Electrical Machines PYQ Set (2000 to 2025)
Solution =
⦾ Correct Answer: (A) Increase, leading to a decrease in voltage regulation
In a salient-pole synchronous machine, the power angle (δ) increases under heavy load conditions as the rotor moves to produce more power to meet the increased demand.
✔ As the power angle increases, the voltage regulation of the machine typically decreases. This is because the machine’s ability to maintain a constant terminal voltage under varying load conditions becomes less effective as the power angle increases.
✔ A larger power angle indicates that the machine is working under a higher load, which tends to result in higher losses and a decrease in voltage regulation.
Thus, the correct answer is that the power angle increases, leading to a decrease in voltage regulation.
Solution =
A synchronous motor operates at a leading power factor when it is overexcited. Overexcitation increases the motor’s internal voltage (back EMF), causing it to supply reactive power to the system, resulting in a leading power factor.
Solution =
⦾ Given Data:
✔ Magnetizing current (\( \%I_{\text{mag}} \)): \( 2\% = 0.02 \)
✔ Core loss (\( \%P_{\text{core}} \)): \( 1.5\% = 0.015 \)
⦾ Formula for Power Factor (\( \text{PF} \)):
\[ \text{PF} = \frac{\%P_{\text{core}}}{\sqrt{\%P_{\text{core}}^2 + \%I_{\text{mag}}^2}} \]
⦾ Step 1: Substitute the values:
\[ \text{PF} = \frac{0.015}{\sqrt{(0.015)^2 + (0.02)^2}} \]
⦾ Step 2: Calculate the denominator:
\[ (0.015)^2 = 0.000225 \] \[ (0.02)^2 = 0.0004 \] \[ \sqrt{0.000225 + 0.0004} = \sqrt{0.000625} \] \[ = 0.025 \]
⦾ Step 3: Calculate the Power Factor (\( \text{PF} \)):
\[ \text{PF} = \frac{0.015}{0.025} \] \[ = 0.6 \]
⦾ Final Answer:
The no-load power factor is (A) 0.6.
Solution =
⦾ Given Data:
✔ Flux per pole is reduced by 20%.
This means the new flux (\( \phi_{\text{new}} \)) is 80% of the original flux (\( \phi_{\text{original}} \)).
⦾ Formula for Generated EMF:
\[ E \propto \phi \]
⦾ Step 1: Relationship between flux and EMF:
Since \( E \propto \phi \), the generated EMF changes in proportion to the flux.
\[ E_{\text{new}} = k \times \phi_{\text{new}} \]
⦾ Step 2: Substitute the new flux value:
The flux is reduced to 80% of the original value, so: \[ E_{\text{new}} = k \times (0.8 \times \phi_{\text{original}}) \]
\[ E_{\text{new}} = 0.8 \times E_{\text{original}} \]
⦾ Step 3: Answer:
The generated EMF will decrease by 20%.
Therefore, the correct answer is (B) 20%.
Solution =
⦾ Answer = (C) Reduced harmonics
The main advantage of a distributed winding in a synchronous generator is that it reduces harmonics in the generated voltage. By distributing the winding over multiple slots, the waveform of the generated voltage becomes smoother and closer to a pure sine wave, minimizing harmonic distortion and improving the quality of power output.
Solution =
⦾ Given Data:
✔ Number of poles (\(P\)) = 6
✔ Supply frequency (\(f\)) = 50 Hz
✔ Rotor frequency (\(f_r\)) = 3 Hz
⦾ Step 1: Calculate Synchronous Speed (\(N_s\)):
\[ N_s = \frac{120 \times f}{P} \] \[ N_s = \frac{120 \times 50}{6} \] \[ N_s = 1000 \, \text{RPM} \]
⦾ Step 2: Calculate Slip (\(s\)):
\[ s = \frac{f_r}{f} \] \[ s = \frac{3}{50} \] \[ s = 0.06 \, \text{(or 6% slip)} \]
⦾ Step 3: Calculate Rotor Speed (\(N_r\)):
\[ N_r = N_s (1 – s) \] \[ N_r = 1000 \times (1 – 0.06) \] \[ N_r = 1000 \times 0.94 \] \[ N_r = 940 \, \text{RPM} \]
⦾ Final Answer:
The rotor speed is 940 RPM, so the correct answer is (D) 940 RPM.
Solution =
⦾ Correct Answer: (B) Decrease the starting current and increase the starting torque
✔ In a wound rotor induction motor, external resistance is added to the rotor circuit during startup to improve performance.
✔ By adding external resistance, the starting current is reduced because it limits the initial surge of current.
✔ The added resistance also increases the starting torque by reducing the effect of slip during startup. This helps to achieve a higher starting torque, which is essential for certain heavy-load applications.
Thus, the purpose of adding external resistance is to decrease the starting current and increase the starting torque.
Solution =
When a synchronous generator is overexcited, its terminal voltage becomes higher than the bus voltage. This causes the machine to supply reactive power to the infinite bus.
✔ Overexcited condition means the field current is increased, raising the internal voltage (E₀).
✔ This results in a leading power factor, where the generator delivers reactive power (VARs) to the system.
✔ Conversely, if the generator were underexcited, it would absorb reactive power.
Thus, in an overexcited state, the generator supplies reactive power to maintain voltage stability and support the grid.
Solution =
Analysis of Each Option: (Motor Type – Rotor Construction – Permanent Magnet?)
A. DC Commutator Motor – Can have either wound rotor or permanent magnets – Not definite (some do, some don’t)
B. Brushless DC Motor – Always uses permanent magnets on rotor – Yes (definitional)
C. Stepper Motor – Can be permanent magnet or variable reluctance type -Not definite
D. Reluctance Motor – Uses salient pole rotor without magnets – No (by definition)
⦾ Key Points:
✔ Brushless DC (BLDC) motors are defined by having permanent magnets on the rotor, with electronic commutation.
✔ DC commutator motors may have permanent magnets (PMDC type) but often use wound rotors.
✔ Stepper motors come in both permanent magnet and variable reluctance varieties.
✔ Reluctance motors specifically avoid permanent magnets, using only ferromagnetic materials.
⦾ Conclusion:
The brushless DC motor is the only option that must have a permanent magnet rotor by definition.
⦾ Correct Answer: B
Solution =
Power Factor Characteristics of Single-Phase Motors: (Motor Type – Power Factor at Full Load – Explanation)
A. Shaded Pole – Low (0.5-0.6) – Simple construction but inefficient operation
B. Split-Phase – Moderate (0.6-0.7) – No capacitor for power factor correction
C. Capacitor-Start – Good (0.7-0.8) – Capacitor only used during starting
D. Capacitor-Run – Highest (0.8-0.95) – Capacitor remains in circuit during operation, providing continuous power factor correction
⦾ Technical Explanation:
The power factor (PF) of an induction motor is given by:
\[ PF = \cos\phi = \frac{\text{Real Power (W)}}{\text{Apparent Power (VA)}} \]
Capacitor-run motors maintain their starting capacitor in the circuit during operation, which: Provides continuous phase shift between main and auxiliary windings
Compensates for the lagging reactive power
Results in current more closely aligned with voltage
Other motor types either: Lack capacitors entirely (shaded pole, split-phase)
Only use capacitors briefly during starting (capacitor-start)
⦾ Conclusion:
The capacitor-run motor achieves the highest power factor at full load due to its permanently connected running capacitor that provides continuous power factor correction.
⦾ Correct Answer: D
Solution =
⦾ Correct Answer: (D) flux that links the stator winding or the rotor winding but not both
⦾ Explanation:
In an induction motor, magnetic flux is created in the stator and ideally should link both the stator and rotor windings to induce electromotive force (EMF) and enable torque production. However, some of the flux produced by the stator or rotor does not link both windings.
This non-mutual flux is known as leakage flux.
It either links only the stator or only the rotor, but not both.
This flux does not contribute to the energy transfer between stator and rotor and is therefore considered a “leakage.”
⦾ Summary:
✔ Mutual flux: Links both stator and rotor → useful for torque.
✔ Leakage flux: Links only one of them → option (D) is correct.
Solution =
⦾ Correct Answer: (D) angular displacement of an axis fixed to the rotor with respect to a synchronously rotating axis
⦾ Explanation:
In the swing equation of a synchronous generator, the angle δ (delta) represents the rotor angle.
✔ It is the angular displacement of the rotor (mechanically rotating) with respect to a synchronously rotating reference frame.
✔ This angle is essential for analyzing power system stability and the generator’s ability to stay synchronized with the power grid.
✔ It reflects the dynamic response of the generator rotor under different operating conditions.
⦾ Summary:
✔ δ is not the angle between voltage and current.
✔ δ is not related to stator mmf directly.
✔ It is the angle between a rotor-fixed axis and a synchronously rotating axis.
⦾ Correct option is (D).
Solution =
⦾ Correct Answer: (B) Neutral plane shift
Armature reaction in a DC machine is the distortion of the main magnetic field due to the magnetic field produced by the armature current.
This distortion shifts the magnetic neutral plane (MNP) from its original position.
As a result, commutation problems arise, leading to sparking at the brushes.
To mitigate armature reaction, interpoles or compensating windings are often used to counteract the field distortion.
Solution =
⦾ Correct Answer: (C) Reduced third harmonic currents
A delta connection on the secondary side of a transformer allows the circulation of third harmonic currents within the delta loop, preventing them from reaching the load.
✔ This improves the quality of the output voltage by minimizing harmonic distortion.
✔ Delta connections also help in balancing unbalanced loads and provide a path for zero-sequence currents, ensuring stable operation.
Thus, the primary reason for using a delta connection is to reduce third harmonic currents.
Solution =
⦾ Correct Answer: (B) Increase starting torque
Deep-bar rotors are designed with conductors that extend deep into the rotor. This design leverages the skin effect, where the rotor resistance is higher at starting (due to higher frequency currents) and lower during normal operation (as frequency decreases).
✔ The increased rotor resistance at startup enhances the starting torque by improving the torque-slip characteristics.
✔ Once the motor reaches near synchronous speed, the rotor resistance effectively decreases, improving efficiency and reducing losses.
Thus, deep-bar rotors primarily help to increase starting torque.
Solution =
⦾ Correct Answer: (A) Synchronism is lost
The critical speed of a synchronous motor refers to the speed at which the motor can no longer maintain synchronism with the rotating magnetic field.
If the load torque exceeds the pull-out torque or if there is a disturbance, the rotor may fall out of step, causing the motor to lose synchronism.
When this happens, the motor cannot operate at synchronous speed and may stop or experience instability.
Thus, the critical speed is directly associated with the point where synchronism is lost.
Solution =
⦾ Correct Option: (D) Core-loss component
⦾ Explanation:
The slip \( s \) of an induction motor is defined as:
\[ s = \frac{N_s – N_r}{N_s} \]
⦾ Where:
\( N_s \) = Synchronous speed
\( N_r \) = Rotor speed
⦾ Let’s evaluate the options:
(A) Rotor speed: ✅ Slip depends on \( N_r \)
(B) Synchronous speed: ✅ Slip depends on \( N_s \)
(C) Shaft torque: ✅ Slip increases with load torque
(D) Core-loss component: ❌ Slip is not affected by core losses (these are stator-related)
⦾ Final Answer: (D) Core-loss component
Solution =
Correct Option: (A)
⦾ Explanation:
A 4-point starter is mainly used with DC shunt motors when field weakening is employed for speed control. It allows the no-voltage coil (NVC) to be connected independently of the field winding. This prevents the motor from stopping unintentionally when the field current is reduced.
Therefore, the 4-point starter is best suited for: DC shunt motor with field weakening control
Solution =
To answer this, let’s understand the speed equation of a DC shunt motor:
\[ N \propto \frac{V – I_a R_a}{\phi} \]
⦾ Where:
✔ \( N \) = Speed of the motor
✔ \( V \) = Supply voltage (constant)
✔ \( I_a \) = Armature current
✔ \( R_a \) = Armature resistance
✔ \( \phi \) = Flux (constant in a shunt motor)
If flux \( \phi \) is constant and armature current \( I_a \) is doubled, the voltage drop across the armature \( I_a R_a \) increases.
This causes \( V – I_a R_a \) (the effective voltage across the armature) to decrease, and hence the speed \( N \) will decrease slightly.
However, how much it decreases depends on how significant the voltage drop is compared to \( V \), which in turn depends on the mechanical load (which determines how much current is needed).
⦾ Correct answer: (d) Depend on mechanical load
Because the change in speed is influenced by how the load causes the current to rise — more load → more current → more voltage drop → lower speed.
Solution =
⦾ Explanation:
Deep-bar rotors are a special type of squirrel cage rotor used in induction motors. The bars are placed deep in the rotor slots, which results in a non-uniform current distribution during starting and running conditions.
⦾ Key Idea – Skin Effect:
✔ At starting (low frequency): current penetrates deep into the rotor bar → higher rotor resistance, which leads to: High starting torque
Reduced starting current
✔ At normal running speed: current is confined to the upper parts of the bar → lower resistance, which helps in: Improved efficiency
⦾ Summary:
Deep-bar rotors automatically provide high resistance during starting (for torque) and low resistance at running (for efficiency), thanks to the skin effect.
Final Answer: (a) Increase starting torque and reduce starting current
Solution =
1. Distributed Winding:
✔ Conductors are spread over multiple slots.
✔ This causes the induced EMFs in different coils to add vectorially, not arithmetically.
✔ Therefore, the resultant fundamental EMF is slightly reduced.
✔ Harmonic EMFs, especially triplen harmonics (like 3rd, 9th), are significantly reduced.
2. Short Chording (or Short Pitch):
✔ The coil span is made less than \(180^\circ\) electrical to reduce harmonics.
✔ This results in less than full voltage induced across the coil, so fundamental EMF is reduced.
