← MES Jr. Grade
Past Questions — 2024
Which type of fault is most frequent fault on a transmission line?
A)
Single-line to ground fault
B)
Double-line to ground fault
C)
Three-phase to ground fault
D)
Three-phase short circuit fault
In power system analysis, faults are categorized based on the number of phases involved and whether they are connected to the ground.
1. **Single-Line-to-Ground (SLG) Fault**: This occurs when one of the three phases comes into contact with the ground or a neutral wire. This is the most common type of fault (occurring roughly 70-80% of the time) due to factors such as lightning strikes, tree branches touching a single conductor, or insulator failure.
2. **Double-Line-to-Ground (DLG) Fault**: This occurs when two phases are shorted together and also touch the ground. This is less frequent than SLG faults.
3. **Three-Phase-to-Ground (3LG) Fault**: This occurs when all three phases touch the ground. While this is the most severe fault in terms of current magnitude, it is the rarest occurrence.
4. **Three-Phase Short Circuit (L-L-L) Fault**: This is a symmetrical fault where all three phases are shorted together without necessarily involving the ground. Like the 3LG fault, it is very rare.
Therefore, based on statistical occurrence in transmission networks, the Single-line to ground fault is the most frequent.
A phase-lag compensation will
A)
improve relative stability
B)
increase the speed of response
C)
increase bandwidth
D)
increase overshoot
A phase-lag compensator adds negative phase and is mainly used to improve steady-state accuracy by increasing low-frequency gain. It generally reduces bandwidth and slows the response, but it can improve relative stability by increasing the gain margin/phase margin characteristics in the low-frequency region. Hence, the correct option is (a) improve relative stability.
Why the others are incorrect:
- (b) increase the speed of response: false. Lag compensation usually makes the system slower, not faster.
- (c) increase bandwidth: false. Lag compensation typically decreases bandwidth.
- (d) increase overshoot: not the intended primary effect. Overshoot may change depending on design, but the standard effect of lag compensation is improved steady-state performance and relative stability, not increased overshoot.
A transformer has negligible resistance and a pure reactance of 0.1. Its voltage regulation on full load with a leading pf angle of 30° leading is
A)
-10%
B)
10%
C)
5%
D)
-5%
For a transformer with negligible resistance, the approximate voltage regulation is
VR ≈ R_eq cosφ ± X_eq sinφ
Since resistance is negligible, R_eq = 0, so
VR ≈ ± X_eq sinφ
Given X_eq = 0.1 pu and the load power factor is 30° leading, so φ = 30° and sin30° = 0.5.
For a leading power factor, the reactance term causes a voltage rise, so regulation is negative:
VR = -0.1 × 0.5 = -0.05 = -5%
Hence the correct option is d) -5%.
Why others are incorrect:
- a) -10%: would require either a larger reactance or a larger angle; not consistent with 0.1 pu and 30°.
- b) 10%: wrong magnitude and wrong sign.
- c) 5%: correct magnitude but wrong sign; leading pf gives negative regulation.
A single-phase step up transformer is having the winding turns ratio of 1:20. The primary side supply frequency is f, the frequency at the secondary side of the transformer will be
A)
f/20
B)
f
C)
2f
D)
20f
A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. One of the fundamental properties of a transformer is that it changes the voltage and current levels of an alternating current (AC) signal, but it does NOT change the frequency. The frequency of the induced electromotive force (EMF) in the secondary winding is determined by the rate of change of the magnetic flux in the core, which is directly driven by the frequency of the primary supply voltage. Therefore, if the primary supply frequency is 'f', the secondary frequency will also be 'f', regardless of the turns ratio (whether it is a step-up or step-down transformer).
Analysis of options:
- a. f/20: Incorrect. The turns ratio affects voltage, not frequency.
- b. f: Correct. Frequency remains constant across the transformer.
- c. 2f: Incorrect. Frequency does not double.
- d. 20f: Incorrect. Frequency does not multiply by the turns ratio.
In force-current analogy, mass is analogous to
A)
Inductance
B)
Current
C)
Voltage
D)
Capacitance
In system modeling, analogies are used to relate mechanical systems to electrical systems. There are two primary types of analogies: Force-Voltage (f-v) and Force-Current (f-i).
