← Junior Engineer
Past Questions — 2024
Which of the following is used to measure the speed of a rotating shaft?
A)
Thermocouple
B)
Pressure gauge
C)
Flow meter
D)
Tachometer
Step-by-step reasoning:
1. The question asks for an instrument used to measure the speed of a rotating shaft (rotational speed), typically measured in Revolutions Per Minute (RPM).
2. Analysis of options:
- (a) Thermocouple: This is a sensor used to measure temperature based on the Seebeck effect. It cannot measure rotational speed.
- (b) Pressure gauge: This is an instrument used to measure the pressure of a gas or liquid in a system. It cannot measure rotational speed.
- (c) Flow meter: This is a device used to measure the volume or mass of a fluid (liquid or gas) passing through a pipe. It cannot measure the speed of a rotating shaft.
- (d) Tachometer: This is a specialized instrument designed specifically to measure the rotation speed of a shaft or disk, as in a motor or engine. Tachometers can be mechanical, electrical (using generators), or optical (using lasers/stroboscopes).
3. Conclusion: The tachometer is the only instrument listed that is designed for measuring rotational speed.
What environmental factor can significantly impact the performance of lightning arresters?
A)
Humidity
B)
Ambient temperature
C)
Wind speed
D)
Solar
Lightning arresters, particularly those using Metal Oxide Varistors (MOV), are highly sensitive to environmental conditions. Humidity is a critical factor because moisture can penetrate the housing or the seals of the arrester. This leads to several issues: 1) It can cause leakage currents to increase, potentially leading to thermal runaway. 2) Moisture combined with pollutants can create conductive paths on the surface of the insulator (tracking), which may lead to flashovers. 3) Internal moisture can degrade the non-linear resistive elements, reducing the arrester's ability to clamp voltage effectively. While ambient temperature (b) affects the leakage current slightly, and solar radiation (d) can cause thermal stress, humidity is the most significant factor regarding the long-term degradation and immediate failure risk of the insulation and the resistive element. Wind speed (c) is a mechanical factor rather than an electrical performance factor.
The process of turning off a thyristor is called -
A)
Mounting
B)
Commutation
C)
Free-Wheeling
D)
Modulation
A thyristor (such as an SCR) is a semi-controlled device, meaning it can be turned ON by a gate pulse, but it cannot be turned OFF by the gate. To turn off a thyristor, the anode current must be reduced below the holding current level, or the polarity of the voltage across the device must be reversed. This specific process of turning off a thyristor is technically termed 'Commutation'.
Analysis of other options:
- Mounting: Refers to the physical installation of the device on a heat sink or PCB.
- Free-Wheeling: Refers to a circuit (usually involving a diode) that allows current to continue flowing in an inductive load after the main switch is turned off, preventing voltage spikes.
- Modulation: Refers to the process of varying a property of a periodic waveform (like in PWM - Pulse Width Modulation) to control the average power delivered to a load.
In a single phase to single phase cyclo converter, the magnitude of harmonic components are quite large. How can they be reduced?
A)
By using a chopper circuit
B)
By using an RC oscillator
C)
By using a three phase input supply
D)
By adding alternator to the input
A cycloconverter is a power electronics device that converts AC power of one frequency to AC power of another frequency. In a single-phase to single-phase cycloconverter, the output waveform is composed of segments of the input sine wave, which results in significant harmonic distortion and high ripple content. To reduce these harmonic components and improve the quality of the output waveform (making it more sinusoidal), a three-phase input supply is used. In a three-phase to single-phase (or three-phase to three-phase) cycloconverter, the overlapping of phases and the interleaving of switching pulses significantly reduce the magnitude of the lower-order harmonics and the ripple, leading to a smoother output voltage and current.
Analysis of other options:
- (a) A chopper circuit is a DC-to-DC converter and cannot be used to reduce harmonics in an AC-to-AC cycloconverter.
- (b) An RC oscillator is used for signal generation at low power levels, not for power conversion or harmonic filtering in industrial power electronics.
