← Junior Engineer
Past Questions
Which of the following additional devices is required in order to measured pressure using LVDT?
A)
Strain gauge
B)
Pilot Tube
C)
Rotameter
D)
Bourdon Tube
An LVDT (Linear Variable Differential Transformer) is a transducer that converts linear displacement into an electrical signal. However, it cannot sense pressure directly because pressure is a scalar quantity, not a displacement. To measure pressure using an LVDT, a primary sensing element is required to convert the pressure into a mechanical displacement. A Bourdon Tube is a C-shaped tube that tends to straighten out as internal pressure increases; this movement (displacement) can be mechanically linked to the core of an LVDT. Thus, the Bourdon Tube acts as the pressure-to-displacement converter, and the LVDT acts as the displacement-to-voltage converter.
Why other options are incorrect:
- a. Strain gauge: This is an independent pressure/force sensing element that changes resistance; it does not provide the linear displacement required to operate an LVDT.
- b. Pilot Tube (Pitot Tube): This is used to measure fluid flow velocity, not static pressure for LVDT conversion.
- c. Rotameter: This is a device used for measuring the flow rate of liquids or gases, not pressure.
Isolators must be open only -
A)
after opening of circuit breaker
B)
after closing of circuit breaker
C)
under closed position of circuit breaker
D)
while on load
An isolator (or disconnecting switch) is a mechanical switch used to isolate a part of an electrical circuit for maintenance purposes. Unlike a circuit breaker, an isolator is NOT designed to interrupt current flow or quench the resulting electric arc.
1. **Operational Sequence**: To safely isolate a circuit, the load must first be interrupted using a circuit breaker (which is designed to handle arcs). Once the circuit breaker is open and the current flow is zero, the isolator is then opened to provide a visible physical gap in the circuit for safety.
2. **Danger of On-Load Operation**: If an isolator is opened while the circuit is under load (current is flowing), a massive electric arc will form between the contacts. Since isolators lack arc-quenching mechanisms, this arc can cause severe equipment damage, lead to phase-to-phase faults, and pose a lethal risk to the operator.
Therefore, the strict sequence is: Circuit Breaker Open $\rightarrow$ Isolator Open. This makes option (a) the only correct and safe procedure.
What is the ideal site of thermal power plant?
A)
In hilly regions
B)
On islands
C)
Near water resources
D)
In centre lands
A thermal power plant requires a massive and continuous supply of water for two primary reasons: 1. Condensation: The steam produced to drive the turbines must be condensed back into water using a condenser, which requires a large volume of cooling water (usually from a river, lake, or sea). 2. Boiler Feed: High-purity water is needed for the steam generation process in the boiler. Therefore, proximity to water resources is the most critical factor for site selection to minimize pumping costs and ensure operational stability.
Why other options are incorrect:
- a. Hilly regions: These are generally unsuitable due to high construction costs, difficulty in transporting heavy machinery (like turbines and boilers), and lack of flat land.
- b. On islands: While islands have water, the logistics of transporting massive amounts of fuel (coal) and connecting the plant to the main electrical grid via undersea cables make it economically unviable.
- d. In centre lands: While land availability is important, 'centre lands' (inland areas far from water) would require expensive piping systems or cooling towers that are less efficient than once-through cooling from a nearby water body.
Trash racks are built for -
A)
discharging the water freely from the turbine exit to tailrace
B)
preventing the turbine from ingress of floating and other materials
C)
creating artificial head to store sufficient potential energy of water
D)
controlling the opening of valves
Step-by-step reasoning:
1. **Definition of Trash Racks**: In a hydroelectric power plant, trash racks are screens or grids made of metal bars placed at the intake of the water conduit (penstock) or the forebay.
2. **Primary Function**: Their main purpose is to act as a filter to prevent debris, such as floating logs, leaves, plastic, and other foreign materials, from entering the penstock and reaching the turbine.
3. **Impact of Debris**: If such materials enter the turbine, they can cause severe mechanical damage to the runner blades, clog the wicket gates, or cause imbalances, leading to plant failure or costly repairs.
4. **Evaluating Options**:
- **Option (a)**: Discharging water from the turbine exit to the tailrace is the function of the **Draft Tube**.
- **Option (b)**: This correctly describes the function of the trash rack (preventing ingress of floating materials).
- **Option (c)**: Creating artificial head to store potential energy is the function of the **Dam or Reservoir**.
