Past Questions

Bulk power transmission over long distances is preferred using HVDC (High Voltage Direct Current) for several technical and economic reasons: 1. **Minimum Line Power Losses (Correct):** In HVDC, there is no skin effect (which increases effective resistance in AC lines) and no reactive power flow (which causes $I^2X$ losses). For a given power rating and distance, HVDC lines have lower conductor losses compared to HVAC lines. 2. **Cost of Terminals (Incorrect):** HVDC terminals (converter stations) are significantly more expensive than AC substations due to the complex power electronics required for rectification and inversion. This is why HVDC is only economical for very long distances (the 'break-even distance'). 3. **Harmonics (Incorrect):** HVDC systems actually introduce significant harmonic distortion into the AC system due to the switching of thyristors/IGBTs in the converters, requiring expensive AC and DC filters to mitigate. 4. **Protection (Incorrect):** Protection of HVDC lines is more complex than AC lines. AC systems benefit from well-established distance and differential protection; HVDC requires specialized control systems to handle faults and converter stability.

Transposition is the process of swapping the positions of the three phases of a three-phase transmission line at regular intervals along the length of the line. 1. Why it is done: In an asymmetrical transmission line (where conductors are not equidistant), the inductance and capacitance of each phase are different. This leads to unbalanced voltage drops and unbalanced currents in the neutral/ground. 2. Impact on Communication Lines: These unbalanced currents create an asymmetrical magnetic field around the transmission line. This magnetic field induces an electromotive force (EMF) in nearby communication lines (telephone lines), causing electromagnetic interference (EMI) or 'noise'. 3. How Transposition helps: By rotating the positions of the conductors, the average inductance and capacitance for all three phases are equalized over the total length of the line. This balances the phase voltages and cancels out the net induced voltage in neighboring communication lines. Analysis of other options: - Skin effect (a): This is a property of the conductor's material and frequency; it is reduced by using stranded conductors or hollow conductors, not transposition. - Ferranti effect (b): This is the rise in voltage at the receiving end of a long line under no-load or light-load conditions; it is mitigated by using shunt reactors. - Transmission loss (c): Losses are primarily $I^2R$ losses; transposition does not significantly reduce the resistance or the current magnitude to lower these losses.

Step-by-step reasoning: 1. **Identify the Characteristic Equation**: The given characteristic equation is (s + 1)(s + 2)(s - 3) = 0. 2. **Find the Poles**: The poles of the system are the roots of the characteristic equation. Setting each factor to zero: - s + 1 = 0 => s₁ = -1 - s + 2 = 0 => s₂ = -2 - s - 3 = 0 => s₃ = +3 3. **Stability Criterion**: In control systems, a linear time-invariant (LTI) system is stable if and only if all the poles of its transfer function (roots of the characteristic equation) lie in the left-half of the s-plane (LHP). This means the real part of every pole must be negative. 4. **Analyze the Poles**: - s₁ = -1 (Real part is negative, stable) - s₂ = -2 (Real part is negative, stable) - s₃ = +3 (Real part is positive, unstable) 5. **Conclusion**: Since there is at least one pole (s = 3) located in the right-half of the s-plane (RHP), the system is unstable. Even a single pole in the RHP causes the system's impulse response to grow exponentially over time. Comparison with other options: - (a) Stable: Incorrect, as not all poles are in the LHP. - (b) Marginally stable: Incorrect, as this requires poles on the imaginary axis (jω axis) with no poles in the RHP. - (c) Not necessarily stable: Incorrect, as the stability is definitively determined to be unstable based on the given equation.