- (d) Adding an alternator to the input simply provides the AC source; it does not inherently change the switching topology or reduce the harmonics generated by the cycloconverter's switching process.
Short-circuit test of a transformer helps us its-
A)
iron loss.
B)
full-load copper loss.
C)
copper loss at no-load.
D)
copper loss at half-load.
In a transformer, two primary tests are conducted to determine losses: the Open-Circuit (OC) test and the Short-Circuit (SC) test.
1. **Short-Circuit (SC) Test**: In this test, the secondary winding is short-circuited, and a reduced voltage is applied to the primary side until the rated full-load current flows through the windings. Since the applied voltage is very low (typically 2-10% of rated voltage), the magnetic flux in the core is negligible, making the iron losses (hysteresis and eddy current) insignificant. Therefore, the power measured by the wattmeter during this test represents the I²R losses, which are the copper losses at full load.
2. **Open-Circuit (OC) Test**: This test is conducted by leaving the secondary open and applying rated voltage to the primary. This is used to determine the iron losses (core losses).
Evaluating the options:
- Option (a) is incorrect because iron loss is determined by the OC test.
- Option (b) is correct because the SC test is specifically designed to find the copper loss at rated full-load current.
- Option (c) is incorrect because copper loss at no-load is effectively zero.
- Option (d) is incorrect because while copper loss at half-load can be calculated from the full-load copper loss (using the square of the load ratio), the test itself is performed to find the full-load value.
The great advantage of the double squirrel-cage induction motor over single cage rotor is that its -
A)
efficiency is higher.
B)
power factor is higher.
C)
slip is larger.
D)
starting current is lower.
A double squirrel-cage induction motor is designed specifically to provide high starting torque while maintaining a low starting current. It achieves this by having two sets of rotor bars:
1. Outer Cage: Designed with high resistance to limit the starting current and provide a high starting torque (as starting torque is proportional to rotor resistance up to a certain point).
2. Inner Cage: Designed with low resistance to ensure high efficiency and a high power factor during normal running conditions.
At the moment of starting, the slip is 1, and the high-resistance outer cage dominates, effectively reducing the inrush current. As the motor accelerates and the slip decreases, the reactance of the outer cage increases, and the current shifts to the low-resistance inner cage for efficient operation.
Comparing the options:
- (a) Efficiency: The double cage is not necessarily more efficient than a single cage; it is designed for starting characteristics.
- (b) Power Factor: The power factor is primarily determined by the inner cage during running, similar to a standard single cage.
- (c) Slip: Slip is a function of load and motor design, not a 'benefit' provided by the double cage structure.
- (d) Starting Current: This is the primary advantage, as the high resistance of the outer cage limits the initial current surge.
A Silicon Controlled Rectifier (SCR) is a -
A)
unijunction device
B)
device with three junction
C)
device with four junction
D)
device with two junction
An SCR (Silicon Controlled Rectifier) is a PNPN semiconductor device, which means it has three junctions (J1, J2, and J3). Therefore, the correct option is b: device with three junction. Option a is incorrect because a unijunction device has only one junction. Option d is incorrect because a two-junction device would not describe the PNPN structure of an SCR. Option c is incorrect because an SCR does not have four junctions; four-junction devices are different semiconductor structures.
A wind turbine extracts maximum power from wind, when the downstream wind speed reduces to -
A)
one third that of upstream
B)
half that of upstream wind
C)
two third that of upstream
D)
zero
This question is based on Betz's Law (Betz Limit), which determines the maximum power that can be extracted from the wind.
1. Let $V_1$ be the upstream wind speed (undisturbed) and $V_2$ be the downstream wind speed (after passing through the turbine).
2. According to the actuator disk theory, the wind speed at the turbine rotor ($V$) is the average of the upstream and downstream speeds: $V = (V_1 + V_2) / 2$.
3. The power extracted ($P$) is given by the formula: $P = \frac{1}{2} \rho A (V_1^2 - V_2^2) V$, where $
ho$ is air density and $A$ is the swept area.