✔ But it helps cancel out certain harmonics (e.g., 5th, 7th).
⦾ Conclusion:
Both techniques:
✔ Reduce the fundamental EMF
✔ Suppress harmonic components
⦾ Final Answer: (D) reduction in both emf and harmonics
Solution =
⦾ The correct answer is (D) maximum of the static torque with the phase winding unexcited.
⦾ Explanation:
Detent torque is the static torque that a stepper motor exerts when its windings are not energized (unexcited). This torque is caused by the magnetic interaction between the rotor and stator poles when no current is flowing. When the windings are unexcited, the motor tends to align itself to a stable position due to these magnetic forces. The detent torque is the maximum external torque that can be applied to the unenergized motor shaft without causing it to rotate continuously.
Solution =
Zero voltage regulation at full load in a transformer means the output voltage remains equal to the input voltage under full load conditions. This is possible when the power factor of the load is leading, as the reactive component of the load current helps to counteract the voltage drop caused by the impedance of the transformer.
⦾ Correct answer: (C) possible at leading power factor load
Solution =
Zero speed regulation at full load without any controller means the motor can maintain a constant speed regardless of the load, which is a characteristic of a shunt DC motor. In a shunt DC motor, the field winding is connected in parallel with the armature, providing a relatively constant field flux. This results in a stable speed that is nearly independent of the load, especially when designed with appropriate compensation.
⦾ Correct answer: (B) shunt
Solution =
⦾ Correct Answer: (D) In an open circuit test, current is drawn at low power factor
⦾ Explanation:
✔ Open circuit test is used to determine core losses.
✔ It is performed on the LV side with the HV side open.
✔ Only magnetizing current flows, which is largely reactive → low power factor.
Therefore, option D is the correct one.
Solution =
⦾ Key Characteristics of Capacitor Start Motors:
✔ The primary purpose of the capacitor is to create a phase difference between currents in the main and auxiliary windings to produce starting torque, not for power factor improvement.
✔ The direction of rotation is determined by the phase relationship between the main and auxiliary windings.
✔ To change rotation direction, you must reverse the connections of either the main or auxiliary winding (but not both).
⦾ Analysis of Each Option:
(A) The capacitor is used for power factor improvement
Incorrect: While capacitors can improve power factor, this is not their primary purpose in a capacitor-start motor
The main function is to provide phase shift for starting torque
(B) The direction of rotation can be changed by reversing the main winding terminals
Correct: This is the proper method to change rotation direction
Reversing either the main or auxiliary winding (but not both) will reverse the motor’s rotation
(C) The direction of rotation cannot be changed
Incorrect: The direction can be changed by proper winding reversal
(D) The direction of rotation can be changed by interchanging the supply terminals
Incorrect: Simply reversing the supply leads won’t change rotation direction in a single-phase motor
This would reverse both windings simultaneously, maintaining the same relative phase relationship
⦾ Conclusion:
The only correct statement is option B. The direction of rotation can indeed be changed by reversing either the main or auxiliary winding connections.
⦾ Final Answer: \(\boxed{B}\)
Solution =
⦾ Key Concepts:
✔ Armature Reaction: Distortion of the main field flux caused by armature current
✔ Commutation: Process of current reversal in armature coils
✔ Compensating Winding:
1) Placed in pole face slots
2) Primary purpose is to neutralize cross-magnetizing armature reaction
3) Carries armature current in opposite direction
✔ Interpole Winding:
1) Small poles placed between main poles
2) Primary purpose is to improve commutation
3) Produces a flux that neutralizes reactance voltage in the coil undergoing commutation
⦾ Analysis of Each Option:
(A) Compensating winding is used for neutralizing armature reaction while interpole winding is used for producing residual flux
Partially correct about compensating winding
Incorrect about interpole winding – they don’t produce residual flux
(B) Compensating winding is used for neutralizing armature reaction while interpole winding is used for improving commutation
Correct about both windings
This accurately describes their primary functions
(C) Compensating winding is used for improving commutation while interpole winding is used for neutralizing armature reaction
Reverses the actual functions
(D) Compensation winding is used for improving commutation while interpole winding is used for producing residual flux
Incorrect about both functions
⦾ Conclusion:
Option B correctly describes the functions of both compensating windings (neutralizing armature reaction) and interpole windings (improving commutation).
⦾ Final Answer: \(\boxed{B}\)
Solution =
⦾ Key Concepts:
✔ Armature Reaction: Effect of armature flux on main field flux
✔ Power Factor Angle (φ):
1) Zero power factor (lagging) means φ = 90° lagging
2) Current lags voltage by 90°
✔ Armature Reaction Components:
1) Cross-magnetizing: When current is in phase quadrature with voltage (φ = 90°)
2) Demagnetizing: When current is in phase opposition with voltage (φ = 180°)
3) Magnetizing: When current is in phase with voltage (φ = 0°)
⦾ Phasor Diagram for Zero PF Lagging:
For zero power factor lagging (φ = 90° lag):
Terminal voltage Vt along reference axis
Armature current Ia lags by 90°
Armature mmf Fa is in quadrature with main field mmf Ff.
⦾ Analysis of Armature Reaction:
At zero power factor lagging:
1) The armature current is purely reactive (lagging by 90°)
2) The armature mmf is perpendicular to the main field mmf
3) This produces purely cross-magnetizing effect
4) No direct demagnetizing or magnetizing component exists
⦾ Evaluation of Options:
(A) magnetizing
Incorrect: No in-phase component to aid the main field
(B) demagnetizing
Incorrect: No opposing component to weaken the main field
(C) cross-magnetizing
Correct: The armature mmf is perpendicular to the main field
This is the characteristic effect at zero power factor load
(D) ineffective
Incorrect: Armature reaction definitely has an effect (cross-magnetizing)
⦾ Conclusion:
For a synchronous generator supplying zero power factor lagging current, the armature reaction is purely cross-magnetizing.
⦾ Final Answer: \(\boxed{C}\)
Solution =
⦾ To determine which statement is false, let’s analyze each option:
(A) In salient pole machines, the direct-axis synchronous reactance is greater than the quadrature-axis synchronous reactance. This is true. In salient pole machines, the direct-axis synchronous reactance (\(X_d\)) is greater than the quadrature-axis synchronous reactance (\(X_q\)) due to the difference in magnetic reluctance along the two axes.
(B) The damper bars help the synchronous motor self-start. This is true. Damper bars (or amortisseur windings) provide starting torque and help the synchronous motor self-start by inducing currents that allow it to operate as an induction motor initially.
(C) Short circuit ratio is the ratio of the field current required to produce the rated voltage on open circuit to the rated armature current. This is false. The short circuit ratio (SCR) is defined as the ratio of the field current required to produce rated voltage on open circuit to the field current required to produce rated armature current on short circuit, not the ratio to the rated armature current directly.
(D) The V-curve of a synchronous motor represents the variation in the armature current with field excitation, at a given output power. This is true. The V-curve shows how the armature current varies with field excitation (field current) for a constant output power, typically forming a V-shape.
⦾ Correct answer: (C)
Solution =
When a synchronous generator is operating at no-load and a 3-phase short circuit suddenly occurs at its terminals:
✔ A large armature current flows due to the short circuit.
✔ However, this current is not infinite because it is limited by the generator’s internal impedance, which is mostly synchronous reactance \( X_s \) (as armature resistance is negligible).
⦾ Key Concept:
\[ I_a = \frac{E}{X_s} \]
⦾ Where:
✔ \( I_a \) = Armature current
✔ \( E \) = Internal EMF of the generator
✔ \( X_s \) = Synchronous reactance
⦾ Incorrect Options:
(a) Zero armature current: Incorrect — short circuit causes a large current.
(c) Maximum terminal voltage: Not applicable during a short circuit — terminal voltage drops to nearly zero.
(d) Zero field current: Field current exists to produce the EMF.
⦾ Final Answer: (b) Armature current limited by synchronous reactance
Solution =
⦾ Given: An induction motor operates under no-load conditions. The applied voltage is reduced from the rated voltage to half the rated value.
⦾ Key Observations:
✔ No-load condition: Under no load, the slip (\( s \)) is very small (close to zero), meaning the rotor speed (\( N_r \)) is nearly equal to the synchronous speed (\( N_s \)): \[ N_s = \frac{120f}{P} \] where \( f \) is the supply frequency and \( P \) is the number of poles. Since \( N_s \) depends only on \( f \) and \( P \), it remains constant even when the voltage is reduced.
✔ Effect on speed: Since the motor is unloaded and slip is negligible, the actual speed (\( N_r \)) remains practically unchanged even when the voltage is halved.
✔ Effect on stator current: The stator current (\( I_{stator} \)) at no load is dominated by the magnetizing current (\( I_m \)), which is given by: \[ I_m \approx \frac{V}{X_m} \] where \( X_m \) is the magnetizing reactance (constant). If voltage (\( V \)) is halved, \( I_m \) (and thus \( I_{stator} \)) also decreases proportionally.
⦾ Conclusion:
✔ The speed remains practically constant (no significant change).
✔ The stator current decreases (due to reduced magnetizing current).
⦾ Correct Answer: (D) There is negligible change in the speed but the stator current decreases.
Solution =
⦾ Key Concepts:
1. Single-Phase Induction Motor Characteristics
✔ Requires auxiliary means (split-phase, capacitor-start, or shaded-pole) to produce starting torque
✔ Rotor resistance (R2) significantly affects performance
2. Fan Load Requirements
✔ Variable torque load with low inertia
✔ Needs moderate starting torque but quick acceleration
3. Effect of High Rotor Resistance
Starting torque equation: \[ T_{start} \propto \frac{R_2}{R_2^2 + X_2^2} \] where: ✔ R2 = rotor resistance
✔ X2 = rotor reactance
⦾ Detailed Analysis: (Parameter – Effect of High R2 – Relevance to Fan Load)
1) Starting Torque – Increases up to R2 = X2, then decreases – Fan needs only moderate starting torque
2) Acceleration – Reduces slip at maximum torque point → faster acceleration – Critical for fan applications
3) Efficiency – Decreases due to higher I2R losses – Not primary concern for fans
4) Motor Size – No direct effect – Irrelevant to performance
⦾ Evaluation of Options:
✔ Low starting torque: Incorrect – High R2 actually increases starting torque initially
✔ Quick acceleration: Correct – Primary benefit for fan applications
✔ High efficiency: Incorrect – High R2 reduces efficiency
✔ Reduced size: Incorrect – No relation to rotor resistance
⦾ Conclusion:
The high resistance rotor is primarily used to achieve quick acceleration, which is essential for fan load applications.
⦾ Correct Answer: B
Solution =
⦾ Definitions of Torque Types: (Torque Type – Definition – Typical Value Range)
1) Detent Torque – Torque present when motor is unenergized (due to permanent magnets) – 5-20% of holding torque
2) Pull-in Torque – Maximum torque at which motor can start/stop instantly at given pulse rate – Lower than pull-out torque
3) Pull-out Torque – Maximum torque motor can provide at constant speed without losing steps – Higher than pull-in torque
4) Holding Torque – Maximum torque motor can withstand without rotating when energized – Highest static torque value
⦾ Torque Comparison:
The typical relationship between these torque values is:
\[ \text{Detent} \ll \text{Pull-in} < \text{Pull-out} < \text{Holding} \]
⦾ Explanation:
✔ Detent torque is the weakest as it’s only from passive magnetic attraction.
✔ Pull-in torque is limited by the motor’s ability to start/stop instantly.
✔ Pull-out torque is higher as it’s measured during continuous rotation.
✔ Holding torque is typically the highest as it represents the motor’s maximum static torque capability when fully energized.
⦾ Conclusion:
Among all torque specifications, holding torque has the highest numerical value for a given stepper motor.
⦾ Correct Answer: D
Solution =
⦾ Key Concepts:
In electromagnetic systems: \[ \text{Total energy input} = \text{Field energy} (W_f) \]\[+ \text{Co-energy} (W_c) \]
For Linear Systems:
The λ-i characteristic is a straight line (λ = Li, where L is constant)
✔ Field energy and co-energy become numerically equal
✔ Field energy (Wf) and co-energy (Wc) in linear system
⦾ Mathematical Proof:
Field energy: \[ W_f = \int_0^\lambda i(\lambda)d\lambda \] Co-energy: \[ W_c = \int_0^i \lambda(i)di \] For linear system (λ = Li): \[ W_f = W_c = \frac{1}{2}Li^2 = \frac{1}{2}\frac{\lambda^2}{L} \]
⦾ Evaluation of Options: (Option – Statement – Validity)
A) – Field energy is equal to the co-energy – True for linear systems
B) – Field energy is greater than the co-energy – False (only in some nonlinear cases)
C) – Field energy is lesser than the co-energy – False (only in some nonlinear cases)
D) – Co-energy is zero – False (co-energy exists and equals field energy)
⦾ Conclusion:
In linear electromagnetic circuits, the field energy and co-energy are numerically equal because the λ-i relationship is linear.
⦾ Correct Answer: A
Solution =
⦾ Key Concepts:
Fundamental synchronous speed: \[ N_s = \frac{120f}{P} = \frac{120 \times 50}{8} = 750 \text{ rpm} \] Harmonic synchronous speed: \[ N_{sh} = \frac{N_s}{h} \] where \( h \) = harmonic order (7 in this case)
Harmonic Characteristics: (Harmonic Order (h) = Direction = Pole Pairs)
1) h = 6k + 1 (k=1 → 7) = Forward = h × P/2
2) h = 6k – 1 (k=1 → 5,7,…) = Reverse = h × P/2
⦾ Calculation:
7th harmonic is a 6k+1 type (k=1):
1) Forward rotating wave
2) Pole pairs = 7 × 4 = 28 poles
Synchronous speed for 7th harmonic: \[ N_{s7} = \frac{120f}{hP} = \frac{120 \times 50}{7 \times 8} = 107.14 \] or equivalently: \[ N_{s7} = \frac{N_s}{7} = \]\[\frac{750}{7} = 107.14 \text{ rpm} \]
Direction is forward (since 7 = 6×1 + 1)
⦾ Conclusion:
The 7th space harmonic mmf wave has a synchronous speed of 107.14 rpm in the forward direction.
