charge equals electric charge
D
Magnetic flux cannot penetrate ferromagnetic substances
💡
Step-by-Step Explanation & Concept Rationale
div B = 0 states that the divergence of magnetic flux density everywhere is zero. This means there are no sources or sinks of magnetic field; isolated magnetic monopoles (North or South alone) do not exist, and magnetic lines always close upon themselves.
Q. 23
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Biot-Savart Law allows calculating the magnetic field produced by a current element. The magnetic field dB at a distance r is proportional to:
A
r
B
1 / r
C
1 / r^2
✓ Correct
D
1 / r^3
💡
Step-by-Step Explanation & Concept Rationale
dB = (mu0 / (4*pi)) * (I * dl x r_hat) / r^2 = (mu0 / (4*pi)) * (I * dl * sin(theta)) / r^2. It follows an inverse-square law with respect to distance from the current element.
Q. 24
Electrical Engineering
Difficulty: Medium
(1 Mark)
Two parallel conductors placed 1 meter apart in vacuum carry currents of 1 A in the same direction. What is the nature and magnitude of the force per unit length between them?
A
Repulsive force of 2 * 10^-7 N/m
B
Attractive force of 2 * 10^-7 N/m
✓ Correct
C
Attractive force of 4*pi * 10^-7 N/m
D
Zero force
💡
Step-by-Step Explanation & Concept Rationale
Parallel conductors carrying currents in the same direction attract each other with force per unit length F/L = (mu0 * I1 * I2) / (2 * pi * d) = (4*pi*10^-7 * 1 * 1) / (2 * pi * 1) = 2 * 10^-7 N/m. This formula officially defines the SI base unit Ampere.
Q. 25
Electrical Engineering
Difficulty: Easy
(1 Mark)
Two parallel conductors carrying currents in opposite directions will:
A
Attract each other
B
Repel each other
✓ Correct
C
Experience zero magnetic interaction
D
Rotate until they are perpendicular
💡
Step-by-Step Explanation & Concept Rationale
When currents flow in opposite directions, the magnetic field between the conductors reinforces while the field outside weakens, creating a lateral pressure that repels the two conductors.
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Ferranti Effect in electrical power transmission lines refers to the phenomenon where:
A
The receiving-end voltage is higher than the sending-end voltage under no-load or lightly loaded conditions on long transmission lines
✓ Correct
B
The line voltage drops significantly during heavy inductive loading
C
Current concentrates on the outer surface of conductors
D
Corona discharge ionizes ambient air at high altitudes
💡
Step-by-Step Explanation & Concept Rationale
On long, lightly loaded or open-ended transmission lines, line charging current flowing through the line's distributed series inductance causes a voltage rise along the line, resulting in receiving-end voltage exceeding sending-end voltage (Vr > Vs). Shunt reactors are installed to absorb this reactive power.
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
Surge Impedance (or characteristic impedance Zc) of a lossless overhead transmission line with inductance L and capacitance C is given by:
A
Zc = sqrt(L / C)
✓ Correct
B
Zc = sqrt(C / L)
C
Zc = 1 / sqrt(L * C)
D
Zc = sqrt(L * C)
💡
Step-by-Step Explanation & Concept Rationale
For a lossless line (R=0, G=0), the surge impedance is Zc = sqrt(L / C). For typical overhead transmission lines, Zc ranges from 400 to 450 ohms, while for underground cables, Zc is around 40 to 50 ohms.
Q. 3
Electrical Engineering
Difficulty: Medium
(1 Mark)
Surge Impedance Loading (SIL) of a transmission line with nominal line-to-line voltage V_LL and surge impedance Zc is:
A
SIL = V_LL^2 / Zc
✓ Correct
B
SIL = Zc / V_LL^2
C
SIL = sqrt(3) * V_LL * Zc
D
SIL = V_LL * Zc
💡
Step-by-Step Explanation & Concept Rationale
Surge Impedance Loading represents the power delivered by a transmission line to a load equal to its surge impedance: SIL = V_LL^2 / Zc (MW). At this loading, reactive power generated by line capacitance exactly balances reactive power absorbed by line inductance, yielding flat voltage along the line.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
Corona discharge on high-voltage overhead transmission lines is accompanied by:
A
A violet-colored luminous glow, hissing noise, ozone gas production, and power loss
✓ Correct
B
Sudden freezing of line conductors
C
Complete cancellation of series line inductance
D
Zero power loss
💡
Step-by-Step Explanation & Concept Rationale
Corona occurs when the electrostatic field gradient at the conductor surface exceeds the dielectric breakdown strength of air (approx. 30 kV/cm peak or 21.2 kV/cm RMS). Ionization of air produces a faint violet glow, hissing acoustic noise, chemical production of ozone (O3), and power loss.
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
Bundle conductors are widely employed in Extra High Voltage (EHV) and Ultra High Voltage (UHV) transmission lines primarily to:
A
Increase the effective conductor diameter (GMR), thereby reducing surface electric field gradient, corona loss, and line reactance
✓ Correct
B
Make tower erection easier and cheaper
C
Eliminate the need for lightning arresters
D
Prevent mechanical sag in hot summers
💡
Step-by-Step Explanation & Concept Rationale
Using bundles of two, three, or four sub-conductors per phase increases the Geometric Mean Radius (GMR) of the phase conductor. This lowers the electrostatic surface stress below the critical disruptive threshold, substantially reducing corona loss, radio interference, and series inductive reactance.
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
The critical disruptive voltage of a transmission line decreases under which environmental condition?
A
High atmospheric pressure and dry air
B
Stormy, foul weather (rain, fog, or snow) and high altitude with low barometric pressure
✓ Correct
C
Sub-zero temperatures with dry breeze
D
Nighttime with clear skies
💡
Step-by-Step Explanation & Concept Rationale
Critical disruptive voltage Vd = m0 * g0 * delta * r * ln(D/r). At high altitudes or stormy weather, the air density factor delta decreases and the conductor irregularity factor m0 drops from 1.0 to ~0.8, causing corona to initiate at much lower voltages.
Q. 7
Electrical Engineering
Difficulty: Medium
(1 Mark)
The proximity effect in transmission conductors refers to:
A
Current redistribution due to the magnetic fields of nearby conductors carrying current, increasing effective AC resistance
✓ Correct
B
Mutual capacitance between tower body and ground wire
C
The reflection of lightning traveling waves
D
Thermal heating of insulators
💡
Step-by-Step Explanation & Concept Rationale
Proximity effect occurs when alternating flux from an adjacent parallel conductor distorts the current distribution across the cross-section of a conductor. When currents flow in the same direction, current shifts away from adjacent sides; when in opposite directions, it concentrates on facing sides, raising AC resistance.