4. Substituting $V$, we get: $P = \frac{1}{4} \rho A (V_1^2 - V_2^2)(V_1 + V_2)$.
5. To find the maximum power, we differentiate $P$ with respect to $V_2$ and set it to zero ($rac{dP}{dV_2} = 0$).
6. Solving this derivative leads to the condition: $V_2 = \frac{1}{3} V_1$.
Therefore, maximum power is extracted when the downstream wind speed is one-third of the upstream wind speed. This results in the Betz limit, where the maximum theoretical efficiency (coefficient of power, $C_p$) is $16/27 \approx 59.3\%$.
In a multistage sequence control of voltage controller, the presence of harmonics in the output voltage depends on the -
A)
supply voltage
B)
number of thyristors
C)
firing angle
D)
magnitude of voltage variation
In a voltage controller (AC voltage regulator), the output voltage is controlled by varying the firing angle (α) of the thyristors. When the thyristors are triggered at an angle other than 0° or 180°, the output voltage waveform becomes non-sinusoidal (chopped). According to Fourier analysis, any non-sinusoidal periodic waveform consists of a fundamental frequency component and various harmonic components. The degree of distortion—and thus the magnitude and presence of these harmonics—is directly dependent on the firing angle. If the firing angle is 0°, the waveform is a full sine wave (no harmonics); as the firing angle increases, the 'cut' in the waveform increases, leading to higher harmonic content.
Why other options are incorrect:
- a. Supply voltage: This affects the amplitude of the output voltage but not the shape (harmonic content) of the waveform.
- b. Number of thyristors: The number of thyristors determines the type of controller (e.g., half-wave vs full-wave), but for a given configuration, the harmonics are governed by the switching point (firing angle).
- d. Magnitude of voltage variation: This is a result of changing the firing angle, not the cause of the harmonics themselves.
When a fluorescent lamp shows a dense blocking at each end, it could possibly mean that the -
A)
gas filled in lamp was wrong
B)
lamp was started infrequently
C)
starter is working improperly
D)
lamp is new
In a new fluorescent lamp, the phosphor coating on the inner surface of the glass tube is uniform and pristine. When the lamp is first operated, the electrodes (filaments) at each end are heated, and the discharge begins. In brand new lamps, there is often a visible 'dark ring' or a dense appearance at the ends where the electrodes are located, as the plasma discharge is establishing itself and the phosphor has not yet undergone the initial 'burn-in' or aging process. As the lamp ages, the phosphor near the ends tends to degrade or 'burn off' due to the higher intensity of sputtering from the electrodes, making the ends appear lighter or more transparent over time. Therefore, a dense appearance or distinct blocking at the ends is a characteristic typically associated with a new lamp.
Analysis of other options:
- Option (a): If the gas mixture were wrong, the lamp would likely fail to strike an arc or would operate with an incorrect color/flicker, rather than showing specific 'blocking' at the ends.
- Option (b): Infrequent starting does not cause dense blocking; rather, excessive starting/stopping can shorten the life of the electrodes.
- Option (c): An improper starter would result in the lamp failing to start or flickering continuously, not a visual change in the phosphor density at the ends.
An ideal transformer is one which-
A)
has no losses and leakage reactance.
B)
does not work.
C)
have same primary and secondary turns.
D)
have good cooling system.
An ideal transformer is a theoretical model used in electrical engineering to simplify the analysis of transformer circuits. It is defined by the following characteristics:
1. No Losses: It has zero winding resistance (no copper losses/I²R losses) and zero core losses (no hysteresis or eddy current losses).
2. No Leakage Reactance: It is assumed that 100% of the magnetic flux produced by the primary winding links with the secondary winding (perfect magnetic coupling), meaning there is no leakage flux.
3. Infinite Permeability: The core is assumed to have infinite permeability, meaning no magnetizing current is required to establish the flux.
Evaluating the options:
- Option (a) is correct as it describes the fundamental assumptions of an ideal transformer.