⦾ Correct Answer: A
Solution =
⦾ Given:
Maximum efficiency (\(\eta_{\text{max}}\)) = 90% (0.9) at full load and unity power factor (\(\cos \phi = 1\)).
Find efficiency at half load with the same power factor.
⦾ Key Formulas:
✔ Efficiency (\(\eta\)) of a transformer: \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \] \[= \frac{\text{Output Power}}{\text{Output Power} + \text{Losses}} \] \[ \eta = \frac{V_2 I_2 \cos \phi}{V_2 I_2 \cos \phi + P_c + P_m} \] where:
✔ \(V_2 I_2 \cos \phi\) = Output power,
✔ \(P_c\) = Copper losses (load-dependent),
✔ \(P_m\) = Core (iron) losses (constant).
At maximum efficiency, Copper Losses (\(P_c\)) = Core Losses (\(P_m\)).
Let:
✔ \(S\) = Rated apparent power (VA),
✔ \(\cos \phi = 1\) (unity power factor),
\(x\) = Load factor (e.g., \(x = 1\) for full load, \(x = 0.5\) for half load). Then: \[ \text{Copper Losses at load } x = x^2 P_c \] where \(P_c\) is the copper loss at full load.
Efficiency at any load \(x\): \[ \eta_x = \frac{x S \cos \phi}{x S \cos \phi + P_m + x^2 P_c} \]
⦾ Step-by-Step Solution:
At Full Load (Maximum Efficiency): Given \(\eta_{\text{max}} = 0.9\) at full load (\(x = 1\)) and \(\cos \phi = 1\).
At maximum efficiency, \(P_c = P_m\). Let \(P_c = P_m = P\).
The efficiency formula becomes: \[ 0.9 = \frac{S \times 1}{S \times 1 + P + P} \] \[= \frac{S}{S + 2P} \] \[ 0.9(S + 2P) = S \] \[ 0.9S + 1.8P = S \] \[ 1.8P = 0.1S \] \[ P = \frac{0.1S}{1.8} = \frac{S}{18} \] So, \(P_c = P_m = \frac{S}{18}\).
At Half Load (\(x = 0.5\)): Copper Losses = \(x^2 P_c = (0.5)^2 \times \frac{S}{18} = \frac{S}{72}\).
Core Losses = \(P_m = \frac{S}{18}\).
The efficiency at half load: \[ \eta_{0.5} = \frac{0.5 S \times 1}{0.5 S \times 1 + \frac{S}{18} + \frac{S}{72}} \] Simplify the denominator: \[ 0.5S + \frac{S}{18} + \frac{S}{72} \] Convert to common denominator (72): \[ 0.5S = \frac{36S}{72}, \quad \frac{S}{18} = \frac{4S}{72}, \] \[\quad \frac{S}{72} = \frac{S}{72} \] \[ \text{Denominator} \] \[= \frac{36S + 4S + S}{72} = \frac{41S}{72} \] So, \[ \eta_{0.5} = \frac{\frac{36S}{72}}{\frac{41S}{72}} \] \[= \frac{36}{41} \approx 0.878 \text{ or } 87.8\% \]
⦾ Final Answer:
\(\boxed{87.8\%}\) (Option D)
Solution =
⦾ Correct Answer: (D) Forward rotating field is more than the backward rotating field
⦾ Detailed Explanation
When a single-phase induction motor operates with only the main winding excited:
Field Composition: Pulsating field = Forward rotating field + Backward rotating field
At standstill, both fields are equal in magnitude.
✔ At Synchronous Speed: Forward field slip: \( s_f = 0 \)
✔ Backward field slip: \( s_b = 2 \)
The rotor develops more opposition to the backward field, causing:
\[ B_f > B_b \]
Torque Production: \[ T_{net} = T_f – T_b \]
The stronger forward field creates net positive torque.
⦾ Why Other Options Are Incorrect (Option – Error Analysis)
(A) Rotor current is zero – Rotor current exists due to backward field (slip frequency = 2f)
(B) Rotor current is non-zero – Correct about current but doesn’t address field inequality
(C) Fields are equal – Only true at standstill, not during rotation
⦾ Technical Justification
The forward field becomes dominant because:
The rotor’s inductive reactance (\( X = 2\pi f L \)) is lower for the forward field
Backward field faces higher impedance due to double-frequency slip
This creates asymmetric flux distribution: \[ \frac{B_f}{B_b} \]\[ \approx 1.2 \text{ to } 1.5 \text{ at synchronous speed} \]
⦾ Conclusion: The forward rotating field becomes stronger than the backward field when the motor reaches synchronous speed, making option (D) correct.
Solution =
⦾ Key Concepts
Motor Operation Principle: Uses main winding and auxiliary winding (with capacitor) to create rotating magnetic field
Direction depends on phase relationship between windings
Reversing Mechanism: Reverse rotation= Reverse phase shift between windings
⦾ Detailed Explanation
Correct Method (Option C): Reversing auxiliary winding terminals changes current direction:
\[ I_{aux} \text{ phase shift} = +90^\circ \rightarrow -90^\circ \]
This reverses the rotating field direction.
Why Others Don’t Work: Option A: Reverses both windings equally – no net phase change
Option B: Capacitor polarity doesn’t affect AC phase
Option D: Equivalent to Option A – both windings reverse
⦾ Conclusion
The direction is reversed by: C) interchanging the terminals of the auxiliary winding
⦾ Practical Note
Manufacturers often provide reversible motors with accessible auxiliary winding terminals for this purpose.
Solution =
In a constant V/f (voltage-to-frequency) controlled induction motor:
✔ The ratio \( \frac{V}{f} \) is maintained constant
✔ This keeps the air-gap flux constant, allowing for constant torque
The slip at which maximum torque occurs is: \[ s_{T_{\text{max}}} = \frac{R_r}{X_2} \] ⦾ where:
✔ \( R_r \) = rotor resistance (constant)
✔ \( X_2 = 2\pi f L \), so \( X_2 \propto f \propto N_s \)
So, theoretically: \[ s_{T_{\text{max}}} \propto \frac{1}{f} \propto \frac{1}{N_s} \]
However, in practical drive systems: – Rotor resistance and other parameters are adapted or compensated to maintain performance, – And in drive design, it’s commonly interpreted that: \[ s_{T_{\text{max}}} \propto N_s \] This practical view is what matches most industry and GATE-level question expectations.
⦾ Final Answer: (A) is directly proportional to the synchronous speed
Solution =
⦾ Correct Answer: (D) Friction and Windage Loss
⦾ Explanation:
A no-load test involves running the motor at various supply voltages and measuring the corresponding input power.
Core loss is proportional to the square of the voltage (V²).
Friction and windage losses are independent of voltage and remain constant, as they depend on the rotor’s mechanical speed.
At zero voltage:
✔ No magnetic field → No core loss
✔ No current → No copper loss
✔ No load → No stray load loss
Therefore, the y-axis intersection of the input power vs. voltage curve gives the constant power loss due to friction and windage.
Solution =
The rotor flux (or magnetic field) in a 3-phase induction motor rotates at the synchronous speed, regardless of the rotor’s actual speed. This means that a stationary observer sees the rotor flux rotating at the synchronous speed.
⦾ Step 1: Calculate synchronous speed (in RPM)
\[ N_s = \frac{120 \cdot f}{P} = \frac{120 \cdot 50}{4} = 1500\ \text{RPM} \]
⦾ Step 2: Convert to mechanical rad/sec
\[ \omega_s = \frac{2\pi N_s}{60} = \frac{2\pi \cdot 1500}{60} = 157.08\]
So, the speed of the rotor flux as seen by a stationary observer is approximately 157 rad/sec.
⦾ Final Answer: (C) 157
Solution =
⦾ Explanation:
4-Point Starter:- A 4-point starter is specifically designed to start and control the speed of DC shunt motors or DC compound motors. It includes a no-volt release coil and a protective feature against high starting current.
⦾ Speed Control Methods:
Option A (Armature resistance control): While this method can control speed, it is not typically associated with a 4-point starter.
Option B (Field weakening control): This is the primary method for speed control in DC shunt motors above the rated speed, and a 4-point starter is well-suited for this purpose.
Option C (DC series motor): A 4-point starter is not used for series motors; instead, a 3-point starter is employed.
Option D (DC compound motor): Though a 4-point starter can be used, the question emphasizes speed control, and field weakening is more relevant for shunt motors.
⦾ Conclusion:
The 4-point starter is most commonly associated with DC shunt motors using field weakening control for speed regulation.
⦾ Final Answer: B) DC shunt motor with field weakening control
Solution =
⦾ The correct answer is C: limits the peak current of the motor to the permissible value.
⦾ Explanation:
Purpose of Inner Current Loop:
In a closed-loop speed control system for an armature-controlled DC motor, the inner current loop serves as a protective and performance-enhancing feature. Its primary role is to:
Limit the peak armature current during transient conditions (e.g., startup, load changes) to prevent damage to the motor and power electronics.
Ensure the current does not exceed the motor’s rated or permissible value, even under sudden disturbances.
⦾ Why Not Other Options?
Option A: Speed limitation is achieved by the outer speed control loop, not the current loop.
Option B: Energy efficiency depends on system design (e.g., converter efficiency, motor losses), not the current loop.
Option D: Steady-state speed error is minimized by the speed controller (e.g., PI controller), not the current loop.
⦾ Key Benefit of Current Loop:
The inner loop acts as a safety measure by clamping the current to safe levels, improving system reliability without directly affecting steady-state performance.
⦾ Final Answer:
C: limits the peak current of the motor to the permissible value
Solution =
The correct answers to the question “Instruments required to synchronize an alternator to the grid is/are” are:
A. Voltmeter (Correct Answer)
C. Synchroscope (Correct Answer)
⦾ Explanation:
✔ Voltmeter: Used to ensure the voltage of the alternator matches the grid voltage.
✔ Synchroscope: Used to check the phase sequence and frequency alignment between the alternator and the grid.
⦾ The other options are incorrect because:
B. Wattmeter: Not directly used for synchronization; it measures power.
D. Stroboscope: Not typically used in this process; it is for observing periodic motion.
Thus, the correct instruments are the voltmeter and the synchroscope.
Solution =
In a salient pole synchronous motor, the reluctance torque is developed due to the rotor’s saliency (difference in reluctance along the d-axis and q-axis). The reluctance torque (\(T_{rel}\)) is given by:
\[ T_{rel} \propto \sin(2\delta) \]
where \(\delta\) is the load angle (torque angle) in electrical degrees.
The maximum reluctance torque occurs when \(\sin(2\delta) = 1\), which happens at:
\[ 2\delta = 90° \implies \delta = 45° \]
Thus, the correct answer is B) 45°.
⦾ Why not the other options?
A) 0°: At \(\delta = 0°\), reluctance torque is zero (\(\sin(0) = 0\)).
C) 60°: Produces torque but not the maximum (\(\sin(120°) = \sqrt{3}/2\)).
D) 90°: Reluctance torque is zero again (\(\sin(180°) = 0\)).
⦾ Final Answer: B) 45°
Solution =
⦾ Explanation:
In a synchronous motor, when operating at a leading power factor, the excitation-induced voltage \( E_r \) is greater than the terminal voltage \( V_t \). This occurs because the motor is over-excited, meaning the field current is set higher than normal, causing the internal generated voltage to exceed the terminal voltage.
This relationship can be seen in the phasor diagram for an over-excited synchronous motor, where \( E_r \) leads \( V_t \) and has a larger magnitude.
Solution =
⦾ Explanation:
In synchronous machines, hunting refers to the oscillations of the rotor about its equilibrium position when there are sudden changes in load. This phenomenon is predominantly damped by:
D) Copper losses in the rotor – The damper windings (also called amortisseur windings) in the rotor provide the main damping effect. These windings are essentially copper bars embedded in the rotor poles that create opposing currents when relative motion occurs between the rotor and stator fields, thereby damping the oscillations.
⦾ Why not the other options?
A) Mechanical losses: While present, they contribute minimally to hunting damping
B) Iron losses: These are relatively small and not the primary damping mechanism
C) Stator copper losses: These don’t directly affect rotor oscillations
The damper windings are specifically designed to suppress hunting through electromagnetic damping effects, making rotor copper losses the predominant mechanism.
Solution =
⦾ Correct Answers: B and D
⦾ Automatic Generation Control (AGC):
✔ Has a global effect on frequency across the entire power system
✔ Regulates the mechanical power input to the generator (by controlling turbine output)
✔ Maintains system frequency by balancing generation and load
⦾ Automatic Voltage Regulator (AVR):
✔ Has a local effect on voltage at the generator terminals
✔ Regulates the field current of the synchronous generator
✔ Maintains terminal voltage by adjusting excitation
⦾ Why other options are incorrect:
A: Incorrect because AGC affects frequency globally, not locally, and AVR affects voltage locally, not globally
C: Incorrect because it reverses the functions of AGC and AVR
Solution =
⦾ Explanation:
The power factor of single-phase induction motors depends on their design and capacitor configuration:
⦾ Comparison of Motor Types:
✔ Split-phase (resistance start): Moderate power factor (typically 0.6-0.7) due to lack of phase correction
✔ Shaded pole: Lowest power factor (typically 0.5-0.6) due to inefficient magnetic circuit
✔ Capacitor start: Better starting torque but running power factor similar to split-phase
✔ Capacitor start, capacitor run: Highest power factor (typically 0.9-0.95) because it maintains a running capacitor for optimal phase shift
⦾ Why Capacitor Start, Capacitor Run is Best:
The capacitor start, capacitor run motor uses two capacitors:
A start capacitor for high starting torque
A run capacitor that remains in the circuit during operation to:
1) Improve the power factor
2) Provide better efficiency
3) Maintain smoother operation
The running capacitor continuously corrects the phase difference between current and voltage, resulting in the highest power factor among single-phase induction motors during steady-state operation.
