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Technical Test SSC JE: Electrical Engineering (EE)- 1 - Electrical Engineering (EE) MCQ


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100 Questions MCQ Test Electrical Engineering SSC JE (Technical) - Technical Test SSC JE: Electrical Engineering (EE)- 1

Technical Test SSC JE: Electrical Engineering (EE)- 1 for Electrical Engineering (EE) 2024 is part of Electrical Engineering SSC JE (Technical) preparation. The Technical Test SSC JE: Electrical Engineering (EE)- 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Technical Test SSC JE: Electrical Engineering (EE)- 1 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Technical Test SSC JE: Electrical Engineering (EE)- 1 below.
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Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 1

Kirchhoff's second law is based on the law of conservation of

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 1

This law is based on the conservation of energy whereby voltage is defined as the energy per unit charge. The total amount of energy gained per unit charge must be equal to the amount of energy lost per unit charge, as energy and charge are both conserved.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 2

The resistance temperature coefficient is defined as

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 2
The resistance temperature coefficient is defined as:
A: Increase in resistance per degree centigrade
- This option suggests that the resistance of a material increases as the temperature increases.
B: Decrease in resistance per degree centigrade
- This option suggests that the resistance of a material decreases as the temperature increases.
C: The ratio of increase in resistance per degree centigrade to the resistance at 0°C
- This option states that the resistance temperature coefficient is a ratio that measures the increase in resistance per degree centigrade relative to the resistance at 0°C. It provides a quantitative measure of the change in resistance with temperature.
D: The ratio of increase in resistance per degree centigrade to the rate of rise of resistance at 0°C
- This option states that the resistance temperature coefficient is a ratio that measures the increase in resistance per degree centigrade relative to the rate of rise of resistance at 0°C. It takes into account the rate at which the resistance changes with temperature.
Answer: C
- The correct definition of the resistance temperature coefficient is the ratio of the increase in resistance per degree centigrade to the resistance at 0°C. This definition provides a standardized way to quantify the change in resistance with temperature.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 3

For the circuit shown the reading in the ammeter A will be

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Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 4

When checked with an ohm meter an open resistor reads

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 4

If you don't connect the probes to anything, you have an open circuit, there is no path from one to the other (except through the air), and the meter will read infinity ohms.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 5

A light dependent resistor is basically a

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 5

A light dependent resistor abbreviated as LDR is an electronic device which senses light. It is basically a variable resistor whose value varies with the intensity of incident light. It consists of a pair of metal contacts between which a curved track of cadmium sulphide is placed

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 6

Voltage dependent resistors are usually made from

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 6

Voltage dependent resistors (VDRs) are electronic components that exhibit a resistance that varies with applied voltage. They are designed to protect circuits from excessive voltage surges by acting as voltage-sensitive switches. VDRs are typically made from materials that have a non-linear resistance characteristic. One commonly used material is silicon carbide (SiC).
Here is a detailed explanation:
1. Voltage dependent resistors: These are electronic components that are used to protect circuits from voltage surges. They have a non-linear resistance characteristic, meaning that their resistance changes with the applied voltage.
2. Material: VDRs are typically made from materials that have a high resistance at low voltages and a low resistance at high voltages. This allows them to act as voltage-sensitive switches, conducting current only when the voltage exceeds a certain threshold.
3. Silicon carbide (SiC): SiC is a compound made up of silicon and carbon atoms. It has excellent electrical properties, including a high resistance to breakdown and a wide bandgap, making it suitable for use in VDRs.
4. Advantages of SiC: SiC offers several advantages as a material for VDRs. It has a high thermal conductivity, allowing it to dissipate heat efficiently. It also has a high voltage withstand capability, making it suitable for applications that require protection from high voltage surges.
5. Other materials: While SiC is a commonly used material for VDRs, other materials such as metal oxide varistors (MOVs) and selenium are also used in certain applications.
In conclusion, voltage dependent resistors (VDRs) are commonly made from silicon carbide (SiC) due to its excellent electrical properties and suitability for protecting circuits from voltage surges.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 7

Resistors across A & B in the circuit shown below is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 7

On rearranging, we get a balanced wheatstone bridge. hence the effective resistance is R

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 8

Voltage dependent resistors are used

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 8

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 9

The power factor of incandescent bulb is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 9

In incandescent bulb there is filament, which is purely restive in nature. and we know that purely resistive element has unity power factor.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 10

The equivalent inductance across 'ab' for the diagram shown below is:

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 10

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 11

If V2 = 1 – e–2t, the value of V1 is given by

 

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Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 12

For a parallel RLC circuit to be overdamped

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 12

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 13

The thevenin resistance across the diode in the circuit is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 13

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 14

Find input resistance of the circuit shown is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 14

 

 

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 15

A 3H – Inductor has 2000 turns. How many turns must be added to increase the inductance to 5H

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 15

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 16

For a series RLC circuit the power factor at the lowest half power frequency is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 16


