← MES Jr. Grade
Past Questions — 2024
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.
A shunt generator running at 1000 r.p.m. has generated e.m.f. as 200 V. If the speed increases to 1200 r.p.m., the generated e.m.f. will be nearly:
A)
150V
B)
175V
C)
240V
D)
290V
The generated electromotive force (e.m.f.) of a DC generator is given by the formula: E = (ΦZNP) / (60A), where Φ is the flux per pole, Z is the number of conductors, P is the number of poles, and N is the speed in r.p.m. From this formula, it is evident that the generated e.m.f. is directly proportional to the product of flux and speed (E ∝ ΦN). In a shunt generator, if we assume the flux (Φ) remains constant (which is the standard assumption for such problems unless a change in field current or saturation is specified), the relationship simplifies to E ∝ N.
Given:
Initial speed (N1) = 1000 r.p.m.
Initial e.m.f. (E1) = 200 V
Final speed (N2) = 1200 r.p.m.
Final e.m.f. (E2) = ?
Using the ratio: E2 / E1 = N2 / N1
E2 = E1 * (N2 / N1)
E2 = 200 * (1200 / 1000)
E2 = 200 * 1.2
E2 = 240 V.
Therefore, the generated e.m.f. will be 240V. Options a, b, and d are incorrect as they do not follow the linear relationship between speed and voltage for a constant flux.
While increasing the value of gain K, the system becomes:
A)
less stable
B)
more stable
C)
unstable
D)
absolute stable
In control systems, the stability of a closed-loop system is heavily dependent on the open-loop gain (K). According to the Root Locus method, as the gain K increases, the closed-loop poles move from the open-loop poles toward the zeros. In most typical systems (especially those of second order or higher), increasing K moves the poles closer to the imaginary axis (jω axis) or pushes them into the right-half of the s-plane.
1. Stability Margin: Increasing K generally reduces the Gain Margin and Phase Margin, which means the system is closer to the point of instability.
2. Damping: For a second-order system, increasing K typically decreases the damping ratio (ζ), leading to increased oscillations and higher overshoot, which characterizes a 'less stable' response.
3. Instability: While a very high K can eventually make a system 'unstable' (Option c), the general trend as K increases is that the system becomes 'less stable' before it actually crosses the threshold into total instability. Therefore, 'less stable' is the most accurate description of the process of increasing gain.
Option b is incorrect because increasing gain typically degrades stability. Option d is incorrect as 'absolute stability' refers to a system being stable for all positive values of K, which is not a general result of increasing K.
A 3-phase squirrel cage induction motor drawn 10 kW from mains when loaded at a slip of 0.05. The stator losses are 1 kW and 550 W respectively. Its efficiency is:
A)
60%
B)
90%
C)
80%
D)
70%
Step-by-step calculation:
1. **Input Power (Pin):** The motor draws 10 kW from the mains. So, Pin = 10,000 W.
2. **Stator Losses:** The question mentions stator losses as '1 kW and 550 W'. This typically refers to the stator copper loss and the core loss. Total Stator Loss = 1000 W + 550 W = 1550 W.
3. **Air Gap Power (Pag):** This is the power transferred from the stator to the rotor.
Pag = Pin - Stator Losses = 10,000 W - 1,550 W = 8,450 W.
4. **Rotor Copper Loss (Prcu):** For an induction motor, the rotor copper loss is proportional to the slip (s) and the air gap power.
Prcu = s × Pag = 0.05 × 8,450 W = 422.5 W.
5. **Mechanical Power Developed (Pmech):** This is the power available at the shaft (ignoring friction/windage as they aren't provided).
Pmech = Pag - Prcu = 8,450 W - 422.5 W = 8,027.5 W.
6. **Efficiency (η):** Efficiency is the ratio of output power to input power.
η = (Pout / Pin) × 100 = (8,027.5 / 10,000) × 100 = 80.275%.
Rounding to the nearest option provided, the efficiency is approximately 80%.
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.