1. **Force-Current (f-i) Analogy**: In this analogy, the force (F) in a mechanical system is analogous to current (i) in an electrical system.
- The governing equation for a mass (M) in a mechanical system is F = M(dv/dt).
- The governing equation for an inductor (L) in an electrical system is v = L(di/dt).
- However, in the f-i analogy, we look at the relationship where the 'effort' (force) is mapped to 'flow' (current). In this specific mapping, Mass (M) is analogous to Inductance (L) because both represent the energy-storing element that opposes changes in the flow (velocity in mechanical, current in electrical).
2. **Comparison Table for Force-Current (f-i) Analogy**:
- Force (F) $\rightarrow$ Current (i)
- Mass (M) $\rightarrow$ Inductance (L)
- Viscous Friction (B) $\rightarrow$ Conductance (1/R)
- Spring Constant (K) $\rightarrow$ Reciprocal of Capacitance (1/C)
- Velocity (v) $\rightarrow$ Voltage (v)
- Displacement (x) $\rightarrow$ Magnetic Flux ($\Phi$)
Therefore, mass is analogous to inductance in the force-current analogy. Option (a) is correct. Options (b), (c), and (d) represent other electrical parameters that do not map to mass in this specific analogy.
The hysteresis power loss in a transformer does not depend upon
A)
applied voltage
B)
applied frequency
C)
transformer oil
D)
material of the transformer core
Hysteresis loss occurs due to the friction between magnetic domains as they align and realign with the alternating magnetic field in the core. The formula for hysteresis power loss is given by: P_h = η * B_max^n * f * V, where η is the Steinmetz coefficient, B_max is the maximum flux density, f is the frequency, and V is the volume of the core material.
1. Applied Voltage (Option a): The maximum flux density B_max is directly proportional to the applied voltage (V ≈ 4.44 * f * N * Φ_max). Therefore, hysteresis loss depends on the applied voltage.
2. Applied Frequency (Option b): As seen in the formula, P_h is directly proportional to the frequency (f).
3. Material of the transformer core (Option d): The Steinmetz coefficient (η) depends on the grade of the core material (e.g., Cold Rolled Grain Oriented - CRGO steel). Better materials have smaller hysteresis loops and lower losses.
4. Transformer Oil (Option c): Transformer oil is used for insulation and cooling purposes. It does not participate in the magnetic flux cycle or the domain switching process within the core; therefore, it has no effect on hysteresis loss.
The series field winding of a series wound DC machine generally
A)
connected in parallel with the armature winding of the machine
B)
made of fine conductor wire
C)
made of large number of turns of wire
D)
carry same amount of current as armature current
In a series wound DC machine, the field winding is connected in series with the armature winding. By the fundamental laws of series circuits, the same current flows through all components connected in series. Therefore, the full load current flowing through the armature must also flow through the series field winding.
Analysis of other options:
- Option (a) is incorrect because if the field winding were in parallel, it would be a shunt wound machine, not a series wound machine.
- Option (b) is incorrect because the series field must carry the full armature current (which is typically high). Fine wire would have high resistance and would burn out due to excessive I²R losses.
- Option (c) is incorrect because a large number of turns would increase the resistance and the voltage drop across the field winding. Series fields are typically made of a few turns of thick wire to keep resistance low while handling high current.
If the reactance ‘X’ of the line could be varied and the resistance ‘R’ remaining constant, the maximum steady state power that could be transmitted over the line would be greatest when
A)
X=√3R
B)
R=√3X
C)
X=R
D)
R=X/√3
The steady-state power transmitted over a lossless or lossy line is given by the formula: P = (V_s * V_r / Z) * sin(θ - α) / (1 - (V_s * V_r / Z^2) * sin^2(θ - α)), where Z is the impedance (Z = √(R² + X²)) and α is the impedance angle (tan α = X/R).
For a line with constant resistance R and variable reactance X, the maximum power transfer occurs when the receiving end voltage is adjusted or the load angle is optimized. Specifically, the condition for maximum power transfer in a transmission line (considering the effect of resistance) is derived by differentiating the power expression with respect to X.
Mathematically, for a fixed R, the maximum power that can be transmitted is maximized when the impedance angle α is 30°.
