← MES Jr. Grade
Past Questions
The Nyquist plot of loop transfer function G(s)H(s) of a closed loop control system passes through the point (-1, j 0) in the G(s)H(s) plane. The phase margin of the system is:
A)
0°
B)
45°
C)
90°
D)
180°
1. **Understanding the Nyquist Plot**: The Nyquist plot is a polar plot of the open-loop transfer function G(s)H(s) as the frequency ω varies from 0 to infinity.
2. **The Critical Point**: In stability analysis, the point (-1, j0) is known as the critical point. This point represents a magnitude of 1 and a phase angle of -180°.
3. **Definition of Phase Margin (PM)**: The Phase Margin is the amount of additional phase lag that can be added to the system before it becomes unstable. Mathematically, it is defined as:
PM = 180° + ∠G(jω_gc)H(jω_gc), where ω_gc is the gain crossover frequency (the frequency at which the magnitude |G(jω)H(jω)| = 1).
4. **Analyzing the Given Condition**: The problem states that the Nyquist plot passes through the point (-1, j0). This means that at some frequency ω_gc, the magnitude |G(jω_gc)H(jω_gc)| = 1 and the phase ∠G(jω_gc)H(jω_gc) = -180°.
5. **Calculation**:
PM = 180° + (-180°) = 0°.
6. **Conclusion**: When the Nyquist plot passes exactly through the critical point (-1, j0), the system is marginally stable, and the phase margin is exactly 0°. Options b, c, and d are incorrect because they imply a degree of stability (positive margin) or a different phase relationship that does not correspond to the point (-1, j0).
A 12-pole, 50 Hz, three phase induction motor runs at 485 rpm. The frequency of the rotor current will be:
A)
2.5 Hz
B)
2Hz
C)
1.5 Hz
D)
3Hz
For a 12-pole, 50 Hz induction motor, synchronous speed is Ns = 120f/P = 120×50/12 = 500 rpm. The rotor runs at 485 rpm, so slip s = (Ns − Nr)/Ns = (500 − 485)/500 = 15/500 = 0.03. Rotor current frequency is fr = s f = 0.03 × 50 = 1.5 Hz. Therefore the correct option is c, not b. The other options are incorrect because 2 Hz, 2.5 Hz, and 3 Hz do not match the slip-frequency relation fr = sf for the given speed and supply frequency.
A 3 phase, 400 V, 50 Hz salient pole synchronous motor is running on no load. If there is break in the excitation winding of the motor:
A)
the motor will stop
B)
the winding will get overheated
C)
the motor will run as reluctance motor at the rated rpm
D)
the motor will run as reluctance motor at lower rpm.
Step-by-step reasoning:
1. **Understanding Salient Pole Construction**: In a salient pole synchronous motor, the air gap is non-uniform. The reluctance is minimum along the axis of the poles (d-axis) and maximum in the direction perpendicular to the poles (q-axis).
2. **Effect of Excitation Break**: When the excitation winding breaks, the DC field current becomes zero. This means the motor no longer has a magnetic field produced by the rotor windings.
3. **Reluctance Torque**: Despite the lack of excitation, the rotating magnetic field (RMF) produced by the 3-phase stator winding interacts with the salient poles of the rotor. The rotor will attempt to align itself with the RMF to minimize the magnetic reluctance of the path. This creates 'reluctance torque'.
4. **Synchronous Speed**: Because the torque is produced by the interaction with the stator's rotating magnetic field, the rotor will be 'locked' to the RMF. Therefore, it will continue to rotate at the synchronous speed (rated rpm), which is determined by the formula Ns = 120f/P.
5. **No-Load Condition**: Since the motor is running on no load, the small amount of reluctance torque produced is sufficient to overcome the friction and windage losses, allowing the motor to maintain its rated synchronous speed.
Why other options are incorrect:
- (a) The motor will not stop because the reluctance torque is sufficient to keep a no-load rotor spinning.
- (b) While there might be some current changes, a break in the excitation winding itself doesn't inherently lead to overheating of the stator windings under no-load conditions.
- (d) Reluctance motors are synchronous motors; they do not run at a 'lower rpm' (slip) like induction motors; they run exactly at the synchronous speed.