Q. 8
Electrical Engineering
Difficulty: Easy
(1 Mark)
ACSR (Aluminum Conductor Steel Reinforced) conductors are standard for overhead transmission because:
A
The inner steel core provides high mechanical tensile strength, while outer aluminum strands provide high electrical conductivity and lighter weight
✓ Correct
B
Steel has higher electrical conductivity than aluminum
C
Aluminum prevents mechanical corrosion of steel completely
D
Steel reduces eddy current losses to zero
💡
Step-by-Step Explanation & Concept Rationale
Pure aluminum has high conductivity and low weight but low tensile strength. Combining a high-tensile galvanized steel core with surrounding aluminum strands gives ACSR high mechanical strength to permit long spans between towers with reduced sag.
Q. 9
Electrical Engineering
Difficulty: Medium
(1 Mark)
String efficiency of a suspension insulator string on an overhead transmission tower is defined as:
A
Total flashover voltage / Breakdown voltage
B
Voltage across the whole string / (Number of insulator discs * Voltage across the disc nearest to the line conductor)
✓ Correct
C
Capacitance of top disc / Capacitance of bottom disc
D
Leakage resistance / Surface resistance
💡
Step-by-Step Explanation & Concept Rationale
Due to shunt capacitance between insulator metal pins and the grounded tower body, voltage distribution across the string is non-uniform, with the bottom disc (nearest the line conductor) experiencing the greatest electrical stress. String efficiency = V_total / (n * V_bottom) * 100%.
Q. 10
Electrical Engineering
Difficulty: Medium
(1 Mark)
A guard ring (static shielding ring) is installed around the bottom disc of an insulator string to:
A
Improve string efficiency by introducing cross-capacitances that equalize the voltage distribution across all discs
✓ Correct
B
Provide mechanical support to line clamps
C
Prevent birds from perching on the line
D
Ground lightning strikes directly to the tower
💡
Step-by-Step Explanation & Concept Rationale
A guard ring introduces capacitance between the line conductor and the insulator pins. These currents counteract the pin-to-ground leakage currents, making the voltage distribution across all discs nearly uniform, raising string efficiency close to 100%.
Q. 11
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a transmission line model, lines classified as 'Short Transmission Lines' have a length and operating voltage of approximately:
A
Length < 80 km and Voltage < 20 kV (where shunt capacitance is completely neglected)
✓ Correct
B
Length 80 to 250 km and Voltage 20 to 100 kV
C
Length > 250 km and Voltage > 100 kV
D
Length < 1 km only
💡
Step-by-Step Explanation & Concept Rationale
Lines up to ~80 km operating below 20-25 kV have negligible charging currents. Their performance is analyzed by representing only series resistance and series inductive reactance, neglecting shunt capacitance entirely.
Q. 12
Electrical Engineering
Difficulty: Medium
(1 Mark)
In the Nominal-T method of representing a medium-length transmission line (80 - 240 km):
A
Total shunt admittance is lumped in the middle, between two half-series impedances (Z/2 each)
✓ Correct
B
Total series impedance is lumped in the middle, between two half-shunt admittances (Y/2 each)
C
Capacitance is neglected entirely
D
Line resistance is doubled
💡
Step-by-Step Explanation & Concept Rationale
In the Nominal-T network, total line impedance is divided into two halves (Z/2 on the sending side and Z/2 on the receiving side), with the entire line capacitance (admittance Y) lumped at the center node.
Q. 13
Electrical Engineering
Difficulty: Medium
(1 Mark)
The ABCD transmission matrix parameters for a symmetric, bilateral, reciprocal transmission network must satisfy which two fundamental identities?
A
A = D (symmetry) and A*D - B*C = 1 (reciprocity)
✓ Correct
B
A = B and C = D
C
A*D + B*C = 1 and B = C
D
A = 1 and D = 0
💡
Step-by-Step Explanation & Concept Rationale
For any passive bilateral two-port network, the reciprocity condition requires the determinant of the transmission matrix to equal unity: AD - BC = 1. If the network is identical viewed from either port (symmetrical), A = D.
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
Transposition of 3-phase overhead transmission line conductors at regular intervals along the route is performed to:
A
Equalize the line inductances and capacitances of all three phases, preventing voltage unbalance and telephone interference
✓ Correct
B
Increase the mechanical tensile strength of the line
C
Prevent bird strikes
D
Reduce corona discharge at terminal bushings
💡
Step-by-Step Explanation & Concept Rationale
When conductors are placed unsymmetrically on towers, mutual inductances and capacitances to ground differ per phase. Transposing (swapping conductor spatial positions at 1/3rd intervals) balances phase impedances and cancels electromagnetic induction into parallel communication lines.
Q. 15
Electrical Engineering
Difficulty: Medium
(1 Mark)
In High Voltage DC (HVDC) transmission systems, which converter configuration is universally employed for bulk power transfer?
A
12-pulse line-commutated converter (LCC) using thyristors or Voltage Source Converter (VSC) using IGBTs
✓ Correct
B
Single-phase half-wave diode rectifiers
C
Rotary mechanical converters
D
Step-down autotransformers
💡
Step-by-Step Explanation & Concept Rationale
Modern HVDC uses 12-pulse Graetz bridge rectifiers/inverters (connecting two 6-pulse bridges in series via Star-Star and Star-Delta transformers to cancel 5th and 7th harmonics) or modern VSC-HVDC utilizing IGBTs for independent real and reactive power control.
Q. 16
Electrical Engineering
Difficulty: Medium
(1 Mark)
The 'Break-Even Distance' beyond which HVDC transmission becomes more economical than HVAC transmission for overhead lines is approximately:
A
20 to 50 km
B
600 to 800 km for overhead lines (and 50 km for submarine cables)
✓ Correct
C
2000 km minimum
D
100 km
💡
Step-by-Step Explanation & Concept Rationale
HVDC terminal converter stations have high capital costs, but DC lines require only 2 conductors (bipolar) instead of 3, have no reactive power charging current limits, and cheaper towers. The cost curves intersect at 600-800 km for overhead lines and ~50 km for underwater cables.
Q. 17
Electrical Engineering
Difficulty: Medium
(1 Mark)
In underground cables, dielectric breakdown stress in the insulation is maximum at:
A
The outer surface of the lead sheath
B
The surface of the inner metallic conductor core (r = r_conductor)
✓ Correct
C
Midway between conductor and sheath
D
The metallic armor outer boundary
💡
Step-by-Step Explanation & Concept Rationale
Electric field gradient in a single-core coaxial cable is g = V / [x * ln(R/r)], where x is radial distance from the center. Electric stress is highest at minimum radius x = r (conductor surface) and lowest at maximum radius x = R (inner surface of lead sheath).