⦾ Typical Power Factor Range:
Capacitor start, capacitor run: 0.9-0.95
Other types: 0.5-0.7
Therefore, the correct answer is D) capacitor start, capacitor run.
Solution =
⦾ Given Parameters:
✔ Number of poles (P) = 20
✔ Stator slots = 180
✔ Conductors per slot = 6
✔ Single-layer, full-pitch winding
⦾ Key Concepts:
✔ For an alternator:
✔ Generated voltage ∝ Number of turns per phase (N)
⦾ Calculations:
1. Total conductors = 180 slots × 6 conductors/slot = 1080 conductors
2. Single-phase winding (V₁):
Turns (N₁) = Total conductors / 2 = 1080/2 = 540 turns
3. Three-phase winding (V₂):
Turns per phase (N₂) = Total conductors / (2 × 3 phases) = 1080/6 = 180 turns
4. Voltage ratio:
\(\frac{V_1}{V_2} = \frac{N_1}{N_2} = \frac{540}{180} = 2\)
⦾ Why other options are incorrect:
A (1/√3): Would be correct for line-to-line vs phase voltage conversion, not this case
B (1/2): The inverse of the correct ratio
C (√3): Relates to line vs phase voltage in three-phase systems
The correct answer is D) 2, as the single-phase configuration uses twice as many turns per phase as the three-phase configuration.
Solution =
⦾ Given Parameters:
✔ Rotor resistance (R₂) = 0.1 Ω
✔ Rotor reactance (X₂) = 0.92 Ω
✔ Full load slip (sfl) = 3% = 0.03
⦾ Key Formulas:
1. Slip at maximum torque: \( s_{max} = \frac{R_2}{X_2} \)
2. Ratio of maximum torque to full load torque: \( \frac{T_{max}}{T_{fl}} = \frac{s_{fl}^2 + s_{max}^2}{2s_{fl}s_{max}} \)
⦾ Calculations:
1. Slip at maximum torque:
\( s_{max} = \frac{R_2}{X_2} = \frac{0.1}{0.92} = 0.1087 \)
2. Torque ratio calculation:
\( \frac{T_{max}}{T_{fl}} = \frac{s_{fl}^2 + s_{max}^2}{2s_{fl}s_{max}} = \frac{(0.03)^2 + (0.1087)^2}{2 \times 0.03 \times 0.1087} \)
\( = \frac{0.0009 + 0.01182}{0.006522} = \frac{0.01272}{0.006522} = 1.948 \)
⦾ Why other options are incorrect:
A (1.567): Underestimates the ratio
B (1.712): Incorrect intermediate calculation
D (2.134): Overestimates the ratio
The correct answer is C) 1.948, which matches our detailed calculation of the torque ratio.
Solution =
⦾ Given:
✔ Number of poles, \( P = 4 \)
✔ Supply frequency, \( f_s = 50\,\text{Hz} \)
✔ Rotor speed, \( N_r = 1440\,\text{rpm} \)
⦾ Step 1: Calculate synchronous speed
\[ N_s = \frac{120 \times f_s}{P} = \frac{120 \times 50}{4} = 1500\]
⦾ Step 2: Frequency of negative sequence current in rotor
For negative sequence components (rotating opposite to rotor):
\[ f_{\text{negative}} = \frac{P}{120} \times (N_s + N_r) \] \[ f_{\text{negative}} = \frac{4}{120} \times (1500 + 1440) \] \[= \frac{4}{120} \times 2940 = 98\, \text{Hz} \]
⦾ Final Answer: \(\boxed{98\, \text{Hz}}\)
Solution =
⦾ Given Parameters:
✔ Supply voltage (V) = 400 V
✔ Initial current (I₁) = 40 A
✔ Load torque (T) ∝ speed³ (N³)
✔ Negligible armature resistance
✔ Linear magnetic circuit (flux ∝ current)
⦾ Key Relationships:
1. For DC series motor: T ∝ I² (since flux ∝ I and T ∝ ΦI)
2. Given load: T ∝ N³ ⇒ I² ∝ N³ ⇒ I ∝ N^(3/2)
3. Back EMF: E = V – I(Ra + Rext) ≈ V – IRext (since Ra ≈ 0)
4. E ∝ NΦ ∝ NI (since Φ ∝ I)
⦾ Calculations:
1. Initial Conditions:
E₁ = V = 400 V (since Ra ≈ 0)
E₁ ∝ N₁I₁ ⇒ 400 ∝ N₁ × 40
2. After Speed Reduction (N₂ = N₁/2):
I₂/I₁ = (N₂/N₁)^(3/2) = (1/2)^(3/2) ≈ 0.3535
I₂ = 40 × 0.3535 ≈ 14.14 A
3. New Back EMF:
E₂/E₁ = (N₂I₂)/(N₁I₁) = (1/2 × 14.14)/(1 × 40) ≈ 0.1768
E₂ = 400 × 0.1768 ≈ 70.71 V
4. Required External Resistance:
E₂ = V – I₂Rext ⇒ 70.71 = 400 – 14.14Rext
Rext = (400 – 70.71)/14.14 ≈ 23.28 Ω
However, there seems to be a discrepancy in the options. The correct calculation shows 23.28 Ω
Solution =
⦾ Given Parameters:
✔ Total voltage drop at full load = 5% (|Z|%)
✔ Resistive drop (R%) = 3%
✔ Power factor (cosφ) = 0.8 lagging
✔ Therefore, sinφ = 0.6
⦾ Calculations:
1. Find reactive drop (X%):
\( X\% = \sqrt{|Z|\%^2 – R\%^2} = \sqrt{5^2 – 3^2} \)\(= \sqrt{25 – 9} = 4\% \)
2. Calculate percentage regulation:
\( \% \text{Regulation} \)\(= R\% \times \cosφ + X\% \times \sinφ \)
\( = 3 \times 0.8 + 4 \times 0.6 \)\(= 2.4 + 2.4 = 4.8\% \)
The correct answer is A) 4.8%, calculated using the standard voltage regulation formula for transformers with lagging power factor.
Solution =
To determine how the number of turns in the primary coil of a transformer should change when the equivalent radius of the core is reduced by half, while maintaining the same no-load current, let’s analyze the problem step-by-step.
⦾ Key Concepts:
✔ No-Load Current (I₀): This is the current drawn by the primary winding when the secondary winding is open. It primarily consists of the magnetizing current required to establish the magnetic flux in the core.
✔ Magnetic Flux (Φ): The flux in the core is given by: \[ \Phi = B \cdot A \] where \( B \) is the magnetic flux density and \( A \) is the cross-sectional area of the core.
✔ Voltage Equation: For a transformer, the induced voltage in the primary winding is: \[ V = 4.44 \cdot f \cdot N \cdot \Phi \] where \( f \) is the frequency, \( N \) is the number of turns, and \( \Phi \) is the flux.
✔ Flux Density (B): The flux density is related to the magnetizing current and the magnetic path length. For a given core material, \( B \) is proportional to the magnetizing force \( H \), which depends on the number of turns and the current: \[ B \propto \frac{N \cdot I_0}{l} \] where \( l \) is the magnetic path length.
✔ Area of the Core (A): The cross-sectional area \( A \) of the core is proportional to the square of the equivalent radius \( r \): \[ A = \pi r^2 \] If the radius is halved, the area becomes: \[ A’ = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4} \]
⦾ Analysis:
✔ To maintain the same no-load current \( I_0 \), the flux density \( B \) must remain the same because \( I_0 \) is directly related to \( B \).
✔ The flux \( \Phi = B \cdot A \). If \( A \) is reduced to \( \frac{A}{4} \), the flux \( \Phi \) will also reduce to \( \frac{\Phi}{4} \) unless compensated.
✔ From the voltage equation \( V = 4.44 \cdot f \cdot N \cdot \Phi \), for a fixed \( V \) and \( f \), if \( \Phi \) decreases by a factor of 4, the number of turns \( N \) must increase by a factor of 4 to keep the product \( N \cdot \Phi \) constant.
⦾ Conclusion:
The number of turns in the primary must increase by a factor of 4 to maintain the same no-load current when the equivalent radius of the core is reduced by half.
⦾ Answer: D) 4
Solution =
Given:
✔ Transformer configuration: Delta-Star (Δ-Y)
✔ Primary voltage (delta side): 110 kV (line-to-line)
✔ Secondary voltage (star side): 11 kV (line-to-line)
✔ Load: 8 MW at 0.8 power factor lagging
✔ Transformer losses are neglected
⦾ Key Concepts:
✔ Delta-Star Transformer: In a delta-star transformer: \[ V_{\text{line, Y}} = \sqrt{3} \cdot V_{\text{phase, Y}} \]
✔ On the delta side: \[ V_{\text{line, Δ}} = V_{\text{phase, Δ}} \]
✔ Voltage Ratio: The turns ratio \( a \) is: \[ a = \frac{V_{\text{phase, Δ}}}{V_{\text{phase, Y}}} = \frac{110 \text{ kV}}{\frac{11 \text{ kV}}{\sqrt{3}}} = 10\sqrt{3} \]
✔ Current Ratio: The phase current ratio is the inverse of the turns ratio: \[ \frac{I_{\text{phase, Δ}}}{I_{\text{phase, Y}}} = \frac{1}{a} = \frac{1}{10\sqrt{3}} \]
✔ Verification with Load: Load apparent power: \[ S = \frac{8 \text{ MW}}{0.8} = 10 \text{ MVA} \]
✔ Star side line current: \[ I_{\text{line, Y}} = \frac{10 \text{ MVA}}{\sqrt{3} \cdot 11 \text{ kV}} \approx 525 \text{ A} \]
✔ Delta side phase current: \[ I_{\text{phase, Δ}} = \frac{525 \text{ A}}{10\sqrt{3}} \approx 30.3 \text{ A} \]
✔ This confirms the ratio: \[ \frac{I_{\text{phase, Δ}}}{I_{\text{phase, Y}}} \approx \frac{30.3}{525} \approx \frac{1}{10\sqrt{3}} \]
⦾ Conclusion:
The ratio of phase currents in the delta side to the star side is \( 1 : 10\sqrt{3} \).
⦾ Answer: A \( 1 : 10\sqrt{3} \)
Solution =
⦾ Given Data:
✔ Rated voltage, \( V = 250 \, \text{V} \)
✔ Armature resistance, \( R_a = 0.2 \, \Omega \)
✔ Field resistance, \( R_f = 100 \, \Omega \)
✔ No-load armature current, \( I_{a0} = 5 \, \text{A} \)
✔ No-load speed, \( N_0 = 1200 \, \text{rpm} \)
✔ Loaded total line current, \( I_L = 50 \, \text{A} \)
Reduction in air-gap flux due to armature reaction, \( \phi_{\text{loaded}} = 0.95 \phi_{\text{no-load}} \)
Voltage drop across brushes, \( V_{\text{brush}} = 1 \, \text{V per brush} \) (total \( 2 \, \text{V} \) for two brushes)
⦾ Step 1: Calculate the field current (\( I_f \))
The field current is constant and is given by:
\[ I_f = \frac{V}{R_f} = \frac{250}{100} = 2.5 \, \text{A} \] ⦾ Step 2: Calculate the armature current under load (\( I_a \))
The total line current is the sum of the armature current and the field current:
\[ I_L = I_a + I_f \] \[ 50 = I_a + 2.5 \implies I_a = 47.5 \, \text{A} \] ⦾ Step 3: Calculate the back EMF under no-load (\( E_{b0} \))
The back EMF under no-load is:
\[ E_{b0} = V – I_{a0} R_a – V_{\text{brush}} \] \[ E_{b0} = 250 – 5 \times 0.2 – 2 \] \[ = 250 – 1 – 2 = 247 \, \text{V} \] ⦾ Step 4: Calculate the back EMF under load (\( E_{b} \))
The back EMF under load is:
\[ E_b = V – I_a R_a – V_{\text{brush}} \] \[ E_b = 250 – 47.5 \times 0.2 – 2 \] \[ = 250 – 9.5 – 2 = 238.5 \, \text{V} \] ⦾ Step 5: Relate the speeds and fluxes
The speed of a DC motor is given by:
\[ N \propto \frac{E_b}{\phi} \] At no-load:
\[ N_0 \propto \frac{E_{b0}}{\phi_0} \] Under load:
\[ N \propto \frac{E_b}{\phi} \] Given \( \phi = 0.95 \phi_0 \), the ratio of speeds is:
\[ \frac{N}{N_0} = \frac{E_b}{\phi} \times \frac{\phi_0}{E_{b0}} \] \[ = \frac{E_b}{0.95 \phi_0} \times \frac{\phi_0}{E_{b0}} = \frac{E_b}{0.95 E_{b0}} \] \[ N = N_0 \times \frac{E_b}{0.95 E_{b0}} \] Substitute the values:
\[ N = 1200 \times \frac{238.5}{0.95 \times 247} \] \[ N = 1200 \times \frac{238.5}{234.65} \] \[ \approx 1200 \times 1.016 \approx 1219.2 \, \text{rpm} \] ⦾ Final Answer: The speed of the motor under the loaded condition is closest to 1220 rpm.