- **Option (d)**: Controlling the opening of valves is the function of **Actuators or Control Valves** (e.g., Guide Vanes/Wicket Gates).
Future expansion in substation is possible for -
A)
underground substation
B)
indoor substation
C)
outdoor substation
D)
pole mounted substation
Future expansion refers to the ability to add more bays, transformers, or switchgear to a substation as the load demand increases over time.
1. Outdoor Substations: These are typically built on open land. Because they are not confined by walls or ceilings, expanding the facility usually only requires additional land or utilizing pre-allocated space within the yard. It is the most flexible and easiest type of substation to expand.
2. Indoor Substations: These are housed within a building. Expansion is severely limited by the physical dimensions of the building (walls and roof). To expand, one would either need to build a new extension to the building or construct a completely new building, which is costly and complex.
3. Underground Substations: These are the most constrained. They are limited by the excavated volume and the structural integrity of the surrounding earth/buildings. Expanding an underground facility is extremely difficult and expensive.
4. Pole Mounted Substations: These are small, distribution-level substations designed for a specific, limited capacity. They are not designed for 'expansion' in the industrial sense; if the load increases beyond the pole's capacity, the entire unit is typically replaced or a new substation is installed nearby.
Modern steam power plant works on which of the following cycle -
A)
carnot cycle
B)
rankine cycle
C)
otto cycle
D)
bell-coleman cycle
Step-by-step reasoning:
1. **Analysis of the Question**: The question asks for the thermodynamic cycle upon which modern steam power plants operate.
2. **Evaluating Option (b) Rankine Cycle**: The Rankine cycle is the idealized thermodynamic cycle for steam power plants. It consists of four main processes: pumping liquid water to high pressure (Pump), heating the water to create superheated steam (Boiler), expanding the steam to produce work (Turbine), and condensing the steam back into water (Condenser). This is the practical basis for all modern thermal power plants.
3. **Evaluating Option (a) Carnot Cycle**: While the Carnot cycle is the most efficient theoretical cycle, it is impossible to implement in a steam plant because it requires isothermal heat addition and removal, and the compression of a two-phase mixture (liquid-vapor) is mechanically impractical.
4. **Evaluating Option (c) Otto Cycle**: The Otto cycle is the theoretical cycle for spark-ignition internal combustion engines (petrol engines), not steam power plants.
5. **Evaluating Option (d) Bell-Coleman Cycle**: Also known as the Reverse Brayton cycle, this is used for air refrigeration systems, not for power generation in steam plants.
Conclusion: The Rankine cycle is the standard cycle for steam power plants.
Penstock in a hydroelectric power plant is -
A)
a pipe connected to runner outlet
B)
nozzle that release high pressure water on turbine blades
C)
a conduit connecting forebay and turbine
D)
a pipe connecting surge tank to dam
In a hydroelectric power plant, the penstock is a high-pressure steel or concrete pipe (conduit) that carries water from the reservoir or forebay (the storage area at the top) down to the turbine. Its primary purpose is to deliver water with high kinetic and potential energy to the turbine runner to generate electricity.
Analysis of options:
- Option (a) is incorrect because the pipe connected to the runner outlet is called the 'Draft Tube', which recovers pressure head.
- Option (b) is incorrect because the nozzle is a specific component (common in Pelton wheels) that converts pressure into a high-velocity jet, but the penstock is the entire conduit leading to that nozzle.
- Option (c) is correct as it accurately describes the penstock as the conduit connecting the forebay (or surge tank) to the turbine.
- Option (d) is incorrect because the surge tank is typically located between the main conduit and the penstock to relieve water hammer pressure; the penstock connects the surge tank to the turbine, not the surge tank to the dam.
A tariff system with fixed rate per unit of energy consumed.
A)
Simple Tariff
B)
Flat-rate Tariff
C)
Block-rate Tariff
D)
Two-part Tariff
Step-by-step reasoning:
1. **Definition of Flat-rate Tariff**: In a flat-rate tariff, the cost of energy is charged at a constant rate per unit (kWh) regardless of the total amount of energy consumed. The formula is: Total Charge = (Units Consumed) × (Fixed Rate per Unit).
2. **Analyzing the Question**: The question specifies a 'fixed rate per unit of energy consumed,' which aligns perfectly with the definition of a flat-rate tariff.
3. **Evaluating Other Options**:
- **Simple Tariff**: This is a general term and not a specific technical classification of a tariff structure in electrical engineering.