Q. 18
Electrical Engineering
Difficulty: Medium
(1 Mark)
The most economical ratio of sheath inner radius (R) to conductor radius (r) in an underground cable to minimize dielectric stress is:
A
R / r = 2.0
B
R / r = e ≈ 2.718
✓ Correct
C
R / r = pi ≈ 3.142
D
R / r = 1.414
💡
Step-by-Step Explanation & Concept Rationale
Maximum electric stress g_max = V / [r * ln(R/r)]. Minimizing g_max with respect to r gives d(g_max)/dr = 0, which occurs when ln(R/r) = 1, meaning R / r = e ≈ 2.718.
Q. 19
Electrical Engineering
Difficulty: Medium
(1 Mark)
Grading of underground cables (capacitance grading or intersheath grading) is performed to:
A
Increase conductor current-carrying capacity
B
Make the radial electrostatic stress distribution uniform throughout the dielectric insulation layer
✓ Correct
C
Reduce cable armor weight
D
Lower the supply voltage
💡
Step-by-Step Explanation & Concept Rationale
In ungraded cables, the dielectric near the conductor is heavily stressed while outer insulation is under-stressed. Capacitance grading uses dielectric layers of decreasing permittivity (epsilon1 > epsilon2 > epsilon3) from inside out, equalizing dielectric stress across the whole insulation.
Q. 20
Electrical Engineering
Difficulty: Easy
(1 Mark)
The metallic lead sheath over the insulation of an underground power cable serves to:
A
Provide mechanical protection against stone impacts
B
Prevent ingress of soil moisture, acids, and gases into the paper or polymeric dielectric insulation
✓ Correct
C
Increase mutual inductance
D
Act as the primary return path for load current
💡
Step-by-Step Explanation & Concept Rationale
High-voltage cable paper insulation is hygroscopic; even tiny traces of moisture ruin its dielectric breakdown strength. An extruded lead or aluminum sheath forms an impervious moisture-proof seal around the insulation.
Q. 21
Electrical Engineering
Difficulty: Easy
(1 Mark)
In power generation economics, the Load Factor of a power station is defined as:
A
Maximum demand / Average load
B
Average load over a given period / Maximum demand during that period (or kWh generated / [Max Demand * Time])
✓ Correct
C
Connected load / Maximum demand
D
Plant capacity / Maximum demand
💡
Step-by-Step Explanation & Concept Rationale
Load Factor = Average Load / Maximum Demand = (Total kWh generated in T hours) / (Maximum Demand in kW * T). A higher load factor means generating units operate near peak capacity consistently, lowering the cost of energy per unit (kWh).
Q. 22
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Diversity Factor of a power supply distribution system is defined as:
A
Sum of individual consumer maximum demands / Coincident maximum demand on the power station
✓ Correct
B
Coincident maximum demand / Connected load
C
Average load / Rated capacity
D
Always less than 1.0
💡
Step-by-Step Explanation & Concept Rationale
Diversity Factor = (Sum of individual maximum demands) / (Coincident maximum demand). Because different consumers reach peak demand at different times of the day, the diversity factor is always greater than 1.0 (typically 1.2 to 1.5).
Q. 23
Electrical Engineering
Difficulty: Easy
(1 Mark)
A high diversity factor in an electrical power system is economically desirable because it:
A
Increases line copper losses
B
Reduces the peak coincident maximum demand on the power plant, thereby minimizing required installed generator capacity and capital investment
✓ Correct
C
Increases fuel consumption
D
Forces transformer oil to degrade faster
💡
Step-by-Step Explanation & Concept Rationale
A higher diversity factor means non-coincident consumer peaks. This flattens the station load curve and lowers peak demand on the generators, allowing a smaller power plant to serve a larger total connected customer base.
Q. 24
Electrical Engineering
Difficulty: Easy
(1 Mark)
Plant Capacity Factor (or Plant Use Factor) of a generating station is defined as:
A
Actual energy produced (kWh) / (Installed Plant Rating * Total Operating Time)
✓ Correct
B
Maximum Demand / Installed Capacity
C
Reserve capacity / Maximum demand
D
Connected load / Plant capacity
💡
Step-by-Step Explanation & Concept Rationale
Plant Capacity Factor = (Actual kWh generated) / (Installed rating in kW * Time). The difference between Installed Capacity and Maximum Demand represents the spinning reserve capacity of the power plant.
Q. 25
Electrical Engineering
Difficulty: Medium
(1 Mark)
Kelvin's Law states that the most economical conductor cross-sectional area for an overhead transmission line occurs when:
A
The annual cost of electrical energy lost in conductor resistance equals the annual interest and depreciation charges on the capital cost of the conductor
✓ Correct
B
Voltage drop along the line is strictly zero
C
Conductor temperature reaches 100 degrees Celsius
D
Insulation cost equals tower cost
💡
Step-by-Step Explanation & Concept Rationale
Kelvin's Law determines conductor sizing: Total Annual Cost = C1 * A + C2 / A. Differentiating with respect to conductor area A yields the minimum cost condition: Annual interest and depreciation on conductor capital outlay (P1 ∝ A) = Annual cost of energy wasted as I^2*R losses (P2 ∝ 1/A).
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All 30 Solved Questions in Power Systems & Transmission Loaded!
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
Which type of fault occurs most frequently on high-voltage overhead power transmission lines?
A
Single line-to-ground (L-G) fault (approx. 70-80% of all faults)
✓ Correct
B
Three-phase symmetrical short-circuit (L-L-L)
C
Line-to-line (L-L) fault
D
Double line-to-ground (L-L-G) fault
💡
Step-by-Step Explanation & Concept Rationale
Single Line-to-Ground (L-G) faults caused by lightning flashover across insulators, tree branches, or tower flashovers account for over 70-80% of all power system faults. Symmetrical 3-phase faults are the rarest (~5%) but the most severe.
Q. 2
Electrical Engineering
Difficulty: Medium
(1 Mark)
In symmetrical component analysis (Fortescue's Theorem), a balanced 3-phase fault contains:
A
Positive-sequence currents only (I1 != 0, I2 = 0, I0 = 0)
✓ Correct
B
Negative-sequence currents only
C
Zero-sequence currents only
D
Equal positive, negative, and zero sequence components
💡
Step-by-Step Explanation & Concept Rationale
A symmetrical 3-phase fault maintains identical phase magnitudes and 120-degree displacements. Therefore, negative-sequence currents (I2) and zero-sequence currents (I0) are zero; only positive-sequence components (I1) exist.