In power system analysis, the classification of a transmission line as short, medium, or long is fundamentally determined by the relationship between the physical length of the line (L) and the wavelength (λ) of the operating frequency. 1. Short Line: When the length is small compared to the wavelength (typically L < 0.1λ or < 80 km), the shunt capacitance is negligible and can be ignored. 2. Medium Line: When the length is between 0.1λ and 0.2λ (typically 80 km to 250 km), the shunt capacitance cannot be ignored and is usually represented as lumped parameters (e.g., nominal-π or nominal-T models). 3. Long Line: When the length exceeds 0.2λ (typically > 250 km), the parameters must be treated as distributed along the entire length of the line, requiring the use of wave equations and hyperbolic functions. While we often use physical distance (km) as a rule of thumb, these distances are derived from the standard power frequency (50Hz or 60Hz). Since wavelength λ = v/f (where v is the velocity of propagation, approximately the speed of light), the classification is physically based on the wavelength. - Option (a) is incorrect because voltage affects insulation and conductor size, not the classification of line models. - Option (b) is a common practical shorthand, but the theoretical basis is the ratio of length to wavelength. - Option (d) is incorrect as power transmission levels do not dictate whether a line is modeled as short, medium, or long.

In a transformer, the insulating oil serves two primary purposes: cooling the core and windings, and providing electrical insulation (dielectric strength) between the live parts and the grounded tank. 1. **Dielectric Strength**: This refers to the maximum electric field that a material can withstand without breaking down (becoming conductive). Pure transformer oil has a very high dielectric strength. 2. **Effect of Moisture**: Water is a polar molecule and is significantly more conductive than transformer oil. When moisture enters the oil, it forms droplets or dissolves, creating conductive paths. This drastically lowers the breakdown voltage of the oil, meaning the oil can no longer effectively prevent electrical arcs or short circuits between the windings or between the windings and the tank. 3. **Analysis of other options**: - (a) Density: While moisture might slightly change the density, this is not the critical failure mechanism for transformer operation. - (c) Lubricating property: Transformer oil is used for insulation and cooling, not primarily for lubrication of moving parts (as there are few to none in a static transformer). - (d) Core rust: While moisture can cause rust over a very long period, the immediate and most dangerous electrical effect is the reduction of dielectric strength, which leads to catastrophic insulation failure.

The Ferranti effect is a phenomenon observed in long transmission lines where the receiving-end voltage (Vr) becomes higher than the sending-end voltage (Vs). 1. **Physical Cause**: Long transmission lines possess a significant shunt capacitance (capacitance between the line and ground). When the line is lightly loaded or open-circuited, the charging current (which flows due to this shunt capacitance) flows through the line's inductance. 2. **Voltage Rise**: This charging current is leading in nature. The voltage drop across the line inductance due to this leading current actually adds to the sending-end voltage at the receiving end, causing the voltage to rise. 3. **Condition**: This effect is most pronounced when the line is 'lightly loaded' or 'no-load' because the inductive voltage drop caused by the load current (which would normally cause a voltage drop) is absent or minimal, allowing the capacitive charging effect to dominate. Analysis of other options: - **b. Power factor is unity**: At unity power factor, the load current is in phase with the voltage, which typically results in a standard voltage drop across the line, not a rise. - **c. Power factor is leading**: While a leading load can exacerbate the voltage rise, the Ferranti effect is a characteristic of the line itself (due to its capacitance) and is specifically identified as occurring during light load or no-load conditions. - **d. Corona effect**: Corona is a loss phenomenon involving the ionization of air; while it may slightly affect the effective capacitance, it is not the cause of the Ferranti effect.

To determine both stability and transient response, we need information about the location of the closed-loop poles in the s-plane. 1. **Root Locus**: This method plots the trajectories of the closed-loop poles as a system parameter (usually the gain K) varies. Since the positions of the poles directly determine both the stability (whether poles are in the left-half plane) and the transient response (damping ratio ζ and natural frequency ωn, which dictate overshoot, settling time, and rise time), the Root Locus is the best tool for analyzing both simultaneously. 2. **Bode Plot**: This is a frequency response method. While it is excellent for determining relative stability (Gain Margin and Phase Margin), it does not provide a direct visualization of the transient response (time-domain behavior) as intuitively as the Root Locus does. 3. **Nyquist Plot**: Like the Bode plot, this is a frequency domain technique. It is primarily used to determine the absolute and relative stability of a closed-loop system by analyzing the open-loop transfer function, but it does not directly provide transient response characteristics. Therefore, because the Root Locus maps the movement of closed-loop poles, it is the most comprehensive method for evaluating both stability and transient performance.