Since tan(α) = X/R, for α = 30°:
tan(30°) = X/R
1/√3 = X/R
Therefore, R = √3X.
At this specific ratio, the line is optimized to deliver the highest possible steady-state power given the constraint of a fixed resistance.
An under-excited alternator operates at :
A)
lagging power factor
B)
leading power factor
C)
unity power factor
D)
leading or unity power factor
In a synchronous alternator, the excitation refers to the DC current supplied to the field winding. The state of excitation determines the reactive power exchange between the alternator and the grid:
1. **Under-excitation**: When the field current is low, the alternator cannot produce enough internal magnetic flux to maintain the terminal voltage. To compensate for this deficit, the alternator absorbs reactive power from the grid. This occurs when the load is inductive, meaning the alternator operates at a **lagging power factor**.
2. **Over-excitation**: When the field current is high, the alternator produces excess reactive power and supplies it to the grid. This occurs when the load is capacitive, meaning the alternator operates at a **leading power factor**.
3. **Normal excitation**: The alternator operates at **unity power factor**, where it neither absorbs nor supplies reactive power.
Therefore, an under-excited alternator operates at a lagging power factor to draw the necessary reactive power from the system to maintain its magnetic field.
A 3-phase, 50 Hz, 4-pole, induction motor runs at 1455 rp.m. Frequency of the current flowing in the rotor coil is
A)
1.5Hz
B)
15Hz
C)
50Hz
D)
150Hz
Step-by-step calculation:
1. **Calculate the Synchronous Speed ($N_s$):**
The synchronous speed of an induction motor is given by the formula:
$N_s = \frac{120 \times f}{P}$
Where $f = 50$ Hz and $P = 4$ poles.
$N_s = \frac{120 \times 50}{4} = \frac{6000}{4} = 1500$ rpm.
2. **Calculate the Slip ($s$):**
Slip is the relative difference between the synchronous speed and the actual rotor speed ($N_r = 1455$ rpm).
$s = \frac{N_s - N_r}{N_s}$
$s = \frac{1500 - 1455}{1500} = \frac{45}{1500} = 0.03$ (or 3%).
3. **Calculate the Rotor Frequency ($f_r$):**
The frequency of the current flowing in the rotor is the product of the slip and the supply frequency:
$f_r = s \times f$
$f_r = 0.03 \times 50 = 1.5$ Hz.
Conclusion: The rotor frequency is 1.5 Hz, which matches option (a). Options (b), (c), and (d) are incorrect as they do not account for the slip of the motor.
The magnitude of voltage at the sending end of a long transmission line becomes higher than the voltage magnitude at the receiving end, this phenomena is called
A)
Hunting effect
B)
Corona effect
C)
Ferranti effect
D)
Voltage instability effect
The phenomenon where the receiving end voltage ($V_r$) becomes higher than the sending end voltage ($V_s$) in a long transmission line under no-load or light-load conditions is known as the Ferranti Effect. This occurs because the line's capacitance creates a charging current that leads the voltage; when the line is lightly loaded, this current flows through the line inductance, causing a voltage rise toward the receiving end.
Analysis of other options:
- a. Hunting effect: This refers to the oscillation of the rotor of a synchronous machine around its equilibrium position due to sudden load changes.
- b. Corona effect: This is the ionization of air surrounding a high-voltage conductor, resulting in a violet glow and power loss.
- d. Voltage instability effect: This is a general condition where the system cannot maintain an acceptable voltage profile under a given load demand, often leading to voltage collapse.
A DC shunt generator delivers 95 A at 250 V and the resistance of the shunt field and armature are 50Ω and 0.05Ω respectively. The generated emf will be
A)
270V
B)
265V
C)
255V
D)
275V
To find the generated emf (Eg) of a DC shunt generator, we follow these steps:
1. **Identify Given Data:**
- Load Current (IL) = 95 A
- Terminal Voltage (V) = 250 V
- Shunt Field Resistance (Rsh) = 50 Ω
- Armature Resistance (Ra) = 0.05 Ω
2. **Calculate Shunt Field Current (Ish):**
In a shunt generator, the field winding is in parallel with the load.