Q. 3
Electrical Engineering
Difficulty: Medium
(1 Mark)
Zero-sequence currents can flow in a power system only if:
A
There is a complete closed path through neutral grounding to return to the source
✓ Correct
B
The system operates at leading power factor
C
The circuit breaker is of vacuum type
D
The generators run below synchronous speed
💡
Step-by-Step Explanation & Concept Rationale
Zero-sequence currents in all three phases are identical in magnitude and phase angle (Ia0 = Ib0 = Ic0). Their sum in the neutral is In = 3 * Ia0. Unless a physical metallic or earth return path exists through grounded neutrals, zero-sequence current cannot flow.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why is SF6 (Sulfur Hexafluoride) gas universally used as an arc-quenching and insulating medium in high-voltage circuit breakers?
A
It is highly flammable and burns the arc away
B
It is an electronegative gas that rapidly captures free electrons to form heavy negative ions, providing extraordinary dielectric strength and arc quenching
✓ Correct
C
It is lighter than air and dissipates immediately
D
It acts as a liquid semiconductor
💡
Step-by-Step Explanation & Concept Rationale
SF6 is strongly electronegative. In an electric arc, SF6 molecules capture free electrons (forming heavy, slow negative ions), deionizing the arc gap within microseconds at current zero-crossing. Its dielectric strength is 2.5 to 3 times that of air at atmospheric pressure.
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
The Making Capacity of a high-voltage AC circuit breaker is related to its Symmetrical Breaking Capacity by which empirical factor?
A
Making Capacity = 1.0 * Symmetrical Breaking Capacity
B
Making Capacity = 2.55 * Symmetrical Breaking Capacity (peak value)
✓ Correct
C
Making Capacity = 1.414 * Symmetrical Breaking Capacity
D
Making Capacity = 0.5 * Symmetrical Breaking Capacity
💡
Step-by-Step Explanation & Concept Rationale
When closing onto a dead short-circuit, maximum DC offset doubles the initial peak current. Accounting for damping, the standard multiplication factor is sqrt(2) * 1.8 ≈ 2.55 times the RMS symmetrical breaking current.
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
Restriking Voltage across circuit breaker contacts during arc interruption is defined as:
A
The normal 50 Hz power frequency voltage
B
The high-frequency transient voltage that appears across circuit breaker contacts at and immediately around current zero
✓ Correct
C
The DC excitation voltage
D
The voltage across the closing spring
💡
Step-by-Step Explanation & Concept Rationale
When the arc extinguishes at AC current zero, the energy stored in system L and C creates high-frequency transient oscillations. The transient voltage across opening contacts is the Restriking Voltage. If the Rate of Rise of Restriking Voltage (RRRV) exceeds dielectric recovery rate, the arc restrikes.
Q. 7
Electrical Engineering
Difficulty: Hard
(1 Mark)
Resistance switching in air-blast and gas circuit breakers is implemented by connecting a shunt resistor across the breaker contacts to:
A
Damp out transient restriking voltage oscillations and reduce the Rate of Rise of Restriking Voltage (RRRV)
✓ Correct
B
Increase full-load line current
C
Preheat the contacts in winter
D
Allow manual tripping
💡
Step-by-Step Explanation & Concept Rationale
Connecting a damping resistor R across the contacts alters the natural frequency of transient oscillation. If R <= 0.5 * sqrt(L / C), the transient becomes critically damped or overdamped, eliminating severe oscillatory restriking overvoltages.
Q. 8
Electrical Engineering
Difficulty: Easy
(1 Mark)
Vacuum circuit breakers (VCB) are most popular and technically superior in which voltage range?
A
Medium Voltage (11 kV to 33 kV)
✓ Correct
B
Extra High Voltage (400 kV to 800 kV)
C
Low Voltage 220 V single-phase
D
Ultra High Voltage > 1000 kV
💡
Step-by-Step Explanation & Concept Rationale
Vacuum interrupters offer maintenance-free operation, compact sealed bottles, high dielectric strength across tiny contact gaps (few mm), and long contact life, making them the industry standard for indoor distribution switchgear at 11 kV to 33 kV.
Q. 9
Electrical Engineering
Difficulty: Hard
(1 Mark)
A Mho relay (admittance relay) is an inherently directional distance relay that operates when:
A
The measured fault impedance falls within a circular operating characteristic that passes through the origin of the R-X diagram
✓ Correct
B
Fault current exceeds a fixed threshold regardless of phase angle
C
Voltage drops to zero on open circuit
D
Reverse power flows into the prime mover
💡
Step-by-Step Explanation & Concept Rationale
A Mho relay has a circular characteristic passing through the origin on the complex R-X plane. Because it covers only the forward quadrant without entering the reverse quadrants, it is inherently directional and ideally suited for long transmission line protection.
Q. 10
Electrical Engineering
Difficulty: Medium
(1 Mark)
An Impedance distance relay has an operating characteristic on the R-X impedance plane shaped as a:
A
Straight line through origin
B
Circle centered at the origin
✓ Correct
C
Hyperbola
D
Ellipse
💡
Step-by-Step Explanation & Concept Rationale
An impedance relay operates whenever the measured impedance magnitude |Z| = sqrt(R^2 + X^2) is less than a preset setting Z_set. On the R-X plane, this forms a circle centered at the origin, meaning it is non-directional unless paired with a separate directional element.
Q. 11
Electrical Engineering
Difficulty: Hard
(1 Mark)
Reactance distance relays are preferred for the protection of:
A
Very long EHV transmission lines
B
Short transmission lines and lines subject to high arc resistance
✓ Correct
C
Underground DC cables
D
Station auxiliary batteries
💡
Step-by-Step Explanation & Concept Rationale
A reactance relay measures only the imaginary component X of line impedance (its characteristic is a horizontal line on R-X plane: X < X_set). High fault arc resistance (R_arc) adds only to the real axis, so it does not affect the reach of a reactance relay, making it ideal for short lines.
Q. 12
Electrical Engineering
Difficulty: Medium
(1 Mark)
In stepped distance protection of transmission lines, Zone 1 typically protects what percentage of the line length, and with what time delay?
A
80% to 85% of line length with instantaneous trip (0 seconds)
✓ Correct
B
100% of line length with 1.0 second delay
C
120% of line length with instantaneous trip
D
50% with 0.5 second delay
💡
Step-by-Step Explanation & Concept Rationale
To prevent overreaching into the adjacent line section due to CT/PT errors, line parameter inaccuracies, and transients, Zone 1 is set to cover 80-85% of the protected line with zero intentional time delay (instantaneous, ~20 ms). Zone 2 covers the remaining 15-20% plus 20-50% of the next line with a delay of ~0.3-0.4 s.