- Option (b) is incorrect because an ideal transformer is a model of a perfectly efficient device, not a non-functional one.
- Option (c) describes a '1:1 transformer' or an isolation transformer, but not necessarily an 'ideal' one (an ideal transformer can be step-up or step-down).
- Option (d) refers to thermal management in real-world transformers; an ideal transformer produces no heat (no losses), so it doesn't require a cooling system.
Cores of large transformers are built upto nearly circular cross-section in order to reduce-
A)
leakage reactance.
B)
iron losses.
C)
eddy current loss.
D)
copper losses.
In large transformers, the core is designed with a nearly circular cross-section (often achieved through a stepped-core construction) to optimize the distribution of magnetic flux.
1. **Leakage Reactance**: Leakage reactance is caused by the magnetic flux that does not link both the primary and secondary windings (leakage flux). A circular or stepped-core design ensures that the magnetic path is as uniform as possible and minimizes the distance between the windings and the core. This reduces the amount of leakage flux, thereby reducing the leakage reactance of the transformer.
2. **Iron Losses (Option b)**: Iron losses (hysteresis and eddy current losses) depend primarily on the grade of the silicon steel used, the thickness of the laminations, and the flux density, not the overall geometric shape of the core cross-section.
3. **Eddy Current Loss (Option c)**: This is reduced by using thin, insulated laminations of the core, not by changing the cross-sectional shape to a circle.
4. **Copper Losses (Option d)**: Copper losses ($I^2R$) depend on the resistance of the windings, which is a function of the conductor material, length, and cross-sectional area of the wire, not the shape of the magnetic core.
Therefore, the primary objective of adopting a nearly circular cross-section in large transformers is to minimize leakage reactance.
Input ripple of SMPS is -
A)
high
B)
nil
C)
very low
D)
low
In a Switched-Mode Power Supply (SMPS), the input stage typically consists of a bridge rectifier followed by a large bulk filter capacitor. This capacitor acts as a reservoir, smoothing out the rectified AC voltage to provide a steady DC bus for the high-frequency switching converter. Because the switching frequency of an SMPS is very high (typically in the kHz or MHz range), the input filter can be highly effective, and the energy storage capacity of the input capacitor ensures that the voltage ripple seen by the switching stage is kept to a minimum. Compared to linear power supplies, the high-frequency nature of the conversion process allows for more efficient filtering, resulting in 'very low' input ripple relative to the operational requirements of the switching transistor. While 'low' is also a characteristic, in the context of technical examinations for SMPS, 'very low' is the preferred term to describe the effectiveness of the input DC bus stabilization.
For which of the following is the frequency is constant from input to output?
A)
AC voltage controller
B)
Chopper
C)
Inverter
D)
Cyclo converter
To determine which device maintains a constant frequency from input to output, let's analyze the function of each power electronic converter:
1. **AC Voltage Controller (a):** This device converts a fixed-frequency AC voltage to a variable-voltage AC voltage of the same frequency. It uses phase control (via thyristors or TRIACs) to change the RMS value of the output voltage, but the fundamental frequency of the input source remains unchanged at the output. Therefore, frequency is constant.
2. **Chopper (b):** A chopper is a DC-to-DC converter. It takes a constant DC input and produces a variable DC output. Since the input and output are DC, the concept of 'frequency' in the sense of a periodic AC waveform does not apply as a constant transfer from input to output.
3. **Inverter (c):** An inverter converts DC to AC. The input has no frequency (DC), while the output is a synthesized AC waveform with a specific frequency. Thus, the frequency changes from zero (DC) to a non-zero value.
4. **Cycloconverter (d):** A cycloconverter converts AC power at one frequency to AC power at a different (usually lower) frequency without an intermediate DC link. By definition, it changes the frequency from input to output.
Conclusion: Only the AC voltage controller maintains the input frequency at the output.
The hospital requires fresh air and that is why air condition system is require.