For a transformer, maximum efficiency occurs when iron loss equals copper loss at that load. Given maximum efficiency = 90% at full load and unity power factor: Let full-load output = P_out and full-load losses = P_i + P_c, where at maximum efficiency P_i = P_c. Efficiency at full load: η = P_out / (P_out + P_i + P_c) = 0.9 Since P_i = P_c, total loss = 2P_i. So, P_out / (P_out + 2P_i) = 0.9 => P_out = 18P_i => P_i = P_c = P_out/18. At half load, copper loss becomes (1/2)^2 = 1/4 of full-load copper loss: P_c(half) = P_c/4 = P_out/72. Output at half load = P_out/2. Thus efficiency at half load: η_half = (P_out/2) / (P_out/2 + P_i + P_out/72) = 36 / (36 + 4 + 1) = 36/41 = 0.8780 ≈ 87.8%. So the correct option is d. Why others are incorrect: - a) 86.7% is too low; it would correspond to larger losses than actually present. - b) 88.26% is slightly higher than the correct value; it does not match the loss calculation. - c) 88.9% is also too high and exceeds the computed half-load efficiency. Note: The numerical result is 87.8%, which corresponds to option d.

The skin effect is the tendency of an alternating electric current (AC) to distribute itself within a conductor such that the current density is largest near the surface of the conductor and decreases with greater depths. The depth to which the current penetrates is characterized by the 'skin depth' (δ), which is mathematically defined as: δ = 1 / √(π f μ σ), where 'f' is the supply frequency, 'μ' is the magnetic permeability of the conductor, and 'σ' is the electrical conductivity. From this formula, it is evident that the skin depth (δ) is inversely proportional to the square root of the frequency (f). Therefore, as the supply frequency increases, the skin depth decreases, meaning the current is pushed further toward the surface (the 'skin') of the conductor. This intensification of the current distribution at the surface is described as an increase in the skin effect. Consequently, higher frequencies lead to a more pronounced skin effect, which increases the effective resistance of the conductor.

In power transmission lines, the line impedance is primarily composed of resistance (R) and inductive reactance (XL). The inductive reactance is the dominant factor that limits the power transfer capability of the line and causes voltage drops. Series capacitors are installed in series with the transmission line to provide a capacitive reactance (XC) that opposes and partially cancels out the inductive reactance (XL). The net reactance becomes X_net = XL - XC. By reducing the overall inductive reactance effect, the power transfer capability (P = (V1*V2/X)*sin(delta)) is increased, and the stability of the system is improved. Why other options are incorrect: - (a) Capacitive reactance: Series capacitors *add* capacitive reactance to the line; they do not 'improve' it in the sense of reducing it. - (c) Voltage: While reducing reactance can indirectly help with voltage drop, the primary physical purpose of a series capacitor is the compensation of the inductive reactance. Voltage support is primarily the role of shunt capacitors. - (d) Regulation: Voltage regulation is a result of the overall impedance; however, the specific mechanism of a series capacitor is the compensation of the inductive reactance effect.

In control systems, specifically for a standard second-order system, the characteristic equation is given by s² + 2ζωn s + ωn² = 0, where ζ is the damping ratio and ωn is the natural frequency. For a closed-loop system with a gain K and a plant G(s) = 1/(s(s+a)), the characteristic equation becomes s² + as + K = 0. Comparing this to the standard form: 1) ωn² = K (Natural frequency increases with gain), and 2) 2ζωn = a. Substituting ωn = √K into the second equation, we get 2ζ√K = a, which simplifies to ζ = a / (2√K). This mathematical relationship shows that the damping ratio ζ is inversely proportional to the square root of the gain K. Therefore, as the gain K increases, the damping ratio ζ decreases, leading to a more oscillatory response (higher overshoot).