The initial slope of bode plot for a transfer function having single pole at origin is:
A)
20 db/decade
B)
-40 db/decade
C)
40 db/decade
D)
-20 db/decade
1. A transfer function with a single pole at the origin is represented as G(s) = K/s.
2. To draw a Bode plot, we substitute s = jω, giving G(jω) = K/(jω).
3. The magnitude of this transfer function is |G(jω)| = K/ω.
4. In a Bode magnitude plot, the magnitude is expressed in decibels (dB) as: Magnitude (dB) = 20 log10(|G(jω)|) = 20 log10(K) - 20 log10(ω).
5. The slope of the magnitude plot is the derivative of the magnitude (dB) with respect to log10(ω).
6. d/d(log10 ω) [20 log10(K) - 20 log10(ω)] = -20.
7. Therefore, a single pole at the origin contributes a constant slope of -20 dB per decade across the entire frequency range.
8. Option (a) and (c) are incorrect because poles cause a decrease in magnitude (negative slope). Option (b) is incorrect because -40 dB/decade corresponds to a double pole at the origin (s^2 in the denominator).
A shunt generator is running at 1000 rpm. If the flux is reduced by half, then what is the new speed such that generated voltage remains the same?
A)
0
B)
500
C)
1000
D)
2000
For a DC generator, the generated emf is proportional to flux and speed: E ∝ ΦN. If the flux is reduced to half and the generated voltage must remain the same, the speed must double to compensate. So, N2 = N1 × (Φ1/Φ2) = 1000 × (1 / 0.5) = 2000 rpm. Hence option d is correct. Option c (1000 rpm) is incorrect because with half flux the emf would also halve at the same speed. Option b (500 rpm) is incorrect because reducing speed would further reduce the emf. Option a (0) is obviously impossible for maintaining the same generated voltage.
Commutator in DC generator is used for:
A)
collecting of current
B)
reduce losses
C)
increase efficiency
D)
convert AC armature current in to DC
In a DC generator, the armature conductors rotate within a magnetic field. According to Faraday's Law of Electromagnetic Induction, the electromotive force (EMF) induced in these conductors is alternating (AC) because the conductors move from a region of North pole flux to South pole flux and vice versa. The commutator consists of a set of segmented copper bars that act as a mechanical rectifier. As the armature rotates, the commutator reverses the connections of the armature coils to the external circuit at the exact moment the current direction reverses in the coil. This process converts the internal AC generated in the armature into a unidirectional DC output at the brushes.
Analysis of other options:
- (a) While the commutator does facilitate the collection of current via brushes, its primary functional purpose in a DC machine is rectification, not merely collection.
- (b) and (c) The commutator actually introduces some losses (voltage drops and friction) and does not inherently increase the efficiency of the machine; it is a necessary component to achieve the desired output type (DC).
Steady-state stability of a power system is the ability of the power system to:
A)
maintain voltage at the rated voltage level.
B)
maintain frequency exactly at 50 Hz.
C)
maintain a spinning reserve margin at all times.
D)
maintain synchronism between machines and on external tie lines.
Steady-state stability is a fundamental concept in power system analysis. It is defined as the ability of the power system to maintain synchronism when subjected to small and gradual changes in load or generation. In a power system, generators must operate at the same electrical frequency and maintain a specific phase relationship (synchronism) to share the load and ensure system stability. If a small increase in load causes the rotor angle of a machine to increase beyond a critical limit where the synchronizing power decreases, the machine will lose synchronism and 'slip poles'.
Why other options are incorrect:
- Option (a): Maintaining voltage is the objective of Voltage Control/Regulation (using AVRs or capacitors), not stability.
- Option (b): Maintaining frequency is the objective of Load Frequency Control (LFC) or Governor control. While related to stability, 'steady-state stability' specifically refers to the synchronism of the machines.
- Option (c): Spinning reserve is a planning and operational requirement for reliability and contingency management, not a definition of stability.
In load flow studies of a power system, a voltage control bus is specified by:
A)
real power and reactive power.
B)
reactive power and voltage magnitude.
C)
voltage and voltage phase angle.
D)
real power and voltage magnitude.