Solution =
⦾ Given Data:
✔ Terminal voltage, \( V = 220 \, \text{V} \)
✔ Rated power, \( P = 10 \, \text{kW} \)
✔ Rated speed, \( N = 900 \, \text{rpm} \)
✔ Armature resistance, \( R_a = 0.02 \, \Omega \)
✔ Electromagnetic torque, \( T = 70 \, \text{Nm} \)
Rotational losses: Negligible
⦾ Step 1: Calculate the Back EMF (\( E_b \))
First, we calculate the angular speed (\(\omega\)):
\[ \omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 900}{60} = 94.25 \, \text{rad/s} \] The electromagnetic power (\( P_{\text{em}} \)) is:
\[ P_{\text{em}} = T \times \omega = 70 \times 94.25 = 6597.5 \, \text{W} \] The back EMF can be expressed as:
\[ E_b = V – I_a R_a \] ⦾ Step 2: Relate Torque and Armature Current
The electromagnetic torque is given by:
\[ T = k \phi I_a \] For a separately excited motor, \( k \phi \) is constant:
\[ k \phi = \frac{T}{I_a} \] The back EMF is also related to \( k \phi \) and speed:
\[ E_b = k \phi \omega = \left( \frac{T}{I_a} \right) \omega \] Setting the two expressions for \( E_b \) equal:
\[ \left( \frac{T}{I_a} \right) \omega = V – I_a R_a \] Substituting known values:
\[ \left( \frac{70}{I_a} \right) \times 94.25 = 220 – I_a \times 0.02 \] \[ \frac{6597.5}{I_a} = 220 – 0.02 I_a \] ⦾ Step 3: Solve the Quadratic Equation
Multiply through by \( I_a \):
\[ 6597.5 = 220 I_a – 0.02 I_a^2 \] Rearrange:
\[ 0.02 I_a^2 – 220 I_a + 6597.5 = 0 \] Divide by 0.02:
\[ I_a^2 – 11000 I_a + 329875 = 0 \] Using the quadratic formula:
\[= \frac{11000 \pm \sqrt{11000^2 – 4 \times 1 \times 329875}}{2} \] \[ I_a = \frac{11000 \pm \sqrt{121000000 – 1319500}}{2} \] \[ I_a = \frac{11000 \pm \sqrt{119680500}}{2} \] \[ I_a = \frac{11000 \pm 10940}{2} \] The two solutions are:
\[ I_a = \frac{11000 + 10940}{2} = 10970 \, \text{A} \] \[ (\text{Unphysical}) \] \[ I_a = \frac{11000 – 10940}{2} = 30 \, \text{A} \] \[ (\text{Valid}) \] ⦾ Final Answer:
The current drawn by the motor from the 220 V supply is 30 A. \[ \boxed{30 \, \text{A}} \]
Solution =
⦾ Given Data:
✔ Condition 1: 440 V, 50 Hz → Total iron loss = 2500 W
✔ Condition 2: 220 V, 25 Hz → Total iron loss = 850 W
⦾ Key Concepts:
Iron losses consist of:
✔ Hysteresis loss (Ph): Proportional to fBmn ≈ fBm1.6
✔ Eddy current loss (Pe): Proportional to f2Bm2
Since V/f is constant in both cases (440/50 = 220/25 = 8.8), the flux density Bm remains constant.
⦾ Step 1: Establish Relationships
At constant flux density:
\[ P_h = k_h f \] \[ P_e = k_e f^2 \] Total iron loss:
\[ P_{total} = P_h + P_e = k_h f + k_e f^2 \] ⦾ Step 2: Set Up Equations
For Condition 1 (50 Hz):
\[ 2500 = 50k_h + 2500k_e \quad \text{(1)} \] For Condition 2 (25 Hz):
\[ 850 = 25k_h + 625k_e \quad \text{(2)} \] ⦾ Step 3: Solve the Equations
Multiply equation (2) by 2:
\[ 1700 = 50k_h + 1250k_e \quad \text{(2a)} \] Subtract equation (2a) from equation (1):
\[ 2500 – 1700 = (50k_h – 50k_h) \] \[+ (2500k_e – 1250k_e) \] \[ 800 = 1250k_e \] \[ k_e = \frac{800}{1250} = 0.64 \] Substitute ke back into equation (2):
\[ 850 = 25k_h + 625 \times 0.64 \] \[ 850 = 25k_h + 400 \] \[ 25k_h = 450 \] \[ k_h = 18 \] ⦾ Step 4: Calculate Losses at Rated Conditions
At 50 Hz:
\[ P_h = k_h f = 18 \times 50 = 900 \, \text{W} \] \[ P_e = k_e f^2 = 0.64 \times 2500 = 1600 \, \text{W} \] ⦾ Final Answer:
At nominal voltage and frequency (440 V, 50 Hz):
Hysteresis loss = 900 W
Eddy current loss = 1600 W
⦾ The correct option is: B) 900 W and 1600 W
Solution =
✔ Self-inductance of primary winding, \( L_1 = 800 \, \text{mH} \)
✔ Self-inductance of secondary winding, \( L_2 = 600 \, \text{mH} \)
✔ Mutual inductance, \( M = 480 \, \text{mH} \)
Secondary winding is short-circuited
Primary is excited with a voltage source
⦾ To Find:
Effective inductance \( L_{\text{eff}} \) seen by the source.
⦾ Formula:
When the secondary is short-circuited, the effective inductance is given by:
\[ L_{\text{eff}} = L_1 – \frac{M^2}{L_2} \]
⦾ Substitute the values:
\[ L_{\text{eff}} = 800 – \frac{480^2}{600} \] \[ = 800 – \frac{230400}{600} \] \[ = 800 – 384 = 416 \, \text{mH} \]
⦾ Answer: Option A: 416 mH
Solution =
⦾ Given:
✔ Type of Machine: 4-pole, separately excited, wave wound DC machine
✔ Rated Voltage (V): 230 V
✔ Rated Power (P): 5 kW
✔ Speed (N): 1200 rpm
✔ Armature Resistance (Ra): Negligible
Reconnection: Armature coils reconnected to form lap winding (field circuit unchanged)
⦾ Key Concepts:
1. Wave Winding vs. Lap Winding:
Wave winding: Number of parallel paths (A) = 2
Lap winding: Number of parallel paths (A) = Number of poles (P) = 4
2. Generated Voltage (E):
The generated voltage in a DC machine is given by:
\[ E = \frac{P \cdot \phi \cdot N \cdot Z}{60 \cdot A} \]
⦾ Where:
✔ \( P \): Number of poles
✔ \( \phi \): Flux per pole
✔ \( N \): Speed in rpm
✔ \( Z \): Total number of armature conductors
✔ \( A \): Number of parallel paths
⦾ Solution:
1. Original Machine (Wave Winding):
Number of parallel paths, \( A_{wave} = 2 \)
Rated voltage, \( V_{wave} = 230 \) V
Rated current, \( I_{wave} = \frac{P}{V} = \frac{5000}{230} \approx 21.74 \) A
2. Reconnected Machine (Lap Winding):
Number of parallel paths, \( A_{lap} = 4 \)
Voltage ratio: \( \frac{E_{lap}}{E_{wave}} = \frac{A_{wave}}{A_{lap}} = \frac{2}{4} = \frac{1}{2} \)
New voltage: \( E_{lap} = \frac{1}{2} \times 230 = 115 \) V
3. Power Calculation:
Since power \( P = V \times I \), and the current capacity increases with more parallel paths:
\[ P_{lap} = V_{lap} \times I_{lap} = 115 \times \left(2 \times 21.74\right) \]\[\approx 5000 \text{ W} = 5 \text{ kW} \]
⦾ Conclusion:
The reconnected machine will have:
✔ Rated Voltage: 115 V
✔ Rated Power: 5 kW
⦾ Correct Option: B) 115 and 5
Solution =
⦾ Given:
✔ Armature resistance \( R = 0.1\, \Omega \)
✔ Rated open-circuit voltage: \( V_{\text{oc}} = 200\, \text{V} \)
✔ Operating at half speed and half field ⇒ terminal voltage \( V_f = \frac{1}{2} \cdot \frac{1}{2} \cdot 200 = 50\, \text{V} \)
✔ Capacitance \( C = 1000\, \mu\text{F} = 1000 \times 10^{-6}\, \text{F} \)
✔ Initial capacitor voltage \( V(0) = 0 \)
✔ We want \( V(t) = 25\, \text{V} \)
The voltage across a charging capacitor in an RC circuit is:
\[ V(t) = V_f \left(1 – e^{-t/RC} \right) \]
Substituting values:
\[ 25 = 50 \left(1 – e^{-t/RC} \right) \]\[\Rightarrow \frac{25}{50} = 1 – e^{-t/RC} \]\[\Rightarrow 0.5 = 1 – e^{-t/RC} \]\[\Rightarrow e^{-t/RC} = 0.5 \]
Taking natural logarithm on both sides:
\[ -\frac{t}{RC} = \ln(0.5) = -\ln(2) \]\[\Rightarrow \frac{t}{RC} = \ln(2) \Rightarrow t = RC \cdot \ln(2) \]
Now calculate:
\[ R = 0.1\, \Omega, \quad C = 1000 \times 10^{-6}\, \text{F} \]\[\Rightarrow t = (0.1)(1000 \times 10^{-6}) \cdot \ln(2) \]\[= 100 \times 10^{-6} \cdot \ln(2) \approx 69.3\, \mu\text{s} \]
⦾ Final Answer: (B) 69.3 microseconds
Solution =
⦾ Step 1: Motor Analysis
✔ Rated voltage is applied to the motor: 1.0 p.u.
✔ Armature resistance of motor: 0.02 p.u.
✔ Rated current is flowing: 1.0 p.u.
So, the voltage drop across the motor’s armature is:
Vdrop = I × Ra = 1.0 × 0.02 = 0.02 p.u.
Hence, the EMF developed by the motor (which becomes mechanical energy) is:
Emotor = V – Vdrop = 1.0 – 0.02 = 0.98 p.u.
⦾ Step 2: Generator Analysis
The generator is identical to the motor, so it has the same armature resistance: 0.02 p.u.
It is mechanically driven by the motor, so it generates the same EMF:
Egenerator = 0.98 p.u.
Rated current is also drawn from the generator: 1.0 p.u.
The voltage at the generator’s terminal (across the load) is:
Vload = Egenerator – I × Ra
Vload = 0.98 – (1.0 × 0.02) = 0.96 p.u.
⦾ Step 3: Load Resistance Calculation
The load is connected across the generator terminals and draws rated current of 1.0 p.u.
So the load resistance is:
Rload = V / I = 0.96 / 1.0 = 0.96 p.u.
⦾ Final Answer: (C) 0.96
⦾ Conclusion:
The per-unit value of the load resistance must be 0.96 p.u. to ensure that both the motor and the generator operate at rated armature current with rated terminal voltage, assuming negligible mechanical and magnetic losses.
Solution =
In a constant V/f control of an induction motor, the ratio V/f is maintained constant from 0 to base frequency to ensure that the stator flux remains constant at its rated value. This is because the stator flux is proportional to V/f. At low frequency operation, if V/f is maintained constant, the stator flux does not increase or decrease from its rated value, nor does the motor saturate.
⦾ The correct answer is: (D) At low frequency, the stator flux remains unchanged at its rated value.
Solution =
⦾ Step 1: Slots per pole per phase (SPP)
Given:
✔ Poles (P) = 4
✔ Phases (m) = 3
✔ Slots (S) = 36
Calculate:
SPP = S / (P × m) = 36 / (4 × 3) = 3
This is an integral slot winding, so it is uniformly distributed and balanced.
⦾ Step 2: Phase spread = 60° electrical
180° electrical = 1 pole pitch
Slots per pole = 36 / 4 = 9
So 60° electrical = (60 / 180) × 9 = 3 slots
This means each phase is spread over 3 slots.
⦾ Step 3: Coil span = 7 slots (short-pitch)
✔ Full pitch = 9 slots (since slots per pole = 9)
✔ Short pitch = 7 slots
✔ Therefore, the winding is short-pitched by 2 slots
⦾ Step 4: Double-layer winding behavior
In a double-layer winding:
Each slot has two coil sides (top and bottom).
One side of the coil goes in one slot (top layer), and the return side goes in another slot (bottom layer), 7 slots apart.
Due to short pitch and 60° phase spread, the two sides may fall into different phase zones.
⦾ Step 5: Determine overlapping of phases
Because:
Each phase occupies 3 slots (due to 60° spread).
Coil span = 7 slots, which is more than the 3-slot phase span.
Many coils will have top and bottom sides in different phases.
Since the pattern is symmetrical across the 36 slots, the number of such slots where top and bottom layers are in different phases is:
36 – 12 = 24 slots
⦾ Final Answer: (A) 24
(A) Synchronous reactance
(B) Magnetizing reactance
(C) Potier reactance
(D) Leakage reactance
Solution =
⦾ In the given diagram:
✔ Curve X represents the Open Circuit Characteristic (OCC) of the synchronous generator.
✔ Curve Y represents the Zero Power Factor (ZPF) characteristic at full load.
✔ Point Q lies on the ZPF curve at 1.0 p.u. voltage.
✔ Point P lies on the OCC at the same field current as Q.
The vertical distance PQ between the OCC and the ZPF characteristic at the same field current represents the Potier Reactance Voltage Drop.
⦾ Correct Answer: (C) Potier Reactance
⦾ Explanation:
✔ The Potier Reactance is an empirical value that accounts for the combined effect of leakage reactance and armature reaction.
✔ The Potier triangle method uses this distance to determine voltage regulation in alternators.
✔ The distance PQ is graphically used to separate the effects of armature reaction and leakage reactance.