In a shaded-pole induction motor, a copper ring (shading coil) is placed around a portion of the pole face. This creates two distinct paths for the magnetic flux: one through the unshaded part and one through the shaded part. 1. **Flux Lagging**: According to Lenz's Law, the current induced in the shading coil opposes the change in flux. This causes the magnetic flux in the shaded portion to lag behind the flux in the unshaded portion in time. 2. **Rotating Field**: Because the flux in the unshaded part reaches its peak first and the flux in the shaded part reaches its peak slightly later, a weak rotating magnetic field is produced that moves from the unshaded portion toward the shaded portion. 3. **Wait, Correction on Direction**: Actually, the rotating field moves from the unshaded to the shaded part. However, the question asks about the rotor's movement relative to the flux phase. Let's re-evaluate: The flux in the shaded part lags the flux in the unshaded part. The resulting rotating field moves from the unshaded part to the shaded part. Wait, looking at the standard theory: The flux in the shaded pole lags the flux in the unshaded pole. This creates a rotating field that moves from the unshaded part to the shaded part. Therefore, the rotor (which follows the field) runs from the unshaded portion to the shaded portion. Let's re-read the options carefully. Option (b) says: 'shaded portion to the unshaded portion... flux in the former lags'. Option (d) says: 'unshaded portion to the shaded portion... flux in the former lags'. Correction: The flux in the shaded portion (former in option b) lags the flux in the unshaded portion. The rotating field moves from unshaded to shaded. Therefore, the rotor moves from unshaded to shaded. Wait, let's check the standard textbook answer for this specific MPSC/Competitive exam question. In many technical contexts for shaded pole motors, the flux in the shaded part lags, and the rotor moves from the unshaded to the shaded part. However, if we look at the phrasing 'flux in the former lags that of the latter' in option (b), 'former' refers to the shaded portion. This is physically correct (shaded flux lags unshaded flux). But the direction of rotation is unshaded $\rightarrow$ shaded. Let's re-examine Option (b) vs (d). (b) shaded $\rightarrow$ unshaded; shaded lags unshaded. (Incorrect direction) (d) unshaded $\rightarrow$ shaded; unshaded lags shaded. (Incorrect phase relationship) Actually, the standard correct physical description is: Flux in shaded part lags flux in unshaded part $\rightarrow$ Rotor moves from unshaded to shaded. Looking at the options provided in the prompt again: If the answer is (b), it implies the rotor moves shaded $\rightarrow$ unshaded. This happens if the flux in the unshaded part lags. But the shading coil is on the shaded part. Let's re-verify the phase: The shading coil increases the inductance of that part of the pole, causing the current/flux to lag. Thus, flux in shaded part lags flux in unshaded part. The field rotates from unshaded to shaded. If the provided official key for this specific exam is (b), it is often due to a common error in question banks or a specific definition of 'former/latter'. But based on physics: Flux(shaded) lags Flux(unshaded). Direction: Unshaded $\rightarrow$ Shaded. Looking at the options again: (a) shaded $\rightarrow$ unshaded; shaded leads unshaded (Wrong) (b) shaded $\rightarrow$ unshaded; shaded lags unshaded (Wrong direction) (c) unshaded $\rightarrow$ shaded; unshaded leads shaded (Correct phase: if unshaded leads, shaded lags) (d) unshaded $\rightarrow$ shaded; unshaded lags shaded (Wrong phase) Wait, Option (c) says: 'unshaded portion to the shaded portion while the flux in the former (unshaded) leads that in the latter (shaded)'. This is physically accurate. Let me double check the MPSC 2022 key. Most sources for this specific question list (b) as the answer, but this is a known error in some test banks. However, following the logic of the rotating field: Flux in shaded part lags $\rightarrow$ Field moves Unshaded to Shaded. If the question is from a source where (b) is the key, it's usually because they misidentified the direction. But based on the options provided here, (c) is the scientifically correct one. Let's re-read (b) again: 'shaded portion to the unshaded portion... flux in the former (shaded) lags that of the latter (unshaded)'. Actually, in some specific motor configurations or terminology, the 'shaded' part is the one that creates the lag, and the rotor follows. Final check: 1. Shaded pole flux lags unshaded pole flux. 2. Rotation is from unshaded to shaded. 3. Option (c) says: Unshaded to shaded, and unshaded leads shaded (which is the same as shaded lags unshaded). Therefore, (c) is the correct technical answer.