In power system load flow analysis, buses are categorized based on the known and unknown variables. There are four primary variables at each bus: Real Power (P), Reactive Power (Q), Voltage Magnitude (|V|), and Voltage Phase Angle (δ).
1. Load Bus (PQ Bus): P and Q are specified; |V| and δ are unknown.
2. Generator Bus / Voltage Control Bus (PV Bus): The real power (P) and the voltage magnitude (|V|) are specified. The reactive power (Q) and the phase angle (δ) are the unknowns to be solved. This is because generators can control the terminal voltage by adjusting the excitation.
3. Slack Bus / Swing Bus: |V| and δ are specified; P and Q are unknown. This bus serves as the reference for the system and accounts for the total system losses.
Therefore, a voltage control bus (PV bus) is specified by real power (P) and voltage magnitude (|V|).
In terms of bode plot, the system is stable if:
A)
P.M=G.M
B)
P.M and GM both are positive
C)
P.M and GM both are negative
D)
P.M negative but G.M positive
In control systems, stability is analyzed using the Bode plot through two primary metrics: Gain Margin (GM) and Phase Margin (PM).
1. **Gain Margin (GM):** This is the amount of gain that can be increased before the system becomes unstable. It is measured at the phase crossover frequency (where the phase angle is -180°). For a stable system, the magnitude at this frequency must be less than 0 dB, meaning the GM is positive.
2. **Phase Margin (PM):** This is the amount of additional phase lag that can be added to the system before it becomes unstable. It is measured at the gain crossover frequency (where the magnitude is 0 dB). For a stable system, the phase angle must be greater than -180°, meaning the PM is positive.
**Stability Criteria:**
- **Stable System:** Both GM > 0 dB and PM > 0° (Both are positive).
- **Marginally Stable System:** GM = 0 dB and PM = 0°.
- **Unstable System:** Either GM < 0 dB or PM < 0° (One or both are negative).
Therefore, for a system to be stable, both the Phase Margin and Gain Margin must be positive.
Ferranti effect happens in transmission line when the line is:
A)
short and loaded
B)
long and loaded
C)
long and unloaded
D)
none of these
The Ferranti effect is a phenomenon in electrical power systems where the voltage at the receiving end of a transmission line is higher than the voltage at the sending end. This occurs due to the capacitive charging current of the line.
1. **Mechanism**: Transmission lines have a shunt capacitance between the conductors and the ground. When the line is long, this capacitance becomes significant.
2. **Condition (Unloaded/Lightly Loaded)**: When the line is unloaded (open-circuited) or very lightly loaded, the current flowing through the line is primarily the capacitive charging current. This current leads the voltage by 90 degrees.
3. **Voltage Rise**: As this charging current flows through the line's inductance, it creates a voltage drop that is additive to the sending end voltage at the receiving end, resulting in $V_{receiving} > V_{sending}$.
4. **Why other options are incorrect**:
- **Short lines (a)**: In short transmission lines (typically < 80km), shunt capacitance is negligible, so the Ferranti effect does not occur.
- **Loaded lines (b)**: When a line is heavily loaded, the inductive voltage drop (due to load current) dominates over the capacitive charging effect, causing the receiving end voltage to be lower than the sending end voltage.
Which point on root locus specifies the meeting or collision of two poles?
A)
Centroid
B)
Break away point
C)
Stability point
D)
Anti-break point
In the context of a Root Locus plot in Control Systems, the 'Break away point' is the specific point on the real axis where two or more branches of the root locus meet and then leave the real axis to move into the complex plane. This occurs when the system has multiple poles on the real axis; as the gain (K) increases, these poles move toward each other and 'collide' before splitting off. Mathematically, these points are found by solving dK/ds = 0.
Analysis of other options:
- a. Centroid: This is the point where the asymptotes of the root locus meet, used to determine the direction of the branches as they go to infinity.
- c. Stability point: This is not a standard term for a specific point on the root locus; stability is generally determined by whether the locus stays in the left-half of the s-plane.
- d. Anti-break point: This is likely a distractor term. The opposite of a break-away point is a 'break-in point', where branches from the complex plane meet and enter the real axis.