Solution =
⦾ Key Concepts
✔ Synchronous Generator Operation: Field current controls excitation (reactive power)
✔ Fuel input controls real power output
✔ Power Factor Correction: Inductive load consumes reactive power (\(Q_L\))
✔ Capacitor supplies reactive power (\(Q_C\))
When \(Q_C = Q_L\), net reactive power = 0 (unity PF)
⦾ Analysis
Initial Condition (Inductive Load):
\[ \text{Complex Power } S = P + jQ_L \]
Generator supplies both real (\(P\)) and reactive (\(Q_L\)) power
Field current provides excitation for \(Q_L\)
Fuel input matches \(P\) requirement
After Adding Capacitor (Unity PF): \[ Q_{\text{net}} = Q_L – Q_C = 0 \]\[ (\text{when } Q_C = Q_L) \]
Generator only supplies real power (\(P\))
No reactive power required from generator
⦾ Impact on Operating Parameters (Parameter – Before Compensation – After Compensation)
1) Field Current – Higher (to supply \(Q_L\)) – Can be reduced
2) Fuel Input – Set for \(P\) – Remains same (still needs \(P\))
⦾ Conclusion
When the capacitor completely nullifies the inductive current:
✔ The field current must be reduced (no reactive power needed)
✔ The fuel input remains unchanged (real power demand constant)
⦾ Final Answer (D) the field current has to be reduced and fuel input left unaltered
Solution =
⦾ Given Data:
✔ Excitation voltage: \( E_f = 1.0 \, \text{p.u.} \)
✔ Terminal voltage: \( V = 1.0 \, \text{p.u.} \) (assumed)
✔ Direct-axis reactance: \( X_d = 1.2 \, \text{p.u.} \)
✔ Quadrature-axis reactance: \( X_q = 0.8 \, \text{p.u.} \)
✔ Power angle: \( \delta = 30^\circ \)
⦾ Formula for Reactive Power (\( Q \)):
\[= \frac{E_f V}{X_q} \sin \delta – \frac{V^2}{2} \left( \frac{1}{X_d} – \frac{1}{X_q} \right) \sin 2\delta \]
⦾ Step 1: Calculate the first term
\[ \text{First term} = \frac{E_f V}{X_q} \sin \delta \] Substitute the values:
\[ \text{First term} = \frac{1.0 \cdot 1.0}{0.8} \cdot \sin 30^\circ \] \[ = \frac{1.0}{0.8} \cdot 0.5 \] \[ = 1.25 \cdot 0.5 \] \[ = 0.625 \]
⦾ Step 2: Calculate the second term
\[ \text{Second term} = \frac{V^2}{2} \left( \frac{1}{X_d} – \frac{1}{X_q} \right) \sin 2\delta \] Substitute the values:
\[ \text{2 nd term} = \frac{1.0^2}{2} \left( \frac{1}{1.2} – \frac{1}{0.8} \right) \sin 60^\circ \] \[ = \frac{1.0}{2} \cdot (0.8333 – 1.25) \cdot \sin 60^\circ \] \[ = 0.5 \cdot (-0.4167) \cdot 0.866 \] \[ = -0.1807 \]
⦾ Step 3: Calculate the total reactive power
\[ Q = \text{First term} – \text{Second term} \] \[ Q = 0.625 – (-0.1807) \] \[ = 0.625 + 0.1807 \] \[ = 0.8057 \]
⦾ Approximation: The closest option is (B) 0.80 p.u.
Solution =
⦾ Given Data:
✔ Percentage resistance (\( \%R \)): \( 2\% = 0.02 \)
✔ Percentage reactance (\( \%X \)): \( 5\% = 0.05 \)
✔ Power factor (\( \cos \phi \)): \( 0.6 \) (lagging)
✔ Calculate \( \sin \phi \):
\[ \sin \phi = \sqrt{1 – \cos^2 \phi} \] \[ = \sqrt{1 – 0.6^2} \] \[ = \sqrt{1 – 0.36} \] \[ = \sqrt{0.64} \] \[ = 0.8 \]
Formula for Voltage Drop (\%):
\[ \text{V %} = \sqrt{(\%R \cdot \cos \phi)^2 + (\%X \cdot \sin \phi)^2} \]
⦾ Step 1: Calculate \( (\%R \cdot \cos \phi) \)
\[ \%R \cdot \cos \phi = 0.02 \cdot 0.6 \] \[ = 0.012 \]
⦾ Step 2: Calculate \( (\%X \cdot \sin \phi) \)
\[ \%X \cdot \sin \phi = 0.05 \cdot 0.8 \] \[ = 0.04 \]
⦾ Step 3: Calculate the Voltage Drop (\%)
\[ \text{V %} = \sqrt{(0.012)^2 + (0.04)^2} \] \[ = \sqrt{0.000144 + 0.0016} \] \[ = \sqrt{0.001744} \] \[ \approx 0.0418 \, \text{or} \, 4.18\% \]
⦾ Answer: The closest option is (D) 4.1%.
Solution =
⦾ Given:
✔ Number of poles (\(P\)) = 8
✔ Flux per pole (\(\phi\)) = 0.06 Wb
✔ Speed (\(N\)) = 250 rpm
✔ Number of coils = 32
✔ Turns per coil = 6
✔ Simplex wave-wound armature (\(A = 2\))
⦾ Step 1: Calculate Total Number of Conductors (\(Z\))
Each coil has 6 turns, and each turn has 2 conductors: \[ Z = \text{Num. of coils} \times \text{Turns per coil} \times 2 \] \[ Z = 32 \times 6 \times 2 = 384 \ \text{conductors} \]
⦾ Step 2: Determine Number of Parallel Paths (\(A\))
For a simplex wave-wound armature: \[ A = 2 \]
⦾ Step 3: Compute Induced Armature Voltage (\(E_a\))
The generated EMF in a DC machine is given by: \[ E_a = \frac{P \cdot \phi \cdot N \cdot Z}{60 \cdot A} \] Substituting the values: \[ E_a = \frac{8 \times 0.06 \times 250 \times 384}{60 \times 2} \] Simplifying step-by-step: \[ E_a = \frac{8 \times 0.06} = 0.48 \] \[ 0.48 \times 250 = 120 \] \[ 120 \times 384 = 46080 \] \[ 60 \times 2 = 120 \] \[ E_a = \frac{46080}{120} = 384 \ \text{V} \]
⦾ Final Answer:
The induced armature voltage is \(\boxed{C}\) (384 V).
Solution =
⦾ Given Data:
✔ Rotor resistance (\( R_r \)): \( 0.2 \, \Omega \)
✔ Rotor standstill reactance (\( X_r \)): \( 0.8 \, \Omega \)
✔ Slip (\( s \)): \( 4\% = 0.04 \)
Formula for Rotor Power Factor (\( \text{PF}_r \)):
\[ \text{PF}_r = \frac{R_r}{\sqrt{R_r^2 + (sX_r)^2}} \]
⦾ Step 1: Calculate \( sX_r \)
\[ sX_r = s \cdot X_r \] Substitute the values:
\[ sX_r = 0.04 \cdot 0.8 \] \[ = 0.032 \, \Omega \]
⦾ Step 2: Calculate \( \sqrt{R_r^2 + (sX_r)^2} \)
\[ \sqrt{R_r^2 + (sX_r)^2} = \sqrt{(0.2)^2 + (0.032)^2} \] \[ = \sqrt{0.04 + 0.001024} \] \[ = \sqrt{0.041024} \] \[ \approx 0.2025 \]
⦾ Step 3: Calculate \( \text{PF}_r \)
\[ \text{PF}_r = \frac{R_r}{\sqrt{R_r^2 + (sX_r)^2}} \] Substitute the values:
\[ \text{PF}_r = \frac{0.2}{0.2025} \] \[ \approx 0.98 \]
⦾ Answer: The rotor power factor is approximately (C) 0.98.
Solution =
⦾ Key Concepts
Synchronous Speed (Nₛ): \[ N_s = \frac{120f}{P} = \frac{120 \times 60}{4} = 1800 \]
⦾ Slip (s) in Induction Generator: \[ s = \frac{f_r}{f} = \frac{5}{60} = -\frac{1}{12}\]\[ (\text{negative for generator operation}) \]
⦾ Rotor Speed (Nᵣ): \[ N_r = N_s (1 – s) \]\[ = 1800 \left(1 – \left(-\frac{1}{12}\right)\right) \]\[ = 1800 \times \frac{13}{12} = 1950 \text{ RPM} \]
⦾ Calculation Steps
1. Calculate synchronous speed
\[ N_s = \frac{120 \times 60}{4} = 1800 \text{ RPM} \]
2. Determine slip
\[ s = \frac{5}{60} = -\frac{1}{12} \]
3. Compute rotor speed
\[ N_r = 1800 \left(1 – (-\frac{1}{12})\right) = 1950 \]
⦾ Conclusion
The mechanical rotor speed is:
\[ \boxed{1950 \text{ RPM}} \quad \text{(Option C)} \]
⦾ Physical Interpretation
Negative slip indicates generator operation (rotor faster than sync speed)
Rotor speed exceeds synchronous speed by: \[ 1950 – 1800 = 150 \text{ RPM} \]
This overspeed produces the 5 Hz rotor current frequency
Solution =
⦾ Key Concepts:
Torque production requires: \[ T \propto \frac{dL}{d\theta} \cdot i^2 \] where \( \frac{dL}{d\theta} \) is the inductance variation with rotor position
⦾ Torque Types Analysis : (Torque Type => Dependence on Position => Possible with Constant L? )
A. Starting Torque => Requires saliency or field interaction => Possible if other mechanisms exist
B. Synchronizing Torque => Requires position-dependent coupling => Possible through other means
C. Hysteresis Torque => Depends on magnetic hysteresis => Independent of inductance variation
D. Reluctance Torque => Requires \( \frac{dL}{d\theta} \neq 0 \) => Impossible with constant L
⦾ Detailed Explanation:
Reluctance torque fundamentally requires: \[ T_r = \frac{1}{2}i^2\frac{dL}{d\theta} \] which is zero when \( L \) is constant
Other torque types can exist without inductance variation: Hysteresis torque from material properties
Synchronizing torque from field alignment
Starting torque from auxiliary mechanisms
The condition specifies constant self-inductances, which specifically eliminates reluctance torque
⦾ Conclusion:
With position-independent inductances, the machine will definitely not develop reluctance torque, as it specifically requires inductance variation with rotor position.
⦾ Correct Answer: D
Solution =
1. Motor Configuration
The cumulative compounded long shunt motor has:
✔ Series field winding in series with the armature
✔ Shunt field winding in parallel with the armature-series field combination
✔ In cumulative compounding, series field flux adds to shunt field flux
2. Effect of Shunting the Series Field
When a resistance (Rsh) is connected in parallel with the series field:
✔ Series field current (Ise) splits: Part flows through series field (Ise,field)
✔ Part flows through shunt resistance (Ish)
✔ Armature current (Ia) remains in series with the parallel combination Effective series field current decreases
3. Torque Equation Analysis
The torque (T) in a DC motor is given by:
\[ T = k \cdot \phi \cdot I_a \]
⦾ Where:
✔ k = motor constant
✔ φ = total flux (shunt + series field)
✔ Ia = armature current
With constant torque and decreased series field flux, the armature current must increase to compensate.
4. Key Observations
✔ Series field current decreases (due to shunting)
✔ Armature current increases (to maintain torque)
✔ Shunt field current remains unchanged
✔ Motor continues to run (does not stall)
5. Evaluating the Options
A) The armature current increases – CORRECT
B) The armature current decreases – Incorrect
C) The series field current remains the same – Incorrect
D) The motor fails to run and comes to a standstill – Incorrect
⦾ Final Answer: The armature current increases
Solution =
⦾ Given Data:
✔ Starting torque ( \( T_\text{starting} \)) = 15 Nm
✔ Full-load torque ( \( T_\text{full-load} \)) = 30 Nm
⦾ Step 1: Formula for torque ratio:
\[ \text{Torque Ratio} = \frac{T_\text{starting}}{T_\text{full-load}} \]
⦾ Step 2: Substitute the values:
\[ \text{Torque Ratio} = \frac{15}{30} \]
⦾ Step 3: Simplify the expression:
\[ \text{Torque Ratio} = 0.5 \]
⦾ Final Answer:
The ratio of starting torque to full-load torque is 0.5.
Correct option: (C) 0.5
Solution =
The slip \(s\) of an induction motor is given by the formula: s = \( \frac{N_s – N_r}{N_s} \)
⦾ where:
✔ \(N_s\) is the synchronous speed
✔ \(N_r\) is the rotor speed
⦾ The synchronous speed \(N_s\) is calculated by:
\(N_s = \frac{120 \times f}{P}\)
where:
✔ \(f = 50 \, \text{Hz}\) (frequency)
✔ \(P = 4\) (number of poles)
So, for a 4-pole motor operating at 50 Hz:
\(N_s = \frac{120 \times 50}{4} = 1500 \, \text{rpm}\)
Now, the slip \(s\) is also approximated by the rotor resistance and reactance:
\(s \propto \frac{R_r}{X_r}\)
If the rotor resistance is doubled, the slip will approximately double:
\(s_{\text{new}} = 2 \times s_{\text{original}} = 2 \times 0.02 = 0.04\)
Solution =
⦾ Explanation:
The overload capacity of a transformer refers to its ability to carry more than its rated load for a short time without damage.
⦾ Key Concept:
Overload causes heating in the transformer.
The ambient temperature (temperature of surrounding air) affects how well the transformer can dissipate this heat.
✔ Higher ambient temperature → less efficient cooling → lower overload capacity
✔ Lower ambient temperature → better cooling → higher overload capacity
Hence, overload capacity is inversely proportional to ambient temperature.