Ish = V / Rsh = 250 V / 50 Ω = 5 A
3. **Calculate Armature Current (Ia):**
The armature current is the sum of the load current and the shunt field current.
Ia = IL + Ish = 95 A + 5 A = 100 A
4. **Calculate Generated EMF (Eg):**
The generated emf must cover the terminal voltage and the voltage drop across the armature resistance.
Eg = V + (Ia × Ra)
Eg = 250 V + (100 A × 0.05 Ω)
Eg = 250 V + 5 V = 255 V
Therefore, the generated emf is 255V, which corresponds to option (c). Options (a), (b), and (d) are incorrect as they do not match the calculated value based on the given parameters.
An induction machine behaves as a generator when the slip ‘s’ is
A)
s=0
B)
0<s<1
C)
s=1
D)
s<0
In an induction machine, the slip 's' is defined as s = (Ns - Nr) / Ns, where Ns is the synchronous speed (speed of the rotating magnetic field) and Nr is the actual rotor speed.
1. When 0 < s < 1: The rotor speed (Nr) is less than the synchronous speed (Ns). The machine consumes electrical power from the supply and converts it into mechanical power. This is the 'Motoring Mode'.
2. When s = 0: The rotor rotates at synchronous speed. No torque is produced, and no power is transferred. This is an ideal state for a motor.
3. When s = 1: The rotor is stationary (Nr = 0). This occurs at the moment of starting.
4. When s < 0: This occurs when the rotor is driven by an external prime mover to a speed greater than the synchronous speed (Nr > Ns). In this state, the direction of torque is reversed, and the machine converts mechanical energy into electrical energy, feeding it back into the grid. This is known as the 'Generating Mode' or 'Induction Generator'.
Therefore, for an induction machine to behave as a generator, the slip must be negative (s < 0).
The relation between electrical angle and mechanical angle for an electrical machine, where P is the number of poles is
A)
Electrical angle = (P/2) x Mechanical angle
B)
(P/2) x Electrical angle = Mechanical angle
C)
Electrical angle = Mechanical angle
D)
Electrical angle = 2 x Mechanical angle
In an electrical machine, the mechanical angle (θm) refers to the actual physical rotation of the shaft. The electrical angle (θe) refers to the angle that describes the change in the magnetic field.
1. A pair of poles (one North and one South) constitutes one electrical cycle (360 electrical degrees).
2. If a machine has P poles, it has P/2 pairs of poles.
3. Therefore, for one full mechanical revolution (360 mechanical degrees), the magnetic field completes P/2 electrical cycles.
4. Mathematically, this relationship is expressed as:
Electrical Angle (θe) = (Number of Poles / 2) × Mechanical Angle (θm)
θe = (P/2)θm
Comparing this to the options:
- Option (a) matches the derived formula.
- Option (b) incorrectly suggests that the mechanical angle is larger by a factor of P/2.
- Option (c) is only true for a 2-pole machine (where P/2 = 1).
- Option (d) is only true for a 4-pole machine (where P/2 = 2).
How the torque produced by a DC series motor depends upon the armature current?
A)
directly proportional to the armature current
B)
directly proportional to the square of armature current
C)
inversely proportional to the armature current
D)
inversely proportional to the square of armature current
In a DC motor, the torque (T) produced is proportional to the product of the magnetic flux (Φ) and the armature current (Ia). The general formula is T ∝ ΦIa. In a DC series motor, the field winding is connected in series with the armature winding. Therefore, the field current is equal to the armature current (If = Ia). Since the flux (Φ) is produced by the field current, Φ ∝ Ia (assuming the magnetic circuit is not saturated). Substituting this into the torque equation: T ∝ (Ia) * (Ia) = Ia². Thus, the torque produced by a DC series motor is directly proportional to the square of the armature current. This characteristic is why series motors provide very high starting torque. Option (a) is incorrect because it describes a shunt motor (where flux is constant); options (c) and (d) are incorrect as torque increases with current, it does not decrease.
A 6-poles synchronous motor is supplied by a 50 Hz, 440 V supply. At what speed the motor does run?
A)
750 rp.m.
B)
1000 rp.m.
C)
1500 rp.m.
D)
3000 rp.m.