Q. 13
Electrical Engineering
Difficulty: Medium
(1 Mark)
Biased (percentage) differential protection relays are used for large power transformers and generators instead of simple overcurrent differential relays to:
A
Prevent false tripping caused by external through-fault currents, CT saturation, and off-nominal transformer tap positions
✓ Correct
B
Measure frequency variations
C
Eliminate the need for current transformers
D
Protect against high line power factor
💡
Step-by-Step Explanation & Concept Rationale
During heavy external through-faults, unequal CT saturation can produce a false differential spill current. A percentage biased differential relay introduces restraining coils carrying average through-current (I1 + I2)/2, requiring differential operating current (I1 - I2) to exceed a fixed percentage slope before tripping.
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
Merz-Price circulating current protection scheme is a form of:
A
Directional overcurrent protection
B
Differential protection comparing currents entering and leaving the protected zone
✓ Correct
C
Distance protection
D
Thermal overload protection
💡
Step-by-Step Explanation & Concept Rationale
Merz-Price protection uses identical CTs on both sides of a generator or transformer winding connected in series opposition. Under normal conditions or external faults, secondary currents circulate harmlessly. An internal fault creates an unbalance that spills through the relay operating coil, initiating an immediate trip.
Q. 15
Electrical Engineering
Difficulty: Hard
(1 Mark)
A Peterson Coil (arc suppression coil) is connected between the neutral of a 3-phase power system and ground to:
A
Provide solid zero-resistance earthing
B
Produce an inductive current that neutralizes the capacitive earth fault current during single line-to-ground faults, extinguishing the arc automatically
✓ Correct
C
Step down generator voltage
D
Damp mechanical turbine vibrations
💡
Step-by-Step Explanation & Concept Rationale
In an ungrounded system, a single line-to-ground fault creates arcing ground transients due to line-to-ground capacitance. A tunable iron-cored reactor (Peterson coil) connected at the neutral is tuned to L = 1 / (3 * omega^2 * C0), canceling capacitive fault current and extinguishing the arc without tripping the line.
Q. 16
Electrical Engineering
Difficulty: Medium
(1 Mark)
The phenomenon of 'Arcing Grounds' in an ungrounded 3-phase system results from:
A
Repeated restriking of the intermittent arc across an earth fault, producing destructive cumulative overvoltages (up to 4-5 times normal voltage)
✓ Correct
B
Direct lightning hits on substations
C
Corroded ground rods
D
Transformer oil leakage
💡
Step-by-Step Explanation & Concept Rationale
When an arc to ground extinguishes at current zero, the healthy phases hold high trapped charges. The arc restrikes at voltage peak, progressively ratcheting up system phase voltages to 4 to 5 times normal phase-to-neutral voltage, puncturing insulation throughout the network.
Q. 17
Electrical Engineering
Difficulty: Medium
(1 Mark)
Zinc Oxide (ZnO) gapless surge arresters have superseded silicon carbide (SiC) arresters because:
A
They exhibit highly non-linear volt-ampere characteristics, drawing virtually zero leakage current under normal voltage and conducting heavily during overvoltage surges without needing series spark gaps
✓ Correct
B
They are soluble in water
C
They eliminate the need for grounding conductors
D
They act as thermal fuses
💡
Step-by-Step Explanation & Concept Rationale
ZnO varistors possess an extremely steep non-linear I-V curve (I ∝ V^alpha, where alpha ≈ 30-50). At normal operating voltage, ZnO acts as an insulator (microamps leakage). During surge overvoltages, resistance drops instantaneously to milliohms, discharging the surge safely to ground.
Q. 18
Electrical Engineering
Difficulty: Medium
(1 Mark)
The Basic Impulse Insulation Level (BIL) of electrical substation equipment is verified using a standard lightning impulse test voltage wave defined by:
A
1.2 / 50 microseconds wave (1.2 μs front time, 50 μs tail time to half value)
✓ Correct
B
8 / 20 microseconds wave
C
250 / 2500 microseconds wave
D
50 Hz sinusoidal wave
💡
Step-by-Step Explanation & Concept Rationale
Under IEC 60060 and IEEE standards, the standard lightning impulse test wave rises from zero to peak in 1.2 microseconds (+- 30%) and decays to 50% of peak value in 50 microseconds (+- 20%), denoted as a 1.2/50 μs wave.
Q. 19
Electrical Engineering
Difficulty: Hard
(1 Mark)
Switching impulse test waves (used to test EHV insulation above 300 kV) have a standard wave shape of:
A
1.2 / 50 microseconds
B
250 / 2500 microseconds
✓ Correct
C
8 / 20 microseconds
D
100 / 1000 microseconds
💡
Step-by-Step Explanation & Concept Rationale
Switching surges caused by breaker operations on long lines have much slower rise times and longer durations than lightning strikes. Under IEC standards, the standard switching impulse wave is 250/2500 μs.
Q. 20
Electrical Engineering
Difficulty: Medium
(1 Mark)
Marx Generator circuits are used in high-voltage testing laboratories to generate:
A
Continuous high-current DC for plating
B
High-voltage lightning and switching impulse voltages by charging capacitors in parallel and discharging them in series through spark gaps
✓ Correct
C
Pure 50 Hz high-power AC
D
Microwave radio frequencies
💡
Step-by-Step Explanation & Concept Rationale
Invented by Erwin Otto Marx, multiple capacitor stages are charged in parallel from a DC supply through high-value charging resistors. When the first spark gap triggers, all inter-stage sphere gaps fire simultaneously, connecting the capacitors in series to multiply the voltage up to megavolts.
Q. 21
Electrical Engineering
Difficulty: Medium
(1 Mark)
Cockcroft-Walton voltage multiplier circuits are employed to produce:
A
High-voltage direct current (HVDC) using a ladder network of diodes and capacitors
✓ Correct
B
Variable frequency 3-phase AC
C
Impulse current surges for arrester testing
D
High magnetic fields for MRI
💡
Step-by-Step Explanation & Concept Rationale
The Cockcroft-Walton multiplier is a cascade voltage doubler circuit consisting of a ladder network of rectifying diodes and capacitors that steps up relatively low AC input voltage into very high, ripple-free DC voltages for particle accelerators and X-ray tubes.
Q. 22
Electrical Engineering
Difficulty: Easy
(1 Mark)
Sphere gaps are widely used in high-voltage laboratories for:
A
Direct measurement of peak high voltages (AC, DC, and impulse) based on calibrated breakdown distance in air
✓ Correct
B
Filtering harmonics from generator supplies
C
Measuring power factor
D
Current amplification
💡
Step-by-Step Explanation & Concept Rationale
Standardized metal sphere gaps have an extremely uniform electric field. The breakdown sparkover voltage depends strictly on sphere diameter, spacing, and air density factor, making sphere gaps the primary international calibration standard for measuring peak high voltages.