⦾ Final Answer: (a) Ambient temperature
Solution =
⦾ Explanation:
A universal motor is a type of electric motor that can operate on both AC and DC supply. The key reason it works on both is due to the series connection of its armature and field windings.
⦾ Why series connection matters:
✔ When supplied with AC, both the armature current and field current reverse direction simultaneously every half-cycle.
✔ This means the direction of torque remains unchanged, allowing the motor to continue rotating in the same direction.
✔ This behavior is identical to how it works under DC supply.
If they weren’t in series:
The motor would not produce unidirectional torque under AC.
⦾ Incorrect Options:
(a) Compensating winding: Used to reduce reactance and improve performance, but not the reason for AC/DC compatibility.
(c) Permanent magnets: Not used in universal motors; they require an electromagnet field.
(d) Non-inductive rotor: Not true; rotors do have inductance.
⦾ Final Answer: (b) Armature and field windings are in series
Solution =
⦾ Key Concepts:
✔ Phase Shift in Transformers: Star-Star and Delta-Delta connections produce no phase shift
✔ Star-Delta and Delta-Star connections introduce a 30° phase shift
Zigzag connections can produce various phase shifts but are primarily used for other purposes
Standard Connections:
✔ Star (Y): Phase voltage lags line voltage by 30°
✔ Delta (Δ): Phase and line voltages are in phase
Phase Shift Mechanism:
In Star-Delta connection:
✔ Primary (Star): Phase voltage lags line voltage by 30°
✔ Secondary (Delta): No phase shift between phase and line voltages
Result: Net 30° phase shift between input and output line voltages
⦾ Analysis of Each Option:
(A) Star-Star
No phase shift between input and output line voltages
Both sides have same phase relationships
(B) Star-Delta
Introduces 30° phase shift
Star side phase voltages lag line voltages by 30°
Delta side has no additional shift
Net result: 30° phase difference between input and output line voltages
(C) Delta-Delta
No phase shift between input and output line voltages
(D) Delta-Zigzag
Zigzag connection can produce phase shifts, but not standard 30°
Primarily used for special applications, not for controlled 30° shift
⦾ Conclusion:
The Star-Delta connection is the standard configuration used to introduce a 30° phase difference between input and output line voltages in three-phase transformers.
⦾ Final Answer: \(\boxed{B}\)
(A) both machines act as generator
(B) the dc machine acts as a generator, and the induction machine acts as a motor
(C) the dc machine acts as a motor, and the induction machine acts as a generator
(D) both machines act as motors
Solution =
For a 4-pole, 50 Hz induction machine, the synchronous speed is given by:
\( N_s = \frac{120 \times f}{P} = \frac{120 \times 50}{4} = 1500 \, \text{rpm} \)
The machines are rotating at 1600 rpm, which is higher than the synchronous speed, indicating that the induction machine is operating as a generator (since it is running above synchronous speed when connected to the supply).
The DC machine is energized first and drives the system to 1600 rpm. When the induction machine is connected to the 50 Hz supply with the correct phase sequence, it will act as a generator if it is over-speeded. Since the DC machine is already driving the system, it must be acting as a motor to maintain the speed.
Thus, the DC machine acts as a motor, and the induction machine acts as a generator.
⦾ Correct answer: (C)
(A) A is stable, B is unstable
(B) A is unstable, B is stable
(C) Both are stable
(D) Both are unstable
Solution =
For a 3-phase squirrel-cage induction motor driving a mechanical load, the stability of equilibrium points on the torque-speed characteristics depends on the slope of the motor torque curve relative to the load torque curve. At equilibrium points (A and B), the motor torque equals the load torque.
✔ Point A is on the rising part of the motor torque curve (before the peak), where the slope is positive. A positive slope indicates stability because any small increase in speed reduces the motor torque below the load torque, causing the speed to decrease back to the equilibrium point.
✔ Point B is on the falling part of the motor torque curve (after the peak), where the slope is negative. A negative slope indicates instability because any small increase in speed increases the motor torque above the load torque, causing the speed to deviate further from the equilibrium point.
Thus, point A is stable, and point B is unstable.
⦾ Correct answer: (A) A is stable, B is unstable
(A) double
(B) remain same
(C) be halved
(D) become one quarter
Solution =
Mutual inductance in a transformer depends on the magnetic flux linkage between the windings, which is influenced by the cross-sectional area of the core and the length of the magnetic path. In the given transformer, the windings are initially on the vertical arms with a cross-sectional area of 20 cm², and the horizontal arms have a cross-sectional area of 10 cm².
When the windings are moved to the opposite horizontal arms, the effective cross-sectional area for the magnetic flux linkage becomes 10 cm² (the area of the horizontal arms). Since mutual inductance is proportional to the cross-sectional area of the core (assuming the permeability and number of turns remain constant), reducing the area to half (from 20 cm² to 10 cm²) will halve the mutual inductance.
⦾ Correct answer: (C) be halved
(A) 1.41 A
(B) 2 A
(C) 2.24 A
(D) 3 A
Solution =
Step 1: Reflect Secondary Current to Primary
Turns ratio is \( N_1:N_2 = 1:2 \), so the current reflected to the primary is: \[ I_2′ = I_2 \cdot \frac{N_2}{N_1} = 1 \cdot \frac{2}{1} = 2\,\text{A} \] This current is in phase with the supply voltage (since the load is purely resistive).
Step 2: Use Vector (Phasor) Addition
The magnetizing current \( I_m = 1\,\text{A} \) lags the supply voltage by 90°, so it is perpendicular to \( I_2′ \). Therefore, the primary current is: \[ I_1 = \sqrt{I_2’^2 + I_m^2} = \sqrt{2^2 + 1^2} \]\[= \sqrt{4 + 1} = \sqrt{5} \approx 2.24\,\text{A} \]
⦾ Final Answer: (C) 2.24 A
Solution =
⦾ Given Parameters:
Voltage: \( V = 440 \, \text{V} \)
Poles: \( P = 6 \)
Frequency: \( f = 50 \, \text{Hz} \)
Slip: \( s = 5\% = 0.05 \)
⦾ Step 1: Calculate Synchronous Speed (\(N_s\))
The synchronous speed of the stator’s rotating magnetic field is:
\[ N_s = \frac{120 \times f}{P} \]\[= \frac{120 \times 50}{6} = 1000 \, \text{rpm} \] ⦾ Step 2: Calculate Rotor Speed (\(N_r\))
The actual rotor speed accounting for slip:
\[ N_r = N_s (1 – s) \]\[= 1000 \times (1 – 0.05) = 950 \, \text{rpm} \] ⦾ Step 3: Relative Speed Between Fields
The stator and rotor magnetic fields rotate synchronously when viewed from the stator frame, so their relative speed is:
Relative speed between fields = 0 rpm
⦾ Step 4: Rotor Speed Relative to Stator Field
The slip speed represents how much slower the rotor is compared to the stator field:
\[ \text{Slip speed} = s \times N_s \]\[ = 0.05 \times 1000 = 50 \, \text{rpm} \] Since the rotor is slower, we express this as:
Rotor speed relative to stator field = -50 rpm
⦾ Analysis of Options:
The correct values are:
✔ Stator field relative to rotor field: \(0 \, \text{rpm}\)
✔ Rotor relative to stator field: \(-50 \, \text{rpm}\)
⦾ Final Answer: \(\boxed{A}\)
Solution =
⦾ Given:
✔ Supply voltage, \( V = 220 \, \text{V} \)
✔ Initial speed, \( N_1 = 1440 \, \text{rpm} \)
✔ Armature resistance, \( R_a = 1.0 \, \Omega \)
✔ Armature current, \( I_{a1} = 10 \, \text{A} \)
✔ Excitation reduced by 10%, so new flux \( \phi_2 = 0.9 \phi_1 \)
⦾ Step 1: Calculate initial back EMF (\( E_{b1} \))
\[ E_{b1} = V – I_{a1} R_a \]\[= 220 – 10 \times 1.0 = 210 \, \text{V} \] ⦾ Step 2: Relate speed, flux, and back EMF
Speed is proportional to back EMF and inversely proportional to flux:
\[ N \propto \frac{E_b}{\phi} \] For the same speed (\( N_1 = N_2 \)):
\[ \frac{E_{b1}}{\phi_1} = \frac{E_{b2}}{\phi_2} \] Given \( \phi_2 = 0.9 \phi_1 \):
\[ \frac{210}{\phi_1} = \frac{E_{b2}}{0.9 \phi_1} \] \[ E_{b2} = 210 \times 0.9 = 189 \, \text{V} \] ⦾ Step 3: Determine new armature current (\( I_{a2} \))
Torque is proportional to flux and armature current:
\[ T \propto \phi I_a \] For the same torque:
\[ \phi_1 I_{a1} = \phi_2 I_{a2} \] \[ \phi_1 \times 10 = 0.9 \phi_1 \times I_{a2} \] \[ I_{a2} = \frac{10}{0.9} \approx 11.11 \, \text{A} \] ⦾ Step 4: Calculate total resistance (\( R_{total} \))
Back EMF equation after adding resistance \( R \):
\[ E_{b2} = V – I_{a2} (R_a + R) \] \[ 189 = 220 – 11.11 (1.0 + R) \] \[ 11.11 (1.0 + R) = 31 \] \[ 1.0 + R = \frac{31}{11.11} \approx 2.79 \] \[ R = 2.79 – 1.0 = 1.79 \, \Omega \] ⦾ Final Answer: The extra resistance to be added is \(\boxed{1.79 \, \Omega}\), which corresponds to option (A).
Solution =
⦾ Given Parameters:
✔ Number of poles, \( P = 4 \)
✔ Rated frequency, \( f_1 = 50 \, \text{Hz} \)
✔ Rated voltage, \( V_1 = 400 \, \text{V} \)
✔ Rated speed, \( N_1 = 1440 \, \text{rpm} \)
✔ New frequency, \( f_2 = 30 \, \text{Hz} \)
Constant V/f ratio maintained
Constant load torque
⦾ Step 1: Calculate Synchronous Speed at 50 Hz
\[ N_{s1} = \frac{120 \times f_1}{P} \]\[= \frac{120 \times 50}{4} = 1500 \, \text{rpm} \] ⦾ Step 2: Determine Slip at Rated Condition
\[ s_1 = \frac{N_{s1} – N_1}{N_{s1}} \]\[= \frac{1500 – 1440}{1500} = 0.04 \, \text{(4%)} \] ⦾ Step 3: Calculate Synchronous Speed at 30 Hz
\[ N_{s2} = \frac{120 \times f_2}{P} \]\[= \frac{120 \times 30}{4} = 900 \, \text{rpm} \] ⦾ Step 4: Determine New Slip
For constant torque operation with constant V/f ratio, the slip speed remains constant:
\[ \text{Slip speed} = N_{s1} \times s_1 \]\[= 1500 \times 0.04 = 60 \, \text{rpm} \] This same slip speed is maintained at 30 Hz:
\[ s_2 = \frac{\text{Slip speed}}{N_{s2}} \]\[= \frac{60}{900} = 0.0667 \, \text{(6.67%)} \] ⦾ Step 5: Calculate Actual Motor Speed at 30 Hz
\[ N_2 = N_{s2} \times (1 – s_2) \]\[= 900 \times (1 – 0.0667) \]\[= 900 \times 0.9333 = 840 \, \text{rpm} \] The calculated speed at 30 Hz is 840 rpm, which corresponds to option C.
Solution =
For transformers to share load in proportion to their kVA ratings, their per-unit impedances must be inversely proportional to their kVA ratings.
⦾ Given:
✔ Transformer 1: \( S_1 = 500 \, \text{kVA}, \, Z_{1(pu)} = 0.05 \, \text{pu} \)
✔ Transformer 2: \( S_2 = 250 \, \text{kVA}, \, Z_{2(pu)} = ? \)
⦾ Condition for Proportional Load Sharing:
The load sharing ratio equals the kVA rating ratio when:
\[ \frac{S_1}{S_2} = \frac{Z_{2(pu)}}{Z_{1(pu)}} \] Rearranging to find \( Z_{2(pu)} \):
\[ Z_{2(pu)} = Z_{1(pu)} \times \frac{S_1}{S_2} \] ⦾ Calculation:
\[ Z_{2(pu)} = 0.05 \times \frac{500}{250} \]\[= 0.05 \times 2 = 0.10 \, \text{pu} \] ⦾ Conclusion:
The second transformer must have a per-unit leakage impedance of 0.10 pu to ensure proportional load sharing according to their kVA ratings.
⦾ Final Answer: \(\boxed{B}\)
Solution =
⦾ Correct Answer: (B) leading
⦾ Explanation:
✔ The motor is initially running at full load and unity power factor, meaning the field current is adjusted to balance the reactive power (neither absorbing nor supplying).
✔ When the shaft load is reduced by half, the motor draws less real power from the supply.
✔ Since the field current is held constant, the internal excitation remains unchanged.
✔ This unchanged excitation now becomes overexcitation for the reduced load, causing the motor to supply reactive power to the system.
Hence, the motor operates at a leading power factor.
This is a well-known behavior of synchronous machines where excitation is kept constant but mechanical load changes.
(A)
(B) 
(C)
(D) 
Solution =
⦾ Correct Answer: (C)
An air-core transformer has a linear B-H curve because it does not contain a ferromagnetic core. This means:
The magnetic flux \( \Phi(t) \) is proportional to the magnetizing current \( i(t) \).
For a sinusoidal voltage input: \[ v(t) = V_m \sin(\omega t) \] the flux is: \[ \Phi(t) \propto \int v(t)\,dt = -\frac{V_m}{\omega} \cos(\omega t) \]
Since \( \Phi(t) \propto i(t) \), the magnetizing current is also sinusoidal: \[ i(t) \propto -\cos(\omega t) \]
Hence, the magnetizing current waveform is perfectly sinusoidal.