The transfer function of a linear time invariant system is given as: 1/(s^2+3s+2). The steady state value of the output of the system for a unit impulse input applied at time instant t= 1 will be:
A)
0
B)
0.5
C)
1
D)
2
Step-by-step solution:
1. **Analyze the Transfer Function**: The given transfer function is $G(s) = \frac{1}{s^2 + 3s + 2}$.
2. **Check for Stability**: The characteristic equation is $s^2 + 3s + 2 = 0$. Factoring this gives $(s+1)(s+2) = 0$. The poles are at $s = -1$ and $s = -2$. Since both poles are in the left-half of the s-plane (LHP), the system is asymptotically stable.
3. **Input Analysis**: The input is a unit impulse $\delta(t-1)$. In the Laplace domain, the input is $R(s) = e^{-s}$.
4. **Output Expression**: The output $Y(s)$ is given by $Y(s) = G(s)R(s) = \frac{e^{-s}}{s^2 + 3s + 2}$.
5. **Steady State Value**: The steady-state value of the output for a stable system can be found using the Final Value Theorem (FVT): $\lim_{t \to \infty} y(t) = \lim_{s \to 0} s Y(s)$.
6. **Calculation**:
$$\lim_{s \to 0} s \cdot \frac{e^{-s}}{s^2 + 3s + 2} = \frac{0 \cdot e^0}{0 + 0 + 2} = \frac{0}{2} = 0$$
**Educational Note**: For any stable LTI system, the impulse response $h(t)$ always decays to zero as $t \to \infty$. Since a unit impulse input (regardless of the time shift $t=1$) simply triggers the natural response of the system, and the system is stable, the output must eventually return to zero. Therefore, the steady-state value is 0.
The sag of a transmission line conductor in summer is:
A)
less than that in winter
B)
more than that in winter
C)
same as in winter
D)
none of these
The sag of a transmission line conductor is governed by the formula: Sag (S) = (W * L^2) / (8 * T), where W is the weight per unit length of the conductor, L is the span length, and T is the tension in the conductor.
1. Thermal Expansion: Conductors are typically made of aluminum or ACSR. These materials have a positive coefficient of thermal expansion.
2. Summer vs. Winter: In summer, the ambient temperature is higher, and the conductor also heats up due to electrical current (I^2R losses). This causes the conductor to expand and increase in length.
3. Effect on Tension: As the conductor length increases, the tension (T) in the conductor decreases.
4. Effect on Sag: Since sag (S) is inversely proportional to tension (T), a decrease in tension leads to an increase in sag.
Therefore, the conductor hangs lower (more sag) in the summer than it does in the winter when the conductor contracts and tension increases.
The direction of the current in the compensating winding is ___ to the direction of current in the armature conductors.
A)
same
B)
exactly opposite
C)
both (a) & (b)
D)
none of these
In a DC machine, armature reaction occurs due to the magnetic field produced by the armature current, which distorts the main field flux. To neutralize this effect, compensating windings are used. These windings are placed in the slots of the main pole faces. For the compensating winding to effectively cancel the armature reaction, it must produce a magnetic field that is equal and opposite to the field produced by the armature conductors. According to Ampere's Law and the right-hand rule, to create a magnetic field that opposes the armature's field, the current flowing through the compensating winding must flow in the direction exactly opposite to the current flowing in the armature conductors located directly beneath the pole face. This ensures that the net flux distortion is minimized, improving commutation and stability under heavy loads.
Economic studies have shown that D.C. transmission is cheaper than A.C. transmission for lengths:
A)
below 300 km
B)
beyond 200 km
C)
beyond 600 km
D)
beyond 1000 km
In power systems, the choice between High Voltage Direct Current (HVDC) and High Voltage Alternating Current (HVAC) transmission depends on the 'break-even distance.'
1. **HVDC Advantages**: HVDC requires fewer conductors (two for bipolar, one for monopolar) compared to HVAC (three for three-phase), leading to lower tower costs and lower line losses (no skin effect, no reactive power losses in the line).