To determine the power factor at the lowest half power frequency in a series RLC circuit, we need to consider the impedance of the circuit at that frequency. Let's break down the solution into the following points:
1. Impedance of a series RLC circuit:
The impedance of a series RLC circuit is given by the formula:
Z = √(R^2 + (Xl - Xc)^2)
where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.
2. Half power frequency:
The half power frequency is the frequency at which the power dissipated in the circuit is half of the power when the circuit is operating at resonance. At this frequency, the impedance of the circuit is equal to the resistance (Z = R).
3. Power factor:
The power factor is defined as the cosine of the phase angle between the voltage and current in an AC circuit. It determines the ratio of real power (active power) to apparent power.
4. Analysis:
At the half power frequency, the impedance of the circuit is equal to the resistance (Z = R). This means that the inductive and capacitive reactances cancel each other out, resulting in a purely resistive circuit.
Since the impedance is equal to the resistance at the half power frequency, the power factor is given by the formula:
Power factor = cos(φ) = R/Z = R/R = 1
5. Conclusion:
Therefore, the power factor at the lowest half power frequency in a series RLC circuit is unity (1), which means it is neither leading nor lagging.
Answer:
D: Unity

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 17

A series RLC circuit resonance at 1MHz at frequency of 1.1 MHz the circuit impedance is

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 18

If each branch of a delta circuit has impedance √3z , then each branch of equivalent Y–circuit
has impedance

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 18

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 19

A delta load is connected to a balanced 400 V, 3-f supply as shown in figure. The total power dissipated in the network is equal to

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 19

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 20

When Q-factor of the circuit is high, then

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 20
Explanation:
When the Q-factor of a circuit is high, it means that the circuit has a high quality factor. The quality factor is a measure of how selective the circuit is in passing a particular frequency and rejecting others. Here's how the different options relate to the Q-factor:
A. Power factor of the circuit is high:
- The power factor is a measure of how effectively the circuit converts electrical power into useful work.
- It is not directly related to the Q-factor of the circuit, so this option is incorrect.
B. Impedance of the circuit is high:
- The Q-factor is inversely proportional to the impedance bandwidth of a resonant circuit.
- A high Q-factor implies a narrow bandwidth and a high impedance at the resonant frequency.
- So, when the Q-factor is high, the impedance of the circuit is also high.
- This option is correct.
C. Bandwidth is large:
- The Q-factor is inversely proportional to the bandwidth of a resonant circuit.
- A high Q-factor implies a narrow bandwidth, not a large one.
- So, when the Q-factor is high, the bandwidth is small, not large.
- This option is incorrect.
D. None of these:
- Since option B is correct, option D is incorrect.
Therefore, the correct answer is B. Impedance of the circuit is high when the Q-factor of the circuit is high.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 21

The dynamic resistance of a parallel resonance circuit is given by

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 22

A parallel resonant circuit can be employed

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 22
Parallel Resonant Circuit
A parallel resonant circuit is a type of electrical circuit that consists of a capacitor and an inductor connected in parallel. It has several applications and can be employed for various purposes. In this response, we will discuss the applications of a parallel resonant circuit and how it can be used.
Applications of Parallel Resonant Circuit:
1. High Impedance: A parallel resonant circuit can be used as a high impedance element in a circuit. This means that it can present a high resistance to the flow of current at a specific resonant frequency. This property makes it useful in applications where a high impedance is required.
2. Frequency Rejection: A parallel resonant circuit can be employed to reject a small band of frequencies. By adjusting the values of the capacitor and inductor, the resonant frequency can be set to the desired value. At this resonant frequency, the parallel resonant circuit presents a high impedance, effectively blocking or attenuating signals at that particular frequency. This property is utilized in filters and selective amplifiers to reject unwanted frequencies.
3. Frequency Amplification: Similarly, a parallel resonant circuit can also be used to amplify certain frequencies. By tuning the circuit to the desired resonant frequency, it can provide a low impedance path to signals at that frequency, effectively amplifying them. This property is utilized in applications such as radio receivers and communication systems.
4. Circuit Tuning: Parallel resonant circuits can also be used for circuit tuning purposes. By adjusting the values of the capacitor and inductor, the resonant frequency of the circuit can be set to a specific value. This allows for precise control over the frequency response of the circuit, which is important in applications such as oscillators and frequency generators.
In conclusion, a parallel resonant circuit can be employed as a high impedance element, to reject a small band of frequencies, to amplify certain frequencies, and for circuit tuning purposes. Its unique properties make it a versatile component in various electrical and electronic applications.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 23

A 3-f star connected symmetrical load consumes p watts of power from a balancedsupply. If the same load is connected in delta to the same supply, the power consumption will be