- **Block-rate Tariff**: In this system, energy is charged in blocks. The rate per unit decreases (or increases) as the consumption moves into higher blocks (e.g., first 100 units at $x, next 100 units at $y). It is not a single fixed rate.
- **Two-part Tariff**: This consists of two components: a fixed charge (based on maximum demand or connected load) and a running charge (based on actual energy consumed). It is not based solely on a fixed rate per unit.
Usually distribution substation are -
A)
underground type
B)
indoor type
C)
outdoor type
D)
dry type
Distribution substations are the final stage of the power system before electricity is delivered to the end consumer. They typically step down voltage from a sub-transmission level (e.g., 33kV or 11kV) to a distribution level (e.g., 415V/230V).
1. Why 'Indoor Type' is correct: Distribution substations are often located in densely populated urban areas or residential neighborhoods. To ensure safety, protect equipment from environmental factors, and prevent unauthorized access or accidents in crowded areas, these substations are usually housed within a building or a secure enclosure, making them 'indoor type'.
2. Why 'Outdoor Type' is incorrect: Outdoor substations are typically used for high-voltage transmission substations where large clearances are required between conductors and equipment to prevent flashovers, and where land is readily available.
3. Why 'Underground Type' is incorrect: While some components (like cables) are underground, the entire substation (transformers, switchgear) is rarely fully underground due to cooling requirements and maintenance difficulties, although 'gas-insulated substations' (GIS) can be placed in basements.
4. Why 'Dry Type' is incorrect: 'Dry type' refers to a specific type of transformer cooling (air-cooled) rather than the overall classification of the substation layout.
In electrical work, the fuse should be placed -
A)
in earth wire
B)
in live or phase wire
C)
in lamp directly
D)
neutral wire
In electrical installations, a fuse is a safety device designed to protect the circuit from overcurrent or short circuits by melting and breaking the circuit. For maximum safety, the fuse must be placed in the live (phase) wire for the following reasons: 1. Isolation: If the fuse blows, the entire circuit downstream of the fuse becomes 'dead' (de-energized). This prevents the appliance or the wiring from remaining energized at a high potential. 2. Safety: If the fuse were placed in the neutral wire and it blew, the appliance would stop working, but the internal circuitry would still be connected to the live wire. If a person touched a fault in the appliance, they would provide a path to the ground, resulting in a severe electric shock. Therefore, placing the fuse in the phase wire ensures that the power source is disconnected immediately upon a fault.
Why other options are incorrect:
- Option (a): The earth wire is a safety path for leakage current; placing a fuse here would prevent the safety path from working, leading to dangerous chassis electrification.
- Option (c): Placing a fuse 'directly in a lamp' is not a standard wiring practice; the fuse must protect the entire branch circuit.
- Option (d): As explained, placing a fuse in the neutral wire leaves the circuit live even after the fuse has blown, creating a significant electrocution hazard.
Primary distribution substation voltage is -
A)
11kV
B)
33kV
C)
66kV
D)
220kV
In a typical electrical power distribution system, the distribution network is divided into two main stages: Primary Distribution and Secondary Distribution.
1. Primary Distribution: This is the stage where power is transmitted from the distribution substation to the distribution transformers. The standard voltage level for primary distribution in most electrical grids (especially in India and similar systems) is 11 kV. This voltage is used to carry power over a moderate distance to local neighborhoods or industrial areas.
2. Secondary Distribution: This is the final stage where the distribution transformer steps down the 11 kV primary voltage to a usable consumer voltage (typically 415V three-phase or 230V single-phase).
Analysis of options:
- Option (a) 11kV: Correct. This is the standard primary distribution voltage.
- Option (b) 33kV: This is typically used for sub-transmission or primary distribution in very large rural areas, but 11kV is the standard answer for 'primary distribution' in general technical exams.
- Option (c) 66kV: This is a sub-transmission voltage level.
- Option (d) 220kV: This is a high-voltage transmission level used for long-distance bulk power transport.
Distance relay is used for -
A)
Transmission line protection
B)
Transformer protection
C)
Bus bar protection
D)
Generator protection
A distance relay is a type of protection relay that operates based on the impedance of the line. It measures the ratio of voltage to current (V/I) at the relay location. Since the impedance of a transmission line is proportional to its length, the relay can determine the distance to the fault.
1. Why Option (a) is correct: Transmission lines are long conductors where faults can occur at various distances from the substation. Distance relays provide fast, selective protection by dividing the line into zones (Zone 1, Zone 2, etc.), ensuring that only the faulted section of the line is isolated.