The closed loop transfer function for a second order system is: T(s) = 4/(s^2 + 4s + 4). The settling time for a 2 percent band will be:
A)
2 seconds
B)
1.5 seconds
C)
2.5 seconds
D)
1 second
Step-by-step solution:
1. **Identify the Standard Form**: The general closed-loop transfer function for a second-order system is given by:
T(s) = ωn² / (s² + 2ζωn s + ωn²)
2. **Compare with the Given Equation**: The given transfer function is:
T(s) = 4 / (s² + 4s + 4)
3. **Determine System Parameters**:
- Comparing the constant terms: ωn² = 4 => ωn = 2 rad/s
- Comparing the coefficients of s: 2ζωn = 4
- Substitute ωn = 2: 2 * ζ * 2 = 4 => 4ζ = 4 => ζ = 1
4. **Analyze the Damping State**: Since ζ = 1, the system is **critically damped**.
5. **Calculate Settling Time (ts)**:
- For a 2% tolerance band, the standard formula for an underdamped system is ts = 4 / (ζωn).
- For a critically damped system (ζ = 1), the settling time for a 2% band is also approximated by the same formula or derived from the time constant (τ = 1/ζωn).
- ts = 4 / (ζωn) = 4 / (1 * 2) = 4 / 2 = 2 seconds.
Conclusion: The settling time for a 2% band is 2 seconds. Therefore, option (a) is correct.
If the fault current is 2000 A, the relay setting is 50% and CT ratio is 400: 5, then plug setting multiplier will be:
A)
10
B)
15
C)
25
D)
50
Plug setting multiplier (PSM) = Fault current / Relay pickup current.
CT ratio = 400:5, so 2000 A primary fault current gives secondary current = 2000 × (5/400) = 25 A.
Relay setting = 50%, so pickup current on the secondary side = 50% of 5 A = 2.5 A.
Therefore, PSM = 25 / 2.5 = 10.
So the correct numerical value is 10, which corresponds to option a. The other options are incorrect because 15, 25, and 50 do not match the standard PSM calculation from the given fault current, CT ratio, and relay setting.
An infinite bus-bar should maintain:
A)
infinite frequency and infinite voltage
B)
constant frequency and constant voltage
C)
constant frequency but variable voltage
D)
variable frequency and variable voltage
In power systems engineering, an 'infinite bus' is a theoretical concept used to simplify the analysis of a machine or a small power system connected to a much larger grid. An infinite bus is defined as a bus whose voltage and frequency remain constant regardless of the amount of real or reactive power delivered to or absorbed from it.
1. Constant Voltage: Because the bus is connected to a massive network of generators, any change in the load or generation of a single connected machine is negligible compared to the total capacity of the system, meaning the voltage magnitude does not change.
2. Constant Frequency: The frequency is determined by the collective rotational speed of all synchronized generators in the large grid. A single machine cannot significantly alter the overall system frequency.
Analysis of options:
- Option (a) is incorrect because 'infinite' voltage and frequency are physically impossible and mathematically meaningless in this context.
- Option (c) and (d) are incorrect because the defining characteristic of an infinite bus is its stability (constancy) in both parameters.
In closed loop control system, with positive value of feedback gain the overall gain of the system will:
A)
decrease
B)
increase
C)
be unaffected
D)
none of these
To determine the overall gain of a closed-loop control system, we use the transfer function formula. Let G(s) be the forward path gain and H(s) be the feedback path gain.
1. For a standard closed-loop system, the overall transfer function (T) is given by: T = G / (1 ± GH).
2. In a negative feedback system (the most common type), the formula is T = G / (1 + GH). Here, the denominator increases, which decreases the overall gain.
3. In a positive feedback system (as specified in the question: 'positive value of feedback gain'), the formula is T = G / (1 - GH).
4. In the case of positive feedback, the term (1 - GH) in the denominator is smaller than 1 (assuming GH < 1 for stability). Dividing the forward gain G by a value smaller than 1 results in an overall gain that is larger than the open-loop gain G.
5. Therefore, positive feedback increases the overall gain of the system, although it often leads to instability.
Comparing the options:
- (a) decrease: This occurs in negative feedback.