A)
50%
B)
70%
C)
90%
D)
100%
Hospitals require a very high percentage of fresh air in air-conditioning systems to maintain hygiene, control infection, and ensure proper ventilation. In standard HVAC practice for hospitals, the fresh air requirement is typically taken as 100% in such exam questions, meaning the system should supply only outdoor air rather than recirculated air. Therefore, option d (100%) is correct. The other options are incorrect because 50%, 70%, and 90% do not meet the full fresh-air requirement expected for hospital environments in this context.
The range of wind speed suitable for wind power generation is -
A)
0 to 5 m/s
B)
5 to 25 m/s
C)
25 to 50 m/s
D)
50 to 75 m/s
In wind energy conversion systems, there are two critical wind speed thresholds: the 'cut-in speed' and the 'cut-out speed'.
1. Cut-in Speed: This is the minimum wind speed at which the wind turbine blades begin to rotate and generate usable electricity. For most commercial wind turbines, this typically ranges from 3 to 5 m/s. Below this speed, the torque generated by the wind is insufficient to overcome the mechanical friction and inertia of the system.
2. Cut-out Speed: This is the maximum wind speed at which the turbine can safely operate. To prevent structural damage or mechanical failure due to extreme centrifugal forces and wind loads, the turbine is automatically braked or feathered (blades turned away from the wind) when speeds reach this limit. This typically occurs between 20 and 25 m/s.
Therefore, the operational range for power generation is generally between 5 m/s and 25 m/s.
Analysis of other options:
- Option (a) 0 to 5 m/s: This is below or at the cut-in speed; insufficient power is generated.
- Option (c) 25 to 50 m/s: This exceeds the cut-out speed; turbines are shut down to prevent destruction.
- Option (d) 50 to 75 m/s: These are storm-level speeds that would cause catastrophic failure to standard wind turbine structures.
A grid interactive solar system -
A)
always supplies power to the grid.
B)
supplies power to as well as receives power from the grid as required.
C)
always receives power from the grid.
D)
works only when the grid fails.
A grid-interactive solar system (also known as a grid-tied system) is designed to operate in parallel with the utility power grid.
1. **Supplying Power:** When the solar panels generate more electricity than the local load requires, the excess energy is exported (supplied) to the utility grid, often resulting in credits via 'Net Metering'.
2. **Receiving Power:** During periods of low sunlight (nighttime or cloudy weather) or when the local demand exceeds the solar production, the system automatically draws (receives) power from the grid to ensure a continuous power supply to the loads.
Analysis of other options:
- Option (a) is incorrect because the system only supplies power when there is a surplus; it cannot supply power at night.
- Option (c) is incorrect because the primary purpose of the solar installation is to generate power and reduce reliance on the grid.
- Option (d) describes a 'Stand-alone' or 'Backup' system with an inverter/battery setup; standard grid-interactive systems actually shut down during a grid failure (anti-islanding) for safety reasons unless they have specific backup capabilities.
A thyristor (SCR) is a -
A)
P-N-P device
B)
N-P-N device
C)
P-N-P-N device
D)
P-N device
A thyristor, specifically a Silicon Controlled Rectifier (SCR), is a four-layer semiconductor switching device. It consists of four alternating layers of P-type and N-type semiconductor materials, forming a P-N-P-N structure. This structure creates three P-N junctions. The device acts as a bistable switch, meaning it can be in either an 'off' state (blocking) or an 'on' state (conducting).
Analysis of options:
- Option (a) P-N-P: This describes a Bipolar Junction Transistor (BJT) with a P-type emitter and collector.
- Option (b) N-P-N: This describes a Bipolar Junction Transistor (BJT) with an N-type emitter and collector.
- Option (c) P-N-P-N: This is the correct structure for a thyristor/SCR.
- Option (d) P-N: This describes a simple diode.
The iron core is used in transformer to-
A)
increase the weight.
B)
provide tight magnetic coupling.
C)
reduce core losses.
D)
increase the overall efficiency.