The current flowing through the armature of a DC motor is governed by the equation: I_a = (V - E_b) / R_a, where V is the supply voltage, E_b is the back e.m.f., and R_a is the armature resistance. 1. Role of Back e.m.f.: The back e.m.f. (E_b) acts as a self-regulating mechanism that opposes the supply voltage. It ensures that the armature current remains at a safe level during normal operation. 2. Scenario without Back e.m.f.: If the back e.m.f. is absent (E_b = 0), the equation becomes I_a = V / R_a. 3. Impact of Low Resistance: Since the armature resistance (R_a) is designed to be very small to minimize copper losses, the resulting current (I_a) becomes extremely high (essentially a short-circuit condition). 4. Result: This excessive current leads to massive I²R heating in the armature windings, which will quickly melt the insulation and cause the motor to burn out. Comparing options: - (a) and (b) are incorrect because speed is a result of rotation; the motor cannot reach a steady speed if it is destroyed by current first. - (c) is incorrect because the motor will actually attempt to start with a massive surge of current, but it will not 'not run' in a stable sense; it will fail catastrophically.

To determine the correct statement, let's analyze the characteristics of Transformer tests: 1. **Open Circuit (OC) Test**: This test is performed to find core losses (iron losses) and magnetizing parameters. The transformer is supplied at rated voltage and frequency on the Low Voltage (LV) side while the High Voltage (HV) side is left open. Because the secondary is open, the current drawn (no-load current $I_0$) is very small (typically 2-5% of rated current). This current is primarily used to magnetize the core and supply losses. Since the core is highly inductive, the no-load current lags the voltage by a large angle (typically 75° to 85°), resulting in a very **low power factor** (cos φ ≈ 0.1 to 0.2). 2. **Short Circuit (SC) Test**: This test is performed to find copper losses and equivalent impedance. The HV side is supplied at a reduced voltage while the LV side is short-circuited. The current drawn is the rated full-load current. Since the transformer is essentially a large inductor at this stage, the current lags the voltage by nearly 90°, meaning it is drawn at a **very low power factor** (close to zero), but not exactly zero. Evaluating the options: - **Option a**: Incorrect. In an OC test, core losses are obtained; in an SC test, copper losses are obtained. The option states the reverse. - **Option b**: Incorrect. In an OC test, the current is drawn at a very low power factor, not high. - **Option c**: Incorrect. While the power factor in an SC test is very low, it is not exactly zero (as there are resistive losses in the windings). - **Option d**: Correct. In an OC test, the no-load current is primarily magnetizing current, which lags the voltage significantly, leading to a low power factor.

For a 4-pole machine at 60 Hz, synchronous speed is Ns = 120f/P = 120×60/4 = 1800 rpm. In an induction generator, rotor current frequency is fr = |s|f, so |s| = 5/60 = 1/12 = 0.0833. Since it is generating, slip is negative, meaning rotor speed is above synchronous speed: N = Ns(1 - s) = 1800(1 + 0.0833) = 1950 rpm. Hence the correct option is d. The other options are incorrect because 1350 rpm and 1650 rpm are below synchronous speed (motor mode), and 2250 rpm would correspond to a much larger slip than 5 Hz rotor frequency.

1. **Nature of Synchronous Motors**: A synchronous motor is not self-starting because the stator's rotating magnetic field rotates at synchronous speed (Ns) immediately upon startup. The rotor, due to its inertia, cannot instantly accelerate to this speed to lock into step with the field. 2. **The Role of Damper Windings**: To overcome this, synchronous motors are equipped with damper windings (Amortisseur windings) embedded in the rotor poles. These are essentially short-circuited bars, similar to those found in a squirrel-cage induction motor. 3. **Starting Process**: When the motor is switched on and the rotor winding is short-circuited (or the damper windings are present), the rotating stator field induces currents in these short-circuited rotor paths. This creates a torque based on the principle of an induction motor. 4. **Operation**: The motor will accelerate as an induction motor. However, it can never reach synchronous speed (Ns) using induction torque alone, as the slip would become zero and torque would vanish. It will continue to run at a speed slightly below synchronous speed (slip speed) as an induction motor unless an external DC excitation is applied to the rotor to pull it into synchronism. 5. **Analysis of Options**: - (a) 'start' is too vague; it specifies how it starts and runs. - (b) 'not start' is incorrect because the short-circuited winding provides induction torque. - (d) 'run as synchronous motor' is incorrect because synchronism requires DC excitation to lock the rotor to the stator field; short-circuiting alone only provides induction action.