- (b) increase: This occurs in positive feedback.
- (c) be unaffected: This would only happen if H = 0 (open loop).
- (d) none of these: Incorrect as (b) is the correct technical result.
A synchronous compensator absorbs inductive reactive power. It is:
A)
overexcited
B)
normally excited
C)
under excited
D)
none of these
A synchronous compensator (or synchronous condenser) is a synchronous motor operating at no-load. Its primary purpose is to regulate the power factor of a system by adjusting its field excitation.
1. **Overexcited Condition**: When the motor is overexcited, it generates reactive power and supplies it to the grid. In this state, it behaves like a capacitor (leading power factor) and is used to counteract inductive loads.
2. **Underexcited Condition**: When the motor is underexcited, it requires reactive power from the grid to maintain its magnetic field. Therefore, it absorbs inductive reactive power from the system. In this state, it behaves like an inductor (lagging power factor).
3. **Normally Excited**: In this state, the motor operates at unity power factor, neither absorbing nor supplying reactive power.
Since the question specifies that the compensator 'absorbs inductive reactive power', it must be operating in the underexcited state.
The induced emf in a dc machine while running at 500 rpm is 180 V. The induced emf when the machine is running at 600 rpm (Assume constant flux) will be:
A)
216V
B)
210V
C)
1666.66 V
D)
None of these
The induced electromotive force (emf) in a DC machine is given by the formula: E = (PΦZN) / (60a), where P is the number of poles, Φ is the flux per pole, Z is the total number of armature conductors, N is the speed in rpm, and 'a' is the number of parallel paths.
From this formula, we can see that if the flux (Φ) and the machine construction (P, Z, a) remain constant, the induced emf is directly proportional to the speed (N): E ∝ N.
Given data:
Initial speed (N1) = 500 rpm
Initial emf (E1) = 180 V
Final speed (N2) = 600 rpm
Final emf (E2) = ?
Using the proportionality ratio:
E2 / E1 = N2 / N1
E2 = E1 * (N2 / N1)
E2 = 180 * (600 / 500)
E2 = 180 * 1.2
E2 = 216 V
Therefore, the induced emf at 600 rpm is 216V. Option (a) is correct. Option (b) is numerically incorrect, and option (c) is an unrealistic value for this scenario.
A 6-pole, wave connected armature has 250 conductors and runs at 1200 rpm. The emf generated on open circuit is 600 V. The useful flux per pole will be:
A)
0.03 Wb
B)
0.04 Wb
C)
0.02 Wb
D)
0.06 Wb
To find the useful flux per pole, we use the standard EMF equation for a DC generator: E = (ΦZNP) / (60A), where:
- E = Generated EMF = 600 V
- Φ = Flux per pole (to be calculated)
- Z = Total number of armature conductors = 250
- N = Speed of rotation = 1200 rpm
- P = Number of poles = 6
- A = Number of parallel paths. For a wave-wound armature, A is always 2, regardless of the number of poles.
Rearranging the formula to solve for Φ:
Φ = (E * 60 * A) / (Z * N * P)
Substituting the given values:
Φ = (600 * 60 * 2) / (250 * 1200 * 6)
Φ = 72,000 / 1,800,000
Φ = 72 / 1800
Φ = 0.04 Wb
Comparing this result with the given options:
- Option (a) 0.03 Wb is incorrect.
- Option (b) 0.04 Wb is correct.
- Option (c) 0.02 Wb is incorrect.
- Option (d) 0.06 Wb is incorrect.
In a three-phase synchronous motor, the magnitude of field flux:
A)
varies with power factor
B)
remains constant at all loads
C)
varies with load
D)
varies with speed
In a three-phase synchronous motor, the field flux is produced by the DC excitation current supplied to the rotor windings. The magnitude of this field flux depends solely on the DC excitation current (I_f) and the number of turns in the field winding. Under normal operating conditions, the DC excitation is kept constant regardless of the mechanical load applied to the shaft. While changing the field excitation will change the power factor (under-excitation or over-excitation), the load itself does not automatically change the field flux; rather, the field flux is a controlled parameter used to regulate the motor's reactive power and power factor. Therefore, for a given setting of the DC source, the field flux remains constant at all loads.