For a synchronous motor, the running speed equals the synchronous speed: Ns = 120f / P. Here f = 50 Hz and P = 6 poles, so Ns = (120 × 50) / 6 = 1000 rpm. Therefore the correct option is b. However, since the question asks for a 6-pole synchronous motor, the motor runs at 1000 rpm, not 750, 1500, or 3000 rpm. Option a (750 rpm) corresponds to an 8-pole machine at 50 Hz, option c (1500 rpm) corresponds to a 4-pole machine, and option d (3000 rpm) corresponds to a 2-pole machine.
The current zero interruption, in oil and air blast circuit breakers is achieved by
A)
lengthening of the gap between the contact terminals
B)
cooling and blast effect
C)
deionizing the oil with forced air
D)
both (a) & (b)
In circuit breakers, 'current zero interruption' refers to the process of extinguishing the arc that forms when contacts separate at the moment the alternating current passes through zero. To prevent the arc from re-striking (re-igniting), the dielectric strength of the gap must be increased faster than the recovery voltage.
1. Lengthening of the gap (Option a): By increasing the distance between the contact terminals, the electric field intensity is reduced, and the physical path for the arc is increased, which helps in breaking the conduction path.
2. Cooling and blast effect (Option b): In Oil Circuit Breakers (OCB), the arc decomposes the oil into hydrogen gas, which creates a high-pressure 'blast' that pushes the arc away and cools it. In Air Blast Circuit Breakers (ABCB), a high-velocity stream of compressed air is used to physically blow the arc away and cool the ionized gases rapidly.
Since both the physical separation (lengthening the gap) and the active removal/cooling of ionized particles (blast effect) are essential mechanisms to achieve successful interruption in these specific types of breakers, option (d) is the correct choice.
What is incorrect statement about a power transformer?
A)
It is designed to have maximum efficiency at or near full load
B)
It is designed to have considerably greater leakage reactance
C)
It isolates the generating plant from the transmission line electrically
D)
It is designed to have maximum efficiency at or near half load
To determine the incorrect statement, let's analyze the characteristics of a power transformer:
1. **Efficiency and Loading (Options a and d):** Power transformers are used in transmission networks and are typically operated at or near their rated capacity (full load) for long periods. Therefore, they are specifically designed to achieve maximum efficiency at or near full load to minimize energy losses during bulk power transfer. Statement (a) is correct, and statement (d) is incorrect.
2. **Leakage Reactance (Option b):** Power transformers are designed to have a higher leakage reactance compared to distribution transformers. This is necessary to limit the short-circuit currents that can occur during faults in the high-voltage transmission system, protecting the equipment from catastrophic failure. Statement (b) is correct.
3. **Electrical Isolation (Option c):** One of the primary functions of a transformer is to provide galvanic isolation between the primary (generating plant) and secondary (transmission line) circuits, while still allowing the transfer of energy via electromagnetic induction. Statement (c) is correct.
Since the question asks for the INCORRECT statement, option (d) is the right choice.
A transfer function of a control system does not have pole-zero cancellation, which one of the following statement is true?
A)
System is neither controllable nor observable
B)
System is completely controllable and observable
C)
System is observable but uncontrollable
D)
System is controllable but unobservable
In control system theory, specifically when dealing with state-space representations and their corresponding transfer functions, there is a fundamental relationship between pole-zero cancellations and the properties of controllability and observability.
1. **Transfer Function Definition**: The transfer function is defined as $G(s) = C(sI - A)^{-1}B + D$. This representation inherently hides any internal dynamics that are not reflected in the input-output relationship.
2. **Pole-Zero Cancellation**: A pole-zero cancellation occurs when a root of the characteristic equation (an eigenvalue of the system matrix $A$) is also a root of the numerator polynomial of the transfer function.
3. **Controllability and Observability**:
- If a system is **completely controllable and completely observable**, the transfer function will be a minimal realization. This means there are no common factors (cancellations) between the numerator and denominator polynomials.
- Conversely, if there is a pole-zero cancellation in the transfer function, it implies that the system is either **uncontrollable**, **unobservable**, or both. The cancelled mode is the 'hidden' part of the system that cannot be reached by the input or seen by the output.