Q. 23
Electrical Engineering
Difficulty: Hard
(1 Mark)
Townsend's criterion describes dielectric breakdown in gases as a result of:
A
Thermal runaway of the electrodes
B
Electron avalanche ionization multiplied by secondary ionization processes at the cathode (gamma mechanism)
✓ Correct
C
Electromechanical puncture
D
Chemical reduction of gas molecules
💡
Step-by-Step Explanation & Concept Rationale
Under Townsend's theory, primary electrons accelerated by the electric field ionize gas molecules by collision (alpha coefficient). Positive ions strike the cathode to release secondary electrons (gamma coefficient). When gamma * [exp(alpha * d) - 1] = 1, current becomes self-sustaining, leading to spark breakdown.
Q. 24
Electrical Engineering
Difficulty: Medium
(1 Mark)
Paschen's Law states that the breakdown voltage (Vb) of a uniform air gap between two parallel electrodes is a function of:
A
Electrode weight only
B
The product of gas pressure and gap distance (Vb = f(p * d))
✓ Correct
C
Supply frequency
D
Electrode temperature exclusively
💡
Step-by-Step Explanation & Concept Rationale
Paschen's Law establishes that breakdown voltage Vb depends on the product (p * d). At very low pressure, collisions are too rare for ionization; at very high pressure, mean free path is too short for electrons to gain ionization energy. Thus, Vb exhibits a distinct minimum (approx. 327 V for air at p*d ≈ 0.5 Torr-cm).
Q. 25
Electrical Engineering
Difficulty: Hard
(1 Mark)
Streamer (or Kanal) mechanism of spark breakdown supersedes Townsend's theory when:
A
Gap length is large (d > several cm) and pressure is atmospheric or higher, where space charge distortion creates rapid self-propagating streamer filaments
✓ Correct
B
The gas is completely evacuated to pure vacuum
C
Voltage is purely DC and below 100 V
D
Electrodes are made of carbon
💡
Step-by-Step Explanation & Concept Rationale
In long air gaps at atmospheric pressure, photo-ionization and positive ion space charge accumulation at the avalanche head exceed 10^8 electrons, transforming the avalanche into an ultra-fast, luminous, self-propagating plasma streamer channel within nanoseconds.
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Q. 1
Electronics Engineering
Difficulty: Medium
(1 Mark)
Light Emitting Diodes (LEDs) are fabricated from direct bandgap compound semiconductors (such as GaAs, GaN, InGaN) rather than indirect bandgap semiconductors like Silicon because:
A
In direct bandgap materials, electrons recombine directly across the bandgap releasing energy predominantly as photons (light) without requiring lattice phonon momentum assistance
✓ Correct
B
Silicon is opaque to electricity
C
Direct bandgap materials have zero thermal resistance
D
Silicon cannot be doped P-type
💡
Step-by-Step Explanation & Concept Rationale
In direct bandgap semiconductors, the conduction band minimum and valence band maximum align at the same crystal momentum (k = 0). Radiative electron-hole recombination is direct and highly efficient, producing photons: lambda = h * c / Eg. In Silicon (indirect bandgap), recombination requires a phonon (lattice vibration), dissipating energy as heat.
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Q. 1
Electrical Engineering
Difficulty: Easy
(1 Mark)
An ideal transformer does not change:
A
Voltage and Current
B
Frequency and Power
✓ Correct
C
Impedance and Voltage
D
Current and Flux
💡
Step-by-Step Explanation & Concept Rationale
A transformer is a static electromagnetic device that transfers electrical energy from one circuit to another without changing the frequency (f1 = f2) and with 100% efficiency in an ideal case (P_in = P_out, V1*I1 = V2*I2).
Q. 2
Electrical Engineering
Difficulty: Easy
(1 Mark)
The EMF per turn in the primary winding of a transformer compared to the EMF per turn in its secondary winding is:
A
Greater than the secondary EMF per turn
B
Less than the secondary EMF per turn
C
Exactly equal to the secondary EMF per turn
✓ Correct
D
Inversely proportional to the transformation ratio
💡
Step-by-Step Explanation & Concept Rationale
From Faraday's law, induced EMF per turn is E1/N1 = E2/N2 = 4.44 * f * phi_m. Since the same mutual magnetic flux links both windings, the induced EMF per turn is identical in both primary and secondary windings.
Q. 3
Electrical Engineering
Difficulty: Easy
(1 Mark)
The open-circuit (OC) test on a transformer is performed to determine:
A
Full-load copper loss and equivalent resistance
B
Core (iron) loss and shunt branch parameters (Rc and Xm)
✓ Correct
C
Leakage reactance of windings
D
Voltage regulation at full load
💡
Step-by-Step Explanation & Concept Rationale
The open-circuit test is conducted at rated voltage and frequency, typically on the Low Voltage (LV) side with the HV side open. Because no-load current is tiny (2-5% of rated), series copper loss is negligible, and the wattmeter reading directly measures the core (hysteresis + eddy current) loss.
Q. 4
Electrical Engineering
Difficulty: Easy
(1 Mark)
The short-circuit (SC) test on a transformer is conducted to measure:
A
Core (iron) losses at rated voltage
B
Full-load copper losses and equivalent series impedance (Req and Xeq)
✓ Correct
C
Dielectric strength of insulation oil
D
Magnetizing current
💡
Step-by-Step Explanation & Concept Rationale
The SC test is performed at reduced voltage (typically 5-10% of rated) on the High Voltage (HV) side with the LV side shorted, circulating full-load rated current. Because the applied voltage and core flux are very low, iron losses are negligible, and the input power represents the full-load copper loss (I^2 * Req).
Q. 5
Electrical Engineering
Difficulty: Medium
(1 Mark)
A 100 kVA transformer has a maximum efficiency of 98% at 80% of full load. What is the ratio of full-load copper loss (P_cu_fl) to iron loss (P_i)?
A
P_cu_fl / P_i = 1.0
B
P_cu_fl / P_i = 1 / 0.8^2 = 1 / 0.64 ≈ 1.56
✓ Correct
C
P_cu_fl / P_i = 0.8
D
P_cu_fl / P_i = 0.64
💡
Step-by-Step Explanation & Concept Rationale
Maximum efficiency occurs when variable copper loss equals constant iron loss: x^2 * P_cu_fl = P_i. At x = 0.8: 0.8^2 * P_cu_fl = P_i -> 0.64 * P_cu_fl = P_i -> P_cu_fl / P_i = 1 / 0.64 = 1.5625.