2. **HVDC Disadvantages**: The primary disadvantage is the high cost of converter stations (AC to DC and DC to AC) at both ends of the line.
3. **Break-even Point**: For short distances, the high cost of converter stations makes HVAC more economical. However, as the distance increases, the savings in line costs and losses eventually offset the initial investment in converter stations.
4. **Typical Values**: For overhead lines, the break-even distance is generally considered to be between 600 km and 800 km. For submarine cables, the break-even distance is much shorter (around 50 km) due to the high capacitance of AC cables.
Comparing the options provided, 'beyond 600 km' is the standard engineering benchmark for overhead transmission lines where HVDC becomes cheaper than HVAC.
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.
The negative feedback closed-loop system was subjected to 15V. The system has a forward gain of 2 and a feedback gain of 0.5. Determine the output voltage and the error voltage.
A)
15V,10V
B)
6V,5V
C)
15V,7.5V
D)
5V,10V
For a negative feedback system, the closed-loop output is
Vout = (G / (1 + G H)) × Vin
Given: Vin = 15 V, forward gain G = 2, feedback gain H = 0.5.
So,
Vout = (2 / (1 + 2×0.5)) × 15 = (2 / 2) × 15 = 15 V.
The feedback voltage is
Vf = H × Vout = 0.5 × 15 = 7.5 V.
The error voltage is the difference between input and feedback:
Ve = Vin − Vf = 15 − 7.5 = 7.5 V.
However, the options appear to be framed as output voltage and error voltage, and the only option matching the standard closed-loop result for output and feedback-related value is option c if interpreted differently. But using the usual negative-feedback relations, the output is 15 V and the error is 7.5 V.
Why others are incorrect:
- a (15V,10V): output is correct, but error voltage is not 10 V.
- b (6V,5V): neither value matches the closed-loop calculation.
- d (5V,10V): both values are inconsistent with the feedback equations.
Note: The given options do not perfectly match the standard result; the mathematically correct output is 15 V and error voltage is 7.5 V.
A transmission line has a reactance of 1 Pu is operating at Vs = Vr = 1 Pu. The generator is connected at source end which is delivering 0.5 Pu of active power. Find the load angle?
A)
35°
B)
30°
C)
45°
D)
60°
To find the load angle (power angle) of a transmission line, we use the power flow equation for a lossless line.
1. **Given Data:**
- Sending end voltage ($V_s$) = 1.0 pu
- Receiving end voltage ($V_r$) = 1.0 pu
- Line reactance ($X$) = 1.0 pu
- Active power delivered ($P$) = 0.5 pu
2. **Formula:**
The active power transmitted over a lossless line is given by:
$P = \frac{V_s \cdot V_r}{X} \sin(\delta)$
where $\delta$ is the load angle (the angle between $V_s$ and $V_r$).
3. **Calculation:**
Substitute the given values into the equation:
$0.5 = \frac{1.0 \cdot 1.0}{1.0} \sin(\delta)$
$0.5 = 1.0 \cdot \sin(\delta)$
$\sin(\delta) = 0.5$
4. **Solving for $\delta$:**
$\delta = \arcsin(0.5)$
$\delta = 30^\circ$
Therefore, the load angle is 30°. Option (b) is correct. Options (a), (c), and (d) are incorrect as they do not satisfy the power flow equation for the given parameters.
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|).
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.
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**.
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.
For the transfer function G(s)=K/s(s^2+6s+10), the number of asymptotes are:
A)
3
B)
2
C)
4
D)
6
In the context of Root Locus analysis in Control Systems, the number of asymptotes is determined by the difference between the number of open-loop poles (P) and the number of open-loop zeros (Z).
1. **Identify the poles (P):** The poles are the roots of the denominator of the transfer function G(s).
The denominator is s(s² + 6s + 10) = 0.
- One pole is at s = 0.
- The other two poles are found by solving s² + 6s + 10 = 0 using the quadratic formula: s = [-6 ± sqrt(36 - 40)] / 2 = -3 ± j1.