⦾ Final Answer: (C)
(A) \( \frac{125}{100} \text{ and } \frac{80}{100} \)
(B) \( \frac{100}{100} \text{ and } \frac{64}{100} \)
(C) \( \frac{100}{100} \text{ and } \frac{100}{100} \)
(D) \( \frac{80}{100} \text{ and } \frac{80}{100} \)
Solution =
⦾ Given:
Transformer turns ratio: \( 1:1.25 \) (Primary = WX, Secondary = YZ)
Attenuator with a factor of 0.8
⦾ Case 1: Input at WX, output open-circuit at YZ
✔ Input voltage: \( V_{WX1} = 100\,\text{V} \)
✔ Transformer output: \( V’_{YZ1} = 100 \times 1.25 = 125\,\text{V} \)
✔ After attenuator: \( V_{YZ1} = 125 \times 0.8 = 100\,\text{V} \)
⦾ Case 2: Input at YZ, output open-circuit at WX
✔ Input voltage: \( V_{YZ2} = 100\,\text{V} \)
✔ After attenuator (before transformer): \( V’_{YZ2} = 100 \times 0.8 = 80\,\text{V} \)
✔ Transformer output: \( V_{WX2} = 80 \times \frac{1}{1.25} = 64\,\text{V} \)
⦾ Calculated Ratios:
\( \frac{V_{YZ1}}{V_{WX1}} = \frac{100}{100} = 1 \)
\( \frac{V_{WX2}}{V_{YZ2}} = \frac{64}{100} = 0.64 \)
⦾ Option: (B) \( \frac{V_{YZ1}}{V_{WX1}} = \frac{100}{100} \), and \( \frac{V_{WX2}}{V_{YZ2}} = \frac{64}{100} \)
Solution =
Locked rotor current is approximately proportional to the ratio \( \frac{V}{f} \):
\[ I_2 = I_1 \times \frac{\left( \frac{V_2}{f_2} \right)}{\left( \frac{V_1}{f_1} \right)} = 50 \times \frac{236 / 57}{230 / 50} \]
\[ I_2 = 50 \times \frac{236 \times 50}{230 \times 57} = 50 \times \frac{11800}{13110} \]\[\approx 50 \times 0.9008 \approx 45.04\,\text{A} \]
⦾ Answer: (B) 45.0 A
Solution =
⦾ Step 1: Find Full-Load Copper Loss (\( P_{cu,\text{rated}} \))
Copper loss varies with the square of the current. Given that at 90% of rated current (\( 0.9I_{\text{rated}} \)), the copper loss is 81 W, we can find the full-load copper loss:
\[ P_{cu,\text{rated}} = \left( \frac{I_{\text{rated}}}{0.9I_{\text{rated}}} \right)^2 \times 81 \]\[= \left( \frac{1}{0.9} \right)^2 \times 81 \] \[ P_{cu,\text{rated}} = \frac{1}{0.81} \times 81 = 100 \text{ W} \] ⦾ Step 2: Condition for Maximum Efficiency
The transformer achieves maximum efficiency when copper loss equals iron loss:
\[ P_{cu} = P_i \] Let \( x \) be the fraction of the rated current where this occurs. The copper loss at this fraction is:
\[ P_{cu} = x^2 \times P_{cu,\text{rated}} \] Setting this equal to the iron loss (\( P_i = 64 \text{ W} \)):
\[ x^2 \times 100 = 64 \] \[ x^2 = \frac{64}{100} = 0.64 \] \[ x = \sqrt{0.64} = 0.8 \] Therefore, the transformer operates at maximum efficiency when the current is 80% of the rated current.
⦾ Final Answer
The correct option is \(\boxed{C}\) (80.0% of the rated current).
Solution =
⦾ Given Data:
✔ Slip \(s = 0.04\) (4%)
✔ Rotor resistance per phase \(R_r = 0.5 \, \Omega\)
✔ Synchronous speed \(N_s = 1500 \, \text{rpm}\)
✔ Efficiency \(\eta = 90\%\) (or 0.9)
✔ Power factor = lagging (not directly needed for this calculation)
⦾ The efficiency equation is:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\)
Rearranging to find \(P_{\text{in}}\):
\(P_{\text{in}} = \frac{P_{\text{out}}}{\eta}\)
Substituting the known values:
\(P_{\text{in}} = \frac{P_{\text{out}}}{0.9}\)
Next, we calculate the power developed by the rotor (\(P_{\text{rotor}}\)):
\(P_{\text{rotor}} = P_{\text{input}} \times s\)
\(P_{\text{rotor}} = 12.5 \, \text{kW} \times 0.04 = 0.5 \, \text{kW}\)
The power developed by the rotor is 12.5 kW, which corresponds to option (C).
Solution =
⦾ Given Data:
✔ Rated voltage: \( V = 220 \, \text{V} \)
✔ Rated power: \( P = 15 \, \text{kW} \)
✔ Rated speed: \( N_1 = 1000 \, \text{rpm} \)
✔ Rated line current: \( I_{L1} = 68 \, \text{A} \)
✔ Rated field current: \( I_{f1} = 2.2 \, \text{A} \)
✔ Armature resistance: \( R_a = 0.25 \, \Omega \)
✔ Armature current at rated conditions:
\[ I_{a1} = I_{L1} – I_{f1} = 68 – 2.2 = 65.8 \, \text{A} \]
⦾ New Operating Conditions:
✔ New speed: \( N_2 = 1600 \, \text{rpm} \)
✔ New line current: \( I_{L2} = 52.8 \, \text{A} \)
✔ New field current: \( I_{f2} = 1.8 \, \text{A} \)
✔ Armature current at new conditions:
\[ I_{a2} = I_{L2} – I_{f2} = 52.8 – 1.8 = 51.0 \, \text{A} \]
⦾ Back EMF Calculation:
\[ E_{b1} = V – I_{a1} R_a = 220 – 65.8 \times 0.25 \]\[= 203.55 \, \text{V} \]
\[ E_{b2} = V – I_{a2} R_a = 220 – 51.0 \times 0.25 \]\[= 207.25 \, \text{V} \]
⦾ DC Motor Speed-Flux Relationship:
\[ N \propto \frac{E_b}{\phi} \]
Taking the ratio of speeds:
\[ \frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \cdot \frac{\phi_1}{\phi_2} \]
Substituting values:
\[ \frac{1600}{1000} = \frac{207.25}{203.55} \cdot \frac{\phi_1}{\phi_2} \]
\[ 1.6 = 1.0182 \cdot \frac{\phi_1}{\phi_2} \Rightarrow \frac{\phi_1}{\phi_2} \]\[= \frac{1.6}{1.0182} = 1.5718 \]
Thus,
\[ \frac{\phi_2}{\phi_1} = \frac{1}{1.5718} = 0.6364 \]
⦾ Percentage Change in Flux:
\[ \text{Flux decrease} = (1 – 0.6364) \times 100 \]\[= 36.36\% \]
⦾ Final Answer: (D) 36.36% decrease
Solution =
⦾ Correct Answer: (C) First decreases and then increases steeply
The motor is operating at no load and is connected to an infinite bus, meaning a constant voltage and frequency source.
In this setup:
✔ The motor draws current based on the field excitation.
✔ At normal excitation, the motor draws only a small current (to overcome internal losses).
✔ As the field excitation is reduced:
1) The internal generated voltage \( E_f \) decreases.
2) The phase difference between bus voltage and generated voltage increases.
3) The motor draws more lagging reactive current.
✔ At zero excitation, the motor behaves like an induction motor and draws maximum lagging current.
✔ As the excitation is increased in reverse direction:
1) The internally generated voltage \( E_f \) now reverses direction.
2) Initially, the vector sum of bus voltage and reversed EMF results in reduced total voltage drop across the synchronous reactance.
3) Therefore, armature current first decreases.
4) As the reverse excitation increases further, the reversed EMF becomes larger, causing armature current to increase steeply again (this time leading).
⦾ In Summary:
✔ At normal excitation → Small armature current.
✔ Reducing excitation → Armature current increases (lagging).
✔ At zero excitation → Armature current is at maximum.
✔ Reverse excitation → Current first decreases, then increases steeply again (leading).
This behavior forms the well-known “V-curve” of synchronous motors.
Solution =
⦾ Given:
✔ 4-pole induction motor
✔ Supply frequency, \( f = 50 \, \text{Hz} \)
✔ Motor speed, \( N_r = 1440 \, \text{rpm} \)
✔ Slightly unbalanced three-phase source
⦾ Step 1: Calculate Synchronous Speed (\( N_s \))
The synchronous speed of an induction motor is given by:
\[ N_s = \frac{120 \times f}{P} \] Where \( P = 4 \) (number of poles). \[ N_s = \frac{120 \times 50}{4} = 1500 \, \text{rpm} \]
⦾ Step 2: Calculate Slip (\( s \))
The slip is the relative difference between synchronous speed and rotor speed: \[ s = \frac{N_s – N_r}{N_s} \] \[ s = \frac{1500 – 1440}{1500} = 0.04 \, \text{(or 4\%)} \]
⦾ Step 3: Negative Sequence Frequency Analysis
✔ An unbalanced supply introduces a negative sequence voltage that rotates in the opposite direction to the positive sequence (normal rotation).
✔ The negative sequence voltage has the same frequency (\( 50 \, \text{Hz} \)), but its effect on the rotor is different because it rotates backward.
⦾ Relative Speed of Negative Sequence Field
The rotor perceives the negative sequence field as rotating at \( -N_s \) (opposite direction). The relative speed between the rotor and the negative sequence field is:
\[ N_{\text{negative}} = N_s + N_r = 1500 + 1440 \]\[= 2940 \, \text{rpm} \] Slip for Negative Sequence (\( s_{\text{negative}} \))
The slip for the negative sequence is:
\[ s_{\text{negative}} = \frac{N_s + N_r}{N_s} \]\[= \frac{1500 + 1440}{1500} = 1.96 \] ⦾ Frequency of Induced Negative Sequence Current
The frequency of the induced negative sequence current in the rotor is: \[ f_{\text{negative rotor}} = s_{\text{negative}} \times f = 1.96 \times 50 \]\[= 98 \, \text{Hz} \]
The electrical frequency of the induced negative sequence current in the rotor is:
\[ \boxed{98} \] Therefore, the correct answer is Option (B) 98.
Solution =
⦾ Given Data:
✔ Initial speed (\(N\)) = 500 rpm
✔ Terminal voltage (\(V\)) = 250 V
✔ Starting resistance (\(R\)) = 0.2 Ω
✔ Load torque doubles.
⦾ Step 1: Key Relationships:
✔ Torque (\(T\)) is proportional to armature current (\(I_a\)) and flux (\(\phi\)).
\[ T \propto \phi I_a \] – Back EMF (\(E\)) is proportional to flux (\(\phi\)) and speed (\(N\)).
\[ E \propto \phi N \]
⦾ Step 2: When torque doubles:
Armature current doubles:
\[ I_a’ = 2I_a \] Flux also doubles because \(\phi \propto I_a\):
\[ \phi’ = 2\phi \]
⦾ Step 3: Relationship between speed and back EMF:
\[ N \propto \frac{E}{\phi} \] Since \(\phi\) doubles:
\[ N’ = \frac{N}{2} \]
⦾ Step 4: Calculate the new speed:
Substitute the initial speed:
\[ N’ = \frac{500}{2} \] \[ N’ = 250 \text{ rpm} \]
⦾ Final Answer:
The new speed of the motor is 250 rpm.
⦾ Correct option: (A) 250 rpm.
Solution =
⦾ The power \(P\) of the motor is given by: \(P = V \times I\)
where:
✔ \(P = 10 \, \text{kW} = 10,000 \, \text{W}\)
✔ \(V = 250 \, \text{V}\) (terminal voltage)
✔ \(I = ?\) (armature current)
⦾ The voltage equation for the motor is: \(V_{\text{terminal}} = E + I_a \times R_a\)
where:
✔ \(V_{\text{terminal}} = 250 \, \text{V}\)
✔ \(E = 240 \, \text{V}\) (back EMF)
✔ \(R_a = 0.2 \, \Omega\) (armature resistance)
⦾ Solving for the armature current \(I_a\):
\(I_a = \frac{V_{\text{terminal}} – E}{R_a}\)
⦾ Substituting the values:
\(I_a = \frac{250 – 240}{0.2} = \frac{10}{0.2} = 50 \, \text{B}\)
Solution =
⦾ Given Data:
✔ Rated power = \( 20 \, \text{kVA} = 20,000 \, \text{VA} \)
✔ Magnetizing current = \( 1\% \) of the rated current
✔ Core loss = \( 300 \, \text{W} \)
⦾ Step 1: Calculate the rated current:
For a single-phase transformer:
\[ I_\text{rated} = \frac{\text{Rated Power}}{\text{Rated Voltage}} \] The magnetizing current is:
\[ I_\text{mag} = 0.01 \cdot I_\text{rated} \]
⦾ Step 2: Calculate apparent power at no-load:
\[ \text{A.P. at No-Load} = V_\text{rated} \cdot I_\text{mag} \] Substitute \( I_\text{mag} \):
\[ \text{A.P. at No-Load} = 0.01 \cdot \text{Rated Power} \] \[ \text{A.P. at No-Load} = 0.01 \cdot 20,000 \] \[ \text{Apparent Power at No-Load} = 200 \, \text{VA} \]
⦾ Step 3: Calculate the no-load power factor:
\[ \text{PF} = \frac{\text{Core Loss}}{\text{Apparent Power at No-Load}} \] Substitute the values:
\[ \text{PF} = \frac{300}{200} \] \[ \text{PF} = 0.15 \]
⦾ Final Answer:
The no-load power factor is approximately 0.15.
⦾ Correct option: (A) 0.15.
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