Q. 6
Electrical Engineering
Difficulty: Medium
(1 Mark)
The condition for zero voltage regulation in a practical power transformer occurs at:
A
Unity power factor
B
Lagging power factor with tan(phi) = Req / Xeq
C
Leading power factor with tan(phi) = Req / Xeq
✓ Correct
D
Zero power factor lagging
💡
Step-by-Step Explanation & Concept Rationale
Voltage regulation is approximate to (I2 * Req * cos(phi) +- I2 * Xeq * sin(phi)) / V2. For zero regulation, Req * cos(phi) - Xeq * sin(phi) = 0 (which requires a leading PF), yielding tan(phi) = Req / Xeq (leading).
Q. 7
Electrical Engineering
Difficulty: Medium
(1 Mark)
The maximum voltage regulation in a transformer occurs at a:
A
Lagging power factor where tan(phi) = Xeq / Req
✓ Correct
B
Leading power factor where tan(phi) = Xeq / Req
C
Unity power factor
D
Zero power factor leading
💡
Step-by-Step Explanation & Concept Rationale
Differentiating the regulation expression with respect to phi and setting to zero yields the maximum regulation condition: tan(phi) = Xeq / Req at a lagging power factor, meaning cos(phi) = Req / Zeq (lagging).
Q. 8
Electrical Engineering
Difficulty: Easy
(1 Mark)
Why are distribution transformers designed to achieve maximum efficiency at 50% to 70% of full load rather than at 100% full load?
A
They operate under heavy overload throughout the entire day
B
They remain connected to the supply 24 hours a day but operate at light or average load (30-60%) for most hours
✓ Correct
C
To keep winding temperatures above ambient at all times
D
To increase leakage reactance
💡
Step-by-Step Explanation & Concept Rationale
Distribution transformers are energized continuously (24 hours/day) with fluctuating consumer loads, averaging 50-70% capacity. Designing core loss to be very low and setting maximum efficiency around average load optimizes all-day energy efficiency.
Q. 9
Electrical Engineering
Difficulty: Easy
(1 Mark)
All-day efficiency of a transformer is defined as the ratio of:
A
Output power in kW to Input power in kW at full load
B
Total energy output in kWh over 24 hours to Total energy input in kWh over 24 hours
✓ Correct
C
KVA output to KVA input
D
Maximum efficiency multiplied by 24
💡
Step-by-Step Explanation & Concept Rationale
All-day efficiency (or commercial energy efficiency) is calculated on an energy basis: eta_allday = (Energy Output in kWh over 24 hours) / (Energy Input in kWh over 24 hours). It is vital for distribution transformers.
Q. 10
Electrical Engineering
Difficulty: Medium
(1 Mark)
In an autotransformer with transformation ratio k = V2 / V1 (where k < 1), the fraction of power transferred inductively through magnetic coupling is:
A
k
B
1 - k
✓ Correct
C
1 / k
D
k^2
💡
Step-by-Step Explanation & Concept Rationale
In an autotransformer, total power is transferred via two mechanisms: conductive power = k * P_total, and inductive power = (1 - k) * P_total. The copper saving is k * (copper of 2-winding transformer).
Q. 11
Electrical Engineering
Difficulty: Medium
(1 Mark)
What is the primary operational hazard of leaving the secondary winding of an energized Current Transformer (CT) open-circuited?
A
The primary winding will immediately burn out
B
The secondary EMF rises to a dangerously high voltage due to core saturation, posing fatal electric shock and insulation breakdown risks
✓ Correct
C
The CT core demagnetizes permanently
D
The primary line current drops to zero
💡
Step-by-Step Explanation & Concept Rationale
In a CT, secondary ampere-turns oppose primary ampere-turns (N2*I2 opposes N1*I1). If the secondary is opened, secondary MMF becomes zero, and the entire large line current acts as magnetizing current. This drives the core deep into saturation, inducing lethal peak voltages (thousands of volts) across the secondary terminals.
Q. 12
Electrical Engineering
Difficulty: Easy
(1 Mark)
The Buchholz relay is a protective device used in oil-immersed transformers to detect:
A
External short circuits on overhead transmission lines
B
Incipient internal faults (such as insulation breakdown, inter-turn faults, and core heating) producing gas accumulation
✓ Correct
C
Overvoltage caused by lightning strikes
D
Excessive load power factor
💡
Step-by-Step Explanation & Concept Rationale
A Buchholz relay is installed in the pipe connecting the main transformer tank to the conservator tank. Gas bubbles produced by slow internal arcing or overheating rise and trigger an alarm. Severe faults produce a violent oil surge that trips the circuit breaker.
Q. 13
Electrical Engineering
Difficulty: Easy
(1 Mark)
The silica gel breather attached to a transformer conservator tank serves to:
A
Filter solid dust particles from the transformer oil
B
Absorb atmospheric moisture from the air entering the conservator during thermal breathing
✓ Correct
C
Cool down the hot circulating oil
D
Neutralize acid formation in the oil
💡
Step-by-Step Explanation & Concept Rationale
As transformer temperature rises and falls, oil expands and contracts, causing air to breathe in and out. The silica gel crystals absorb moisture from incoming air to maintain the high dielectric breakdown strength of the insulating oil.
Q. 14
Electrical Engineering
Difficulty: Easy
(1 Mark)
Fresh, dry silica gel in a transformer breather has which color, and what color does it turn when saturated with moisture?
A
Pink when dry, Deep Blue when wet
B
Deep Blue when dry, Pink (or whitish) when saturated with moisture
✓ Correct
C
Yellow when dry, Green when wet
D
White when dry, Black when wet
💡
Step-by-Step Explanation & Concept Rationale
Dry cobalt chloride-impregnated silica gel is deep blue. When it absorbs moisture, it chemically transitions to hydrated cobalt chloride, turning light pink, indicating it needs reactivation by heating.
Q. 15
Electrical Engineering
Difficulty: Medium
(1 Mark)
Which three-phase transformer connection is capable of converting a 3-phase supply into a balanced 2-phase supply for electric arc furnaces?
A
Open-Delta (V-V) Connection
B
Scott-T (or Teaser) Connection
✓ Correct
C
Star-Zigzag Connection
D
Delta-Star Connection
💡
Step-by-Step Explanation & Concept Rationale
The Scott connection utilizes two single-phase transformers: a main transformer with center-tapped primary (50% tap) and a teaser transformer tapped at 86.6% (sqrt(3)/2) of primary turns, effectively converting 3-phase to balanced 2-phase or vice versa.
Q. 16
Electrical Engineering
Difficulty: Medium
(1 Mark)
In an Open-Delta (V-V) transformer bank where one transformer of a 3-transformer Delta-Delta bank is removed, the remaining two transformers can deliver what percentage of the original bank's rated kVA capacity?