The primary purpose of the iron core in a transformer is to provide a low-reluctance path for the magnetic flux. By concentrating the magnetic field, the core ensures that the maximum amount of flux produced by the primary winding links with the secondary winding. This is known as 'tight magnetic coupling.' Without the core (as in an air-core transformer), the magnetic flux would spread out (leakage), resulting in very poor coupling and extremely low efficiency in power transfer.
Analysis of other options:
- Option (a): Increasing weight is a byproduct of using a core, not the purpose.
- Option (c): The iron core actually introduces core losses (hysteresis and eddy current losses) that would not exist in an air-core transformer; however, we use specialized materials (like silicon steel) to *minimize* these losses, but the core itself is not used *to* reduce them.
- Option (d): While the core increases overall efficiency by improving coupling, 'providing tight magnetic coupling' is the specific physical mechanism that enables that efficiency increase, making (b) the more precise technical answer.
For a practical synchronous motor, the pull-out torque will occur when the torque angle is nearly equal to -
A)
0°
B)
30°
C)
45°
D)
75°
1. Definition of Pull-out Torque: Pull-out torque (also known as maximum torque) is the maximum torque that a synchronous motor can develop before it loses synchronism and begins to slip.
2. Torque Equation: The torque (T) developed by a synchronous motor is given by the formula: T = (3 * V * E * Xs / ωs) * sin(δ), where V is the terminal voltage, E is the excitation voltage, Xs is the synchronous reactance, ωs is the synchronous speed, and δ is the torque angle (load angle).
3. Condition for Maximum Torque: Mathematically, the torque T is a function of sin(δ). The maximum value of sin(δ) occurs when δ = 90°.
4. Practical Considerations: In an ideal (cylindrical rotor) machine without armature reaction or saturation, the theoretical pull-out angle is 90°. However, for a 'practical' synchronous motor, factors such as armature reaction, magnetic saturation, and the fact that the motor is often over-excited or under-excited shift this peak. In practical operation and standard electrical machine textbooks (such as B.L. Theraja or Kothari & Nagrath), it is observed that the stability limit and the pull-out torque typically occur at angles significantly less than 90°, often falling in the range of 45° to 60° depending on the machine's design and loading conditions.
5. Evaluating Options: Among the given options (0°, 30°, 45°, 75°), 45° is the most representative value for the practical pull-out torque angle of a standard synchronous motor, as 0° is no-load, 30° is well within the stable operating region, and 75° is too close to the theoretical limit for most practical industrial machines.
If no load is connected to solar PV system -
A)
it will stop absorbing light.
B)
it will dissipate energy in the panel and increase its temperature.
C)
its voltage will go on increasing till its breakdown.
D)
it will start reflecting the light.
Step-by-step reasoning:
1. **Understanding Open Circuit Condition**: When no load is connected to a solar PV system, it is in an 'Open Circuit' state. In this state, the current (I) is zero.
2. **Energy Conversion**: A solar cell continues to absorb photons from sunlight regardless of whether a load is connected. This absorption creates electron-hole pairs.
3. **Recombination**: Since there is no external path (load) for the generated electrons to flow through, the charge carriers cannot be extracted as electrical current. Instead, they undergo 'recombination' within the semiconductor material of the cell.
4. **Thermal Dissipation**: The process of recombination releases the energy of the absorbed photons as heat (phonons) rather than electrical energy.
5. **Temperature Increase**: Because the absorbed solar energy is converted into heat instead of being exported as electricity, the temperature of the PV panel increases.
Why other options are incorrect:
- **Option a**: The absorption of light is a physical property of the semiconductor material (bandgap) and does not depend on the electrical load.
- **Option c**: The voltage of a solar cell is limited by its built-in potential (the p-n junction barrier). It will reach its Open Circuit Voltage (Voc) and stabilize; it will not increase indefinitely to a breakdown point.
- **Option d**: Reflection is determined by the surface properties (like anti-reflective coatings) and the angle of incidence, not by the electrical load connected to the circuit.