4. **Conclusion**: Since the question explicitly states that the transfer function does **not** have pole-zero cancellation, it implies that the state-space realization is minimal. Therefore, the system must be both completely controllable and completely observable.
The characteristics equation of a closed-loop control system is given as s^2+4s+16 =0. The resonant frequency in radians/sec. of the system is
A)
2
B)
2/3
C)
4
D)
2/2
Step-by-step solution:
1. The standard characteristic equation for a second-order control system is given by: s^2 + 2ζωn s + ωn^2 = 0, where ζ is the damping ratio and ωn is the undamped natural frequency.
2. Comparing the given equation s^2 + 4s + 16 = 0 with the standard form:
- ωn^2 = 16 => ωn = √16 = 4 rad/sec
- 2ζωn = 4 => 2 * ζ * 4 = 4 => ζ = 4/8 = 0.5
3. The resonant frequency (ωr) for a second-order system is the frequency at which the peak magnitude of the frequency response occurs. It is calculated using the formula:
ωr = ωn * √[1 - 2ζ^2]
4. Substituting the values of ωn and ζ:
ωr = 4 * √[1 - 2(0.5)^2]
ωr = 4 * √[1 - 2(0.25)]
ωr = 4 * √[1 - 0.5]
ωr = 4 * √0.5
ωr = 4 * 0.707 ≈ 2.828 rad/sec.
Observation on Options:
Looking at the provided options (a. 2, b. 2/3, c. 4, d. 2/2), the calculated resonant frequency (2.828) is not explicitly listed. However, in many competitive exams, if the question asks for 'resonant frequency' but the options align with the 'natural frequency' (ωn), it is a common nomenclature error in the question paper.
- Natural Frequency (ωn) = 4 rad/sec (Matches Option c).
- Resonant Frequency (ωr) = 2.828 rad/sec (No match).
Given the options provided, the examiner intended to ask for the natural frequency (ωn), making option (c) the intended correct answer.
What material is used in the breather of a transformer to absorb moisture content of air entering into the transformer’s oil contact area?
A)
Calcium carbonate
B)
Silica gel
C)
Sodium chloride
D)
Carbon crystals
In a transformer, the oil expands and contracts due to temperature changes during operation. As the oil level fluctuates, air is drawn in and pushed out of the conservator tank. To prevent moisture from entering the oil (which would lower the dielectric strength and cause insulation failure), a 'breather' is used. The breather contains Silica Gel, a desiccant material that has a high affinity for water molecules. Silica gel absorbs moisture from the incoming air, ensuring that only dry air enters the transformer. When the silica gel becomes saturated with moisture, it changes color (typically from blue to pink), indicating that it needs to be replaced or reheated to remove the moisture.
Why other options are incorrect:
- Calcium carbonate (a) is primarily used in construction or as a filler; it is not a desiccant.
- Sodium chloride (c) is common table salt; it does not possess the porous structure required to act as an effective industrial desiccant for transformer breathers.
- Carbon crystals (d) are conductive materials and are not used for moisture absorption.
The ON-OFF controller is a
A)
linear device
B)
nonlinear device
C)
discontinuous device
D)
none of these
An ON-OFF controller (also known as a bang-bang controller or hysteresis controller) is a system where the output switches between two fixed states (e.g., fully ON or fully OFF) based on whether the input error is positive or negative.
1. Why it is Nonlinear: A linear system must satisfy the principles of superposition and homogeneity. In a linear system, if the input increases by a certain factor, the output increases proportionally. In an ON-OFF controller, the output does not change proportionally with the input; it remains constant until a threshold is hit, then jumps abruptly. This 'step' behavior is a classic characteristic of a nonlinear system.
2. Why it is not a Linear Device (Option a): As explained above, the relationship between the input (error) and output (control action) is not a straight line passing through the origin; it is a step function.
3. Why 'nonlinear' is a better choice than 'discontinuous' (Option c): While the output signal of an ON-OFF controller is indeed discontinuous, in the context of control systems and system classification, 'nonlinear' is the standard categorical term used to describe devices that do not follow linear differential equations. Discontinuity is a specific type of nonlinearity.
Therefore, the ON-OFF controller is fundamentally a nonlinear device.