2. Why other options are incorrect:
- Transformer protection: Typically uses Differential Relays (for internal faults) and Buchholz relays (for incipient faults).
- Bus bar protection: Typically uses Bus Differential Protection schemes because the bus bar is a single point of connection with no 'distance' or length to measure.
- Generator protection: Typically uses Differential Relays, Reverse Power Relays, and Overcurrent/Overvoltage relays to protect the machine windings and rotor.
Which is the most common voltage level of electrical lines that is running along the road sides on poles?
A)
66kV
B)
33kV
C)
11kV
D)
400 V
In a typical electrical power distribution system, power is transmitted from generating stations at very high voltages, stepped down at transmission substations, and then further stepped down at distribution substations. The lines running along roadsides on poles are typically 'Primary Distribution Lines'. In most standard power systems (especially in India and similar grids), the primary distribution voltage is 11kV. These lines carry power from the distribution substation to local distribution transformers (pole-mounted transformers), which then step the voltage down to the secondary distribution level (415V/230V) for end-user consumption.
Analysis of options:
- 66kV and 33kV: These are typically used for sub-transmission lines (connecting main substations to distribution substations) and are usually carried on larger steel towers or specialized high-tension poles, not the common roadside distribution poles.
- 400 V: This is the secondary distribution voltage (Low Tension - LT). While LT lines also run on poles, the 'most common' characteristic voltage associated with the primary distribution network seen on roadside poles is 11kV. In many exam contexts, when asked about the primary lines on poles, 11kV is the standard answer.
What type of relay is used in applications where high frequency switching is required?
A)
Electromechanical relay
B)
Reed relay
C)
Solid state relay
D)
Mercury relay
To determine the correct relay for high-frequency switching, we must analyze the switching mechanism of each option:
1. **Electromechanical Relays (EMR):** These rely on a physical armature moving to close or open contacts. Due to the inertia of the mechanical parts, they have a slow response time and a limited switching frequency. They are also subject to mechanical wear and tear.
2. **Reed Relays:** These are a type of electromechanical switch consisting of two ferromagnetic blades in a glass envelope. While faster than standard EMRs, they still involve physical movement and are not suitable for very high-frequency switching.
3. **Solid State Relays (SSR):** These use semiconductor devices (such as thyristors, TRIACs, or transistors) to switch the load. Because there are no moving parts, the switching speed is extremely fast (microseconds), allowing for very high-frequency operation. They also offer longer lifespans as there is no mechanical wear or contact arcing.
4. **Mercury Relays:** These use a pool of mercury to establish a connection. While they provide low contact bounce and are reliable, they are still mechanical in nature and cannot compete with the switching speeds of semiconductor-based SSRs.
Therefore, for applications requiring high-frequency switching, the Solid State Relay is the only viable choice.
Transmission efficiency of a transmission line increases with the -
A)
decrease in power factor and voltage
B)
increase in power factor and voltage
C)
increase in power factor but decrease in voltage
D)
increase in current
Transmission efficiency (η) is defined as the ratio of the power delivered at the receiving end to the power supplied at the sending end. Mathematically, η = (Power Delivered) / (Power Delivered + Power Losses). The primary source of loss in a transmission line is I²R loss (copper loss).
1. Effect of Voltage: For a constant amount of power (P = VI cosφ), increasing the transmission voltage (V) results in a decrease in the line current (I). Since power loss is proportional to the square of the current (I²R), increasing the voltage significantly reduces the losses, thereby increasing efficiency.
2. Effect of Power Factor: Power factor (cosφ) represents the ratio of real power to apparent power. For a fixed amount of real power to be delivered, a lower power factor requires a higher current to flow through the line (I = P / (√3 * V * cosφ)). By increasing the power factor (bringing it closer to unity), the current required to transmit the same amount of power decreases. This reduction in current leads to lower I²R losses and higher efficiency.
Therefore, both an increase in voltage and an increase in power factor lead to a reduction in line losses and an increase in transmission efficiency.
What environmental factors can affect the performance of circuit breakers?