A synchronous motor is unique because its power factor can be controlled by adjusting the DC excitation of the field winding while the motor is running at a constant synchronous speed. This is explained by the V-curve characteristics of the motor: 1. Under-excitation: When the field current is low, the motor draws reactive power from the grid to maintain the magnetic field, causing it to operate at a lagging power factor (behaving like an inductor). 2. Normal excitation: When the field current is adjusted such that the motor draws only active power, it operates at a unity power factor (PF = 1). 3. Over-excitation: When the field current is increased beyond the normal requirement, the motor supplies reactive power back to the grid, causing it to operate at a leading power factor (behaving like a capacitor). Because of this ability to operate across the entire spectrum of power factors, synchronous motors are often used for power factor correction in industrial plants (referred to as a 'synchronous condenser' when running without a mechanical load). Therefore, options a, b, and d are only partial truths, while option c is the complete and correct answer.

The no-load current (I₀) of a transformer consists of two components: the magnetizing current (Iₘ) and the core-loss current (I꜀). 1. **Magnetizing Current (Iₘ):** This component is responsible for creating the magnetic flux in the core. Since the core is primarily inductive, Iₘ lags the applied voltage by exactly 90°. 2. **Core-loss Current (I꜀):** This component accounts for hysteresis and eddy current losses. It is essentially a resistive component and is in phase with the applied voltage (0° lag). The total no-load current I₀ is the phasor sum of these two components: I₀ = √(Iₘ² + I꜀²). Because I₀ is the resultant of a component at 90° (Iₘ) and a component at 0° (I꜀), the resulting phasor I₀ will lag the voltage by an angle (φ₀) that is less than 90°. Typically, for a power transformer, this angle is very large (around 75° to 80°), but it is always 'somewhat less than 90°'. - Option (a) is incorrect because it ignores the core-loss current. - Option (c) and (d) are incorrect because the transformer is an inductive load, and current in an inductor always lags the voltage.

Restriking voltage is the transient voltage that appears across the contacts of a circuit breaker at the instant an arc is extinguished. In an AC system, the arc is typically extinguished at a current zero. Immediately after the arc is extinguished, the system voltage rises toward its peak value. Because the restriking voltage is a transient phenomenon that can lead to the re-ignition of the arc if it exceeds the dielectric strength of the gap, it is critical to monitor the maximum voltage reached. Therefore, restriking voltage is always expressed and measured as a peak value. - RMS value (a) is used for steady-state power calculations but does not represent the maximum stress on the insulation during a transient. - Instantaneous value (c) refers to the voltage at any specific moment in time, whereas 'restriking voltage' specifically refers to the peak of the transient oscillation. - Average value (d) is not applicable for analyzing transient dielectric breakdown in switchgear.

In a 3-phase induction motor, the direction of rotation is determined by the sequence of the phases (e.g., R-Y-B). This sequence creates a Rotating Magnetic Field (RMF) that rotates in a specific direction around the stator. According to the principles of electromagnetism, if any two of the three supply phases are interchanged (for example, changing R-Y-B to Y-R-B), the direction of the Rotating Magnetic Field is reversed. Since the rotor is dragged along by this magnetic field, the motor will run in the reverse direction. This is the standard method used in industry to reverse the direction of a 3-phase motor. Why other options are incorrect: - (a) The motor cannot run in the same direction because the phase sequence determines the RMF direction. - (b) The motor will not stop; it will still receive power and generate torque, just in the opposite direction. - (d) Interchanging phases does not inherently cause the motor to draw high current, as the electrical characteristics of the windings remain the same.