If the armature current of de series motor has become twice then the torque will become:
A)
twice of the former
B)
four times of the former
C)
one fourth of the former
D)
remains same
In a DC series motor, the field winding is connected in series with the armature winding. Therefore, the field current (If) is equal to the armature current (Ia).
1. The torque (T) produced by a DC motor is proportional to the product of the magnetic flux (Φ) and the armature current (Ia): T ∝ Φ × Ia.
2. In a series motor, the flux (Φ) is produced by the field current, which is the same as the armature current (Φ ∝ Ia), provided the magnetic circuit is not saturated.
3. Substituting this into the torque equation: T ∝ Ia × Ia = Ia².
4. If the armature current becomes twice (Ia_new = 2 × Ia_old), the new torque (T_new) will be: T_new ∝ (2 × Ia_old)² = 4 × (Ia_old²).
5. Therefore, the torque becomes four times the former value.
Option (a) is incorrect because it assumes a linear relationship (like in a shunt motor where flux is constant). Option (c) is mathematically inverse. Option (d) is incorrect as torque is dependent on current.
If the characteristics equation of a system is s^2+2=0, then the system is:
A)
Overdamped
B)
Underdamped
C)
Critically damped
D)
Undamped
Step-by-step reasoning:
1. The standard characteristic equation for a second-order system is given by: s^2 + 2ζωn s + ωn^2 = 0, where ζ (zeta) is the damping ratio and ωn is the natural frequency.
2. The given characteristic equation is s^2 + 2 = 0.
3. Comparing the two equations:
- The coefficient of s^2 is 1 (matches).
- The coefficient of s is 0. Therefore, 2ζωn = 0.
- The constant term is 2. Therefore, ωn^2 = 2, which means ωn = √2 rad/s.
4. Since 2ζωn = 0 and ωn ≠ 0, it follows that ζ = 0.
5. The damping ratio (ζ) determines the nature of the system:
- If ζ > 1: Overdamped
- If ζ = 1: Critically damped
- If 0 < ζ < 1: Underdamped
- If ζ = 0: Undamped
6. Since ζ = 0, the system is undamped. This means the system will oscillate indefinitely with a constant amplitude at its natural frequency (ωn = √2) without any decay.
Transmission lines are transposed to reduce:
A)
ferranti effect
B)
skin effect
C)
proximity effect
D)
interference with neighbouring communication lines
Transposition of transmission lines involves swapping the positions of the three phases at regular intervals along the length of the line.
1. Why it is done: In an asymmetrical arrangement of conductors, the magnetic flux linkages for each phase are different, leading to unequal inductances and unbalanced voltage drops. This imbalance creates an electromagnetic field that induces unwanted currents (interference) in nearby communication lines (like telephone lines).
2. How it works: By transposing the lines, the average inductance of each phase becomes equal over the entire length of the line. This balances the phase voltages and ensures that the net magnetic field produced by the three-phase system at a distance is minimized, thereby reducing interference with communication lines.
Why other options are incorrect:
- a. Ferranti Effect: This is caused by the capacitance of the line under no-load or light-load conditions, causing the receiving end voltage to be higher than the sending end. Transposition does not prevent this.
- b. Skin Effect: This is a phenomenon where current tends to flow on the outer surface of a conductor due to internal flux. It depends on the conductor material, frequency, and diameter, not the position of the phases.
- c. Proximity Effect: This occurs when the magnetic field of one conductor affects the current distribution in a nearby conductor. While transposition changes which phases are adjacent, it does not 'reduce' the proximity effect itself; it merely distributes it evenly across phases.
The efficiency of the transformer will be maximum when:
A)
iron losses is equal to the twice of the copper losses
B)
copper losses is equal to the twice of the iron losses
C)
iron losses is equal to the copper losses
D)
all of these
The efficiency of a transformer is defined as the ratio of output power to input power. The total losses in a transformer consist of two main components:
1. Iron Losses (Core Losses): These are constant losses (hysteresis and eddy current losses) that do not change with the load.