A)
Humidity and dust buildup
B)
Exposure to direct sunlight
C)
Extremely high or low temperature
D)
Exposure to rain
Humidity and dust buildup are the most direct environmental factors that can degrade circuit breaker performance. Moisture can reduce insulation resistance and promote corrosion, while dust can contaminate contacts and moving parts, causing overheating, poor contact, or mechanical sticking. Option c is also an environmental factor in general, but the question asks for the factor that affects circuit breaker performance most characteristically in service conditions; humidity and dust are the standard answer. Option b (direct sunlight) may raise temperature slightly, but it is not a primary factor affecting breaker operation. Option d (rain) can be harmful if equipment is exposed, but circuit breakers are normally enclosed/protected, so rain is not the usual general factor cited compared with humidity and dust.
An insulator is so designed that it should fail only when -
A)
a line breakdown
B)
a flash over occurs
C)
it is punctured
D)
sag is less
In transmission line design, there are two primary ways an insulator can fail: puncture and flashover.
1. **Puncture**: This is the electrical breakdown of the insulating material itself. Once an insulator is punctured, it is permanently damaged and must be replaced. This is a catastrophic failure.
2. **Flashover**: This is an arc that forms through the air around the insulator surface. While a flashover causes a short circuit, it does not necessarily damage the physical structure of the insulator; once the fault is cleared, the insulator can often continue to function.
An ideal insulator is designed such that the **flashover voltage is lower than the puncture voltage**. This ensures that if the system is subjected to an overvoltage (like a lightning strike), the electricity will 'flash over' the surface rather than 'puncturing' through the material. Therefore, the insulator is designed to 'fail' (allow current to pass) via flashover first to protect the integrity of the insulator itself from permanent destruction.
- Option (a) 'line breakdown' is a general term for system failure, not a specific insulator design criterion.
- Option (c) 'it is punctured' is exactly what designers want to avoid, as it leads to permanent failure.
- Option (d) 'sag is less' refers to the mechanical tension of the conductor, not the electrical failure of the insulator.
Power factor can be improved by installing a device in parallel with load which takes -
A)
lagging reactive power
B)
leading reactive power
C)
apparent power
D)
leading apparent power
1. Understanding Power Factor: In most industrial electrical systems, the load is predominantly inductive (e.g., motors, transformers), which causes the current to lag behind the voltage. This results in a 'lagging power factor'.
2. The Problem: Inductive loads consume lagging reactive power (Q_L), which increases the total current drawn from the source for the same amount of real power, leading to higher losses and inefficiency.
3. The Solution: To improve (increase) the power factor, we need to neutralize the lagging reactive power. This is achieved by installing a device (typically a capacitor bank or a synchronous condenser) in parallel with the load.
4. Mechanism: Capacitors supply 'leading reactive power' (Q_C). Because leading reactive power is mathematically the opposite of lagging reactive power, the capacitor provides the reactive current required by the inductor locally, reducing the reactive power demand from the main supply.
5. Analysis of Options:
- (a) Lagging reactive power: Adding more lagging power would further decrease the power factor.
- (b) Leading reactive power: Correct. This cancels the lagging effect of inductive loads.
- (c) Apparent power: Apparent power (S) is the vector sum of real and reactive power; it is not something 'taken' by a device to specifically correct the phase angle.
- (d) Leading apparent power: This is a technically incorrect term; 'leading' or 'lagging' refers specifically to the reactive component, not the apparent power.
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.
The best solar battery performance is obtained in temperature range of -
A)
0-20°C
B)
20-40°C
C)
40-50°C
D)
50-60°C
Solar energy storage systems typically utilize Lead-Acid or Lithium-ion batteries. The performance of these batteries is highly temperature-dependent:
1. **Chemical Reaction Rates:** Battery performance relies on chemical reactions. At very low temperatures (Option a), the internal resistance increases, and the chemical reactions slow down, significantly reducing the capacity and discharge efficiency.
2. **Optimal Range:** The standard design temperature for most battery performance specifications is 25°C. The range of 20-40°C (Option b) provides the ideal balance where the electrolyte maintains optimal viscosity and ion mobility, ensuring maximum efficiency and capacity.
3. **Degradation at High Temperatures:** While higher temperatures (Options c and d) may temporarily increase the discharge capacity due to faster kinetics, they accelerate permanent degradation. High heat leads to increased self-discharge, electrolyte evaporation (in lead-acid), and accelerated aging of the electrodes (SEI layer growth in lithium batteries), which shortens the overall lifespan of the battery.
Therefore, the range of 20-40°C is considered the 'sweet spot' for maintaining high performance without compromising the longevity of the battery.
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.
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 $16/27 \approx 59.3\%$.
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.
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.
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.