The no-load input power to a transformer is practically equal to-
A)
the iron loss in the transformer
B)
the copper loss in the transformer
C)
the eddy current in the transformer
D)
the hysteresis loss in the transformer
Step-by-step reasoning:
1. **Understanding No-Load Condition**: In a transformer, 'no-load' means the secondary winding is open-circuited. In this state, no current flows through the secondary winding, and only a small no-load current (excitation current) flows through the primary winding.
2. **Analyzing Power Consumption**: The power input at no-load is used to maintain the magnetic flux in the core and to compensate for the losses occurring within the core.
3. **Copper Loss**: Copper loss is caused by the resistance of the windings ($I^2R$). Since the no-load current is very small (typically 2-5% of full-load current), the $I^2R$ loss in the primary is negligible. Therefore, copper loss is practically zero at no-load.
4. **Iron Loss (Core Loss)**: Iron loss consists of Hysteresis loss and Eddy current loss. These losses depend on the supply voltage, frequency, and core material, not on the load. Since the transformer is connected to the rated supply voltage, the core is magnetized, and iron losses occur continuously.
5. **Conclusion**: Because copper losses are negligible at no-load, the total input power is almost entirely consumed by the core losses. Thus, no-load input power $\approx$ Iron Loss (Hysteresis Loss + Eddy Current Loss).
Why other options are incorrect:
- Option (b): Copper loss is load-dependent and negligible at no-load.
- Option (c) and (d): While eddy current and hysteresis losses are components of iron loss, neither one alone represents the total no-load power; their sum (the total iron loss) does.
Solar energy can be utilised directly by how many technologies?
A)
1
B)
2
C)
3
D)
4
Solar energy is utilized directly through two primary technological pathways:
1. Photovoltaic (PV) Technology: This converts sunlight directly into electricity using semiconductor materials (the photoelectric effect).
2. Solar Thermal Technology: This converts sunlight directly into heat (thermal energy), which can be used for water heating, space heating, or to drive a steam turbine for power generation.
While there are many variations of these (e.g., Concentrated Solar Power vs. Flat Plate Collectors), they all fall under these two fundamental direct utilization categories: conversion to electricity (PV) or conversion to heat (Thermal). Therefore, the number of primary direct utilization technologies is 2.
Windmill works on the principle of -
A)
rotation
B)
momentum
C)
gravitation
D)
collision
A windmill (or wind turbine) operates based on the principle of momentum. Specifically, it utilizes the change in momentum of the air flowing past the blades. According to Newton's Second Law of Motion, the force exerted by the wind on the blades is equal to the rate of change of momentum of the air. As the wind hits the blades, its velocity changes (direction and/or magnitude), creating a force (lift or drag) that causes the rotor to rotate. This conversion of the kinetic energy of the wind into mechanical energy is fundamentally a momentum transfer process.
Why other options are incorrect:
- Rotation (a): Rotation is the *result* of the wind's force, not the underlying physical principle that causes the movement.
- Gravitation (c): Gravitation is the force of attraction between masses and does not drive the rotation of a windmill.
- Collision (d): While air molecules do collide with the blades, the overall operation of a turbine (especially modern aerodynamic blades) is described by fluid dynamics and the conservation of momentum rather than simple discrete collisions.
A 4-pole, 3-phase, 60 Hz induction motor is operating at a speed of 1500 rpm. The frequency of the rotor current of the motor in Hz is
A)
5
B)
10
C)
2
D)
7
For a 4-pole, 60 Hz induction motor, the synchronous speed is Ns = 120f/P = 120×60/4 = 1800 rpm. The rotor speed is 1500 rpm, so slip s = (Ns − Nr)/Ns = (1800 − 1500)/1800 = 300/1800 = 1/6. Rotor current frequency is fr = s×f = (1/6)×60 = 10 Hz. Therefore, the correct value is 10 Hz. Among the options, 10 Hz corresponds to option b. The other options are incorrect because 7 Hz, 2 Hz, and 5 Hz do not match the rotor frequency obtained from the slip relation fr = sf.