The transient stability of the power system can be effectively improve by
A)
excitation control
B)
phase shifting transformer
C)
single pole switching of circuit breakers
D)
increasing the turbine valve opening
Transient stability refers to the ability of a power system to maintain synchronism when subjected to a severe disturbance (like a fault). To improve transient stability, the goal is to increase the synchronizing power or reduce the acceleration of the generator rotor during the fault.
1. **Excitation Control (Correct):** High-speed excitation control (or Fast Valving/AVR) increases the internal voltage (E) of the generator during and immediately after a fault. According to the power-angle equation P = (EV/X)sinδ, increasing the excitation (E) increases the maximum power transfer capability and the decelerating area in the Equal Area Criterion, thereby preventing the rotor from swinging out of step.
2. **Phase Shifting Transformer (Incorrect):** These are primarily used for steady-state power flow control between parallel lines, not for improving transient stability during faults.
3. **Single Pole Switching (Incorrect):** While single-pole switching can reduce the severity of the disturbance compared to three-pole switching (by maintaining some synchronizing power), it is a method of fault clearing/handling rather than a primary control mechanism to 'improve' the inherent stability limit of the machine in the way excitation control does.
4. **Increasing Turbine Valve Opening (Incorrect):** Increasing the valve opening increases the mechanical power input (Pm). According to the swing equation, increasing Pm during a fault increases the accelerating power (Pm - Pe), which actually worsens the transient stability by causing the rotor to accelerate faster.
For extra high voltage lines which circuit breaker is preferred
A)
Bulk oil circuit breaker
B)
Vacuum circuit breaker
C)
SF6 gas circuit breaker
D)
Minimum oil circuit breaker
In power systems, the choice of circuit breaker depends primarily on the voltage level and the arc-quenching capability of the medium used.
1. **SF6 (Sulfur Hexafluoride) Circuit Breakers**: SF6 gas has exceptional dielectric strength and electronegative properties, meaning it can absorb free electrons quickly to extinguish an arc. This makes it ideal for Extra High Voltage (EHV) and Ultra High Voltage (UHV) applications (typically above 220 kV). It allows for a more compact design compared to oil breakers and offers superior reliability and safety at high voltages.
2. **Bulk Oil Circuit Breakers (a)**: These use a large volume of oil for insulation and arc quenching. They are outdated for EHV due to the risk of fire, high maintenance, and massive size.
3. **Vacuum Circuit Breakers (b)**: These are highly efficient for medium voltage (MV) applications (typically up to 33 kV or 66 kV). However, they are not suitable for EHV because it is technically difficult and expensive to maintain a high-vacuum gap across the large distances required for EHV insulation.
4. **Minimum Oil Circuit Breakers (d)**: These were an improvement over bulk oil breakers to reduce fire hazards, but they are still limited to lower voltage levels compared to SF6 and are largely replaced by SF6 or Vacuum technology in modern grids.
Two alternators A and B are sharing an inductive load equally. If the excitation of alternator A is increased
A)
Alternator B will deliver more current and alternator A will deliver less current
B)
Alternator B will deliver less current and alternator A will deliver more current
C)
Both will continue to share load equally
D)
Both will deliver more current
In parallel operation of alternators, the sharing of the real (kW) and reactive (kVAR) load depends on the excitation and the nature of the load.
1. **Excitation and Reactive Power**: The excitation of an alternator controls the internal generated voltage (E). Increasing the excitation of an alternator increases its internal voltage.
2. **Inductive Load Behavior**: For an inductive load, the current lags the voltage. When the excitation of Alternator A is increased, its internal voltage increases relative to Alternator B. This causes Alternator A to take over a larger share of the reactive power (kVAR) from the load.
3. **Load Redistribution**: Since the total load is constant, if Alternator A increases its output current to supply more of the reactive power, Alternator B must decrease its output current to maintain the balance.
4. **Conclusion**: Therefore, increasing the excitation of Alternator A results in Alternator A delivering more current and Alternator B delivering less current.
Comparing with options:
- Option (a) is the opposite of the physical reality.
- Option (c) is incorrect because excitation directly affects load sharing in inductive loads.
- Option (d) is incorrect because the total load is constant; one cannot increase without the other decreasing.