A
66.7%
B
57.7% (1 / sqrt(3))
✓ Correct
C
50.0%
D
86.6%
💡
Step-by-Step Explanation & Concept Rationale
The V-V bank capacity is sqrt(3) * V * I = sqrt(3) * S_single. Since the original 3-transformer delta bank capacity was 3 * S_single, the ratio is (sqrt(3) * S_single) / (3 * S_single) = 1 / sqrt(3) ≈ 57.7% (operating at 86.6% of the two remaining transformers' combined rating).
Q. 17
Electrical Engineering
Difficulty: Hard
(1 Mark)
The tertiary winding in a large Star-Star (Y-Y) power transformer is connected in Delta primarily to:
A
Provide a path for circulating triplen harmonic (3rd, 9th, etc.) currents, stabilizing the neutral and ensuring sinusoidal phase voltages
✓ Correct
B
Increase primary leakage reactance
C
Eliminate secondary line resistance
D
Reduce the total iron weight of the core
💡
Step-by-Step Explanation & Concept Rationale
In ungrounded Y-Y transformers, 3rd harmonic currents cannot flow, causing severe distortion in phase voltages and neutral oscillation. A closed Delta tertiary winding provides a circulating path for 3rd harmonic magnetizing currents, eliminating distortion.
Q. 18
Electrical Engineering
Difficulty: Medium
(1 Mark)
For two 3-phase transformers to operate successfully in parallel, which condition is ABSOLUTELY ESSENTIAL and non-negotiable?
A
Identical kVA ratings
B
Identical phase sequence, zero relative phase displacement (same vector group), and identical voltage ratios
✓ Correct
C
Identical magnetizing reactance
D
Equal core cross-sectional areas
💡
Step-by-Step Explanation & Concept Rationale
If phase sequence is incorrect or vector groups have phase displacement (e.g., Yd1 vs Yd11), a massive phase difference occurs across the secondary terminals, creating dead short-circuit circulating currents that will immediately destroy the transformers.
Q. 19
Electrical Engineering
Difficulty: Medium
(1 Mark)
When two transformers of different kVA ratings but equal per-unit leakage impedances operate in parallel, how do they share a common load?
A
Equally (50% each)
B
In direct proportion to their individual kVA ratings
✓ Correct
C
Inversely proportional to their kVA ratings
D
The smaller transformer carries the entire load
💡
Step-by-Step Explanation & Concept Rationale
When per-unit impedances based on their own ratings are equal (Z_pu1 = Z_pu2), the ohmic impedances are inversely proportional to their kVA ratings. Consequently, load current divides such that each transformer carries load strictly proportional to its kVA capacity without overloading.
Q. 20
Electrical Engineering
Difficulty: Hard
(1 Mark)
Transformer magnetizing inrush current occurs upon energizing because:
A
The load demands instantaneous reactive power
B
Core flux attempts to double (flux doubling effect) when energized at voltage zero-crossing, driving the core deep into magnetic saturation
✓ Correct
C
The oil temperature is cold
D
Secondary winding is open-circuited
💡
Step-by-Step Explanation & Concept Rationale
Since flux is the integral of voltage, switching on at a voltage zero-crossing requires flux to rise from 0 to 2*phi_m (flux doubling). In saturation, core permeability drops to near air permeability, resulting in inrush currents up to 6-10 times rated full-load current (predominantly 2nd harmonic).
Q. 21
Electrical Engineering
Difficulty: Medium
(1 Mark)
Harmonic analysis shows that transformer magnetizing inrush current is predominantly rich in which harmonic?
A
3rd harmonic
B
2nd harmonic
✓ Correct
C
5th harmonic
D
7th harmonic
💡
Step-by-Step Explanation & Concept Rationale
Transformer inrush current has a strong unidirectional DC offset and asymmetric wave shape rich in the 2nd harmonic. Modern differential protective relays use 2nd Harmonic Restraint to distinguish inrush from internal short-circuit faults.
Q. 22
Electrical Engineering
Difficulty: Easy
(1 Mark)
In a core-type transformer, the low-voltage (LV) winding is placed closest to the iron core, with the high-voltage (HV) winding on the outside. Why?
A
To reduce core eddy current losses
B
To minimize the thickness and cost of insulating materials required between winding and grounded core
✓ Correct
C
To improve cooling efficiency of the LV winding
D
To increase mutual inductance
💡
Step-by-Step Explanation & Concept Rationale
Insulating the winding against the grounded core requires dielectric insulation proportional to operating voltage. Placing the LV winding closest to the core minimizes insulation thickness and clearance distance, substantially lowering material costs and physical size.
Q. 23
Electrical Engineering
Difficulty: Medium
(1 Mark)
What is the standard test method used to determine the dielectric breakdown strength of transformer insulating oil?
A
Megger insulation resistance test at 500 V
B
Oil Breakdown Voltage (BDV) test across a 2.5 mm spark gap using spherical or mushroom electrodes
✓ Correct
C
Kelvin double bridge measurement
D
Open circuit test
💡
Step-by-Step Explanation & Concept Rationale
Under IEC 60156 standards, insulating oil BDV is tested across a calibrated 2.5 mm electrode gap. Good transformer oil must withstand at least 30 kV (and above 50 kV for high-voltage power transformers) without dielectric flashover.
Q. 24
Electrical Engineering
Difficulty: Hard
(1 Mark)
Sumpner's Test (Back-to-Back Test) on two identical transformers is conducted to determine:
A
Only the transformation ratio
B
Full-load temperature rise and total losses under full-load conditions without drawing large net power from the supply
✓ Correct
C
Mechanical strength of transformer tank
D
Insulation resistance of bushings
💡
Step-by-Step Explanation & Concept Rationale
Sumpner's test connects two identical transformers with primaries in parallel and secondaries in series opposition. It allows operating both at rated flux and rated full-load current simultaneously, consuming only the internal losses (core + copper), enabling precise heat run testing.
Q. 25
Electrical Engineering
Difficulty: Hard
(1 Mark)
A 50 Hz transformer is operated on a 60 Hz supply of the same rated voltage. What will be the effect on its core losses?
A
Hysteresis loss decreases, while eddy current loss remains essentially constant
✓ Correct
B
Both hysteresis and eddy current losses increase
C
Both hysteresis and eddy current losses decrease to zero
D
Total core loss increases by 20%
💡
Step-by-Step Explanation & Concept Rationale
With constant voltage V, Bm is proportional to V/f = 1/f. Hysteresis loss P_h ∝ f * Bm^1.6 ∝ f * (1/f)^1.6 = f^-0.6 (decreases). Eddy current loss P_e ∝ f^2 * Bm^2 ∝ f^2 * (1/f)^2 = constant. Thus, total core loss decreases.
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