2. Copper Losses (Ohmic Losses): These are variable losses that vary with the square of the load current (I²R).
Mathematically, Efficiency (η) = Output / (Output + Constant Losses + Variable Losses). To find the maximum efficiency, we differentiate the efficiency expression with respect to the load current and set it to zero. This calculus derivation proves that the maximum efficiency occurs at the specific load where the variable losses (copper losses) are exactly equal to the constant losses (iron losses).
Therefore:
- Option (a) is incorrect because iron losses are not twice the copper losses at max efficiency.
- Option (b) is incorrect because copper losses are not twice the iron losses at max efficiency.
- Option (c) is correct as it satisfies the condition: Iron Loss = Copper Loss.
Bundled conductors are used to:
A)
reduce inductance of the line.
B)
reduce both inductance and capacitance.
C)
reduce corona loss.
D)
reduce corona loss and the line inductance.
Bundled conductors are used in high-voltage transmission lines for two primary reasons:
1. **Reduction of Corona Loss**: Corona occurs when the electric field intensity at the conductor surface exceeds the dielectric strength of air. By bundling conductors (using multiple smaller conductors per phase), the 'effective radius' (or Geometric Mean Radius, GMR) of the conductor is increased. A larger effective radius reduces the maximum electric field intensity at the surface, thereby increasing the critical disruptive voltage and significantly reducing corona loss and radio interference.
2. **Reduction of Line Inductance**: The inductance of a transmission line is inversely proportional to the logarithm of the GMR. Since bundling increases the GMR, the self-inductance of the line decreases. This reduction in inductance leads to a decrease in the inductive reactance, which in turn improves the voltage regulation and increases the Surge Impedance Loading (SIL) capacity of the line.
Regarding other options:
- Option (a) is partially correct but incomplete.
- Option (b) is incorrect because bundling actually *increases* the capacitance of the line (due to the increased effective radius).
- Option (c) is partially correct but incomplete.
The open loop transfer function of a unity feedback control system is given as G(s) = (as+1)/s^2. The value of ‘a’ to give a phase margin of 45° is equal to:
A)
0.141
B)
0.441
C)
0.841
D)
1.141
For unity feedback, the phase margin (PM) is evaluated at the gain crossover frequency \(\omega_{gc}\), where \(|G(j\omega)|=1\).
Given
\[
G(s)=\frac{as+1}{s^2}
\Rightarrow G(j\omega)=\frac{1+j a\omega}{(j\omega)^2}
\]
Magnitude:
\[
|G(j\omega)|=\frac{\sqrt{1+a^2\omega^2}}{\omega^2}
\]
At gain crossover,
\[
\frac{\sqrt{1+a^2\omega^2}}{\omega^2}=1
\Rightarrow 1+a^2\omega^2=\omega^4
\]
Phase of \(G(j\omega)\):
- numerator phase = \(\tan^{-1}(a\omega)\)
- denominator phase = \(180^\circ\) (because \((j\omega)^2=-\omega^2\))
So open-loop phase is
\[
\angle G(j\omega)=\tan^{-1}(a\omega)-180^\circ
\]
Phase margin is
\[
PM=180^\circ+\angle G(j\omega_{gc})=\tan^{-1}(a\omega_{gc})
\]
Given \(PM=45^\circ\),
\[
\tan^{-1}(a\omega_{gc})=45^\circ \Rightarrow a\omega_{gc}=1
\Rightarrow a=\frac{1}{\omega_{gc}}
\]
Using the magnitude condition with \(a\omega_{gc}=1\):
\[
1+a^2\omega_{gc}^2=\omega_{gc}^4
\Rightarrow 1+1=\omega_{gc}^4
\Rightarrow \omega_{gc}^4=2
\Rightarrow \omega_{gc}=2^{1/4}\approx 1.189
\]
Hence
\[
a=\frac{1}{1.189}\approx 0.841
\]
So the correct option is **c (0.841)**.
Why others are incorrect:
- **a (0.141)**: too small; it would not satisfy the gain-crossover and phase-margin conditions.
- **b (0.441)**: also does not satisfy \(a\omega_{gc}=1\) with the required crossover frequency.
- **d (1.141)**: too large; would give a different phase margin than 45°.
Therefore, the correct answer is **c**.