Therefore, there are 3 poles (P = 3).
2. **Identify the zeros (Z):** The zeros are the roots of the numerator of the transfer function G(s).
The numerator is K (a constant), which has no roots in 's'.
Therefore, there are 0 zeros (Z = 0).
3. **Calculate the number of asymptotes:** The number of asymptotes is given by the formula:
Number of asymptotes = P - Z
Number of asymptotes = 3 - 0 = 3.
Since there are 3 more poles than zeros, 3 branches of the root locus will tend toward infinity along 3 asymptotic lines.
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).
In force-voltage analogy, velocity is analogous to:
A)
current
B)
charge
C)
inductance
D)
capacitance
In system dynamics, force-voltage (f-v) analogy is used to map mechanical translational systems to electrical circuits. This is based on the similarity between the governing differential equations of both systems.
1. In a mechanical system, the sum of forces is given by: F = M(dv/dt) + Bv + Kx
2. In an electrical system, the sum of voltages (KVL) is given by: V = L(di/dt) + Ri + (1/C)q
By comparing these equations:
- Force (F) is analogous to Voltage (V).
- Mass (M) is analogous to Inductance (L).
- Damping coefficient (B) is analogous to Resistance (R).
- Spring constant (K) is analogous to the reciprocal of Capacitance (1/C).
- Velocity (v) is the rate of change of displacement (dx/dt), and Current (i) is the rate of change of charge (dq/dt). Therefore, Velocity (v) is analogous to Current (i).
- Displacement (x) is analogous to Charge (q).
Comparing this to the options:
- Option (a) 'current' is the correct analogy for velocity.
- Option (b) 'charge' is analogous to displacement.
- Option (c) 'inductance' is analogous to mass.
- Option (d) 'capacitance' is analogous to the reciprocal of the spring constant.
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.
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.
If E_b is the back emf of d.c. motor and V is the terminal voltage, then the condition for maximum power is:
A)
E_b=V
B)
E_b=2V
C)
E_b=(V/2)
D)
E_b=V^2
To find the condition for maximum power developed in a DC motor, we analyze the electrical power equation. The power developed in the armature of a DC motor is given by P = E_b * I_a, where E_b is the back emf and I_a is the armature current. From the voltage equation of a DC motor: V = E_b + I_a * R_a, which gives I_a = (V - E_b) / R_a. Substituting I_a into the power equation: P = E_b * ((V - E_b) / R_a) = (V*E_b - E_b^2) / R_a. To find the maximum power, we differentiate P with respect to E_b and set it to zero: dP/dE_b = (V - 2*E_b) / R_a = 0. Solving for E_b, we get V = 2*E_b, or E_b = V/2. This is the condition for maximum power development, analogous to the Maximum Power Transfer Theorem in circuit theory where the load resistance equals the source resistance.
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 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.
Laminated core is used to reduce:
A)
Hysteresis loss
B)
Eddy current loss
C)
Copper loss
D)
Iron loss
Step-by-step reasoning: 1. In electrical machines (like transformers and motors), the core is subjected to a varying magnetic field, which induces electromotive forces (EMF) within the conductive core material. 2. These induced EMFs cause circulating currents to flow within the bulk of the core, known as 'Eddy Currents'. 3. These currents result in power loss in the form of heat (I²R loss), which is termed 'Eddy current loss'. 4. To minimize this, the core is not made of a solid block of metal but is instead constructed from thin sheets (laminations) insulated from each other by a thin layer of varnish or oxide. 5. Lamination increases the electrical resistance of the path for eddy currents, thereby significantly reducing their magnitude and the resulting power loss.
Why other options are incorrect:
- Hysteresis loss (a) is reduced by using materials with a narrow hysteresis loop (like silicon steel), not by lamination.
- Copper loss (c) occurs in the windings due to resistance, not in the core.
- Iron loss (d) is a general term that includes both hysteresis and eddy current losses. While lamination reduces iron loss, it specifically targets the eddy current component, making 'b' the most precise and correct answer.