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 23
The
When a 3-phase star-connected symmetrical load is connected to a balanced supply, the power consumption is given by the formula:
P = √3 × VL × IL × cosφ
Where:
P = Power consumed
VL = Line voltage
IL = Line current
φ = Angle between the line current and line voltage
Now, let's consider the load connected in a delta configuration. In a delta connection, the line current is equal to the phase current. Since the load is symmetrical, the line current in the delta connection will be the same as the phase current in the star connection. Hence, IL remains the same.
However, the line voltage in a delta connection is √3 times the phase voltage in a star connection. So, VL in the delta connection becomes √3 × Vph.
Using the same formula as before, the power consumption in the delta connection is given by:
P = √3 × VL × IL × cosφ
Substituting the values of VL and IL, we get:
P = √3 × (√3 × Vph) × IL × cosφ
P = 3 × √3 × Vph × IL × cosφ
We can see that the power consumption in the delta connection is 3 times the power consumption in the star connection.
Therefore, the correct answer is option C: 3p. The power consumption will be 3 times the power consumed when the load is connected in a star configuration.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 24

In two wattmeters method of power measurement, one of the wattmeters will show negative reading when the load power factor angle is strictly

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 24

The reading of a wattmeter is proportional to (current through the current coil)x(voltage across the pressure coil)xCos(angle between this current and this voltage).

So if this angle becomes more than 90 degrees the the reading of the wattmeter will be negative.This could happen in a 3-ph unbalanced circuit.

For a 3-ph balanced load the two wattmeter readings will be,say Vab*Ia*cos (30+phi) and Vcb*Ic* cos(30-phi). So if phi >60 degrees (pf less than 0.5) a negative reading is obtained.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 25

The external characteristics of a shunt generator can be obtained directly from its ____characteristics

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 26

For the voltage build up of a self excited d.c. generator, which of the following is not anessential condition?

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 27

Which of the following d.c. generator cannot build up on open-circuit?

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 28

Which of the following methods of speed control of D.C. machine will offer minimum efficiency?

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 29

Which winding of the transformer has less crosssectional area?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 29

High voltage winding because coil thickness is low comparing to others

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 30

A dc series motor of resistance 1Ω across terminal runs at 1000 rpm at 250V taking acurrent of 20A when an additional resistance 6W is inserted in series and taking the samecurrent, the new speed will be

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Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 31

In dc motors, the condition for maximum power is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 31

The condition for maximum power in case of D.C. motor is

Supply voltage =I × back e.m.f. D.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 32

The speed of a series wound dc motor

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 32
Speed Control of a Series Wound DC Motor

A series wound DC motor is a type of DC motor where the field winding is connected in series with the armature winding. In this configuration, the speed of the motor can be controlled by various methods:


1. Shunt Field Regulator:

  • The speed of a series wound DC motor can be controlled by a shunt field regulator.

  • A shunt field regulator is a device that controls the current flowing through the shunt field winding of the motor.

  • By adjusting the current in the shunt field winding, the flux produced by the field winding can be controlled, thus affecting the speed of the motor.


2. Diverter:

  • A diverter is not typically used to control the speed of a series wound DC motor.

  • A diverter is a device used to divert excess current away from the motor, protecting it from damage.

  • It does not directly control the speed of the motor.


3. Flux and Speed Relationship:

  • The speed of a series wound DC motor increases as the flux decreases.

  • Flux is directly proportional to the field current, so by decreasing the field current, the flux decreases, resulting in an increase in speed.

  • This relationship is due to the back EMF generated in the armature winding, which opposes the applied voltage and affects the speed of the motor.


4. Armature Circuit Resistance:

  • The speed of a series wound DC motor increases as the armature circuit resistance increases.

  • Increasing the armature circuit resistance reduces the armature current, which in turn affects the motor speed.

  • This relationship is based on Ohm's Law, where the speed is inversely proportional to the armature current.


Therefore, the correct answer is option C: The speed of a series wound DC motor increases as the flux decreases.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 33

In dc series motor 1 shaft torque is less than the armature torque due to

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 33
Explanation:
DC series motors are a type of electric motor where the field winding is connected in series with the armature winding. In such motors, the shaft torque is less than the armature torque due to various losses. Here are the explanations for each option:
A. Stray losses: These losses occur due to the leakage of magnetic flux from the core of the motor. Some of the magnetic flux does not pass through the armature, resulting in a decrease in the effective magnetic field and thus reducing the overall torque output.
B. Eddy current losses: Eddy currents are induced in the conducting parts of the motor, such as the armature core. These currents circulate within the conducting material and result in power loss in the form of heat. This loss further reduces the available torque.
C. Hysteresis losses: Hysteresis losses occur due to the magnetic properties of the core material. When the magnetic field changes, the magnetization of the core lags behind, resulting in energy loss in the form of heat. This loss also contributes to the reduction in shaft torque.
Therefore, the correct answer is option A: stray losses. Stray losses, along with eddy current losses and hysteresis losses, all contribute to the reduction in shaft torque in a DC series motor.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 34

A dc shunt motor has rated rpm of 480 contain industrial application require this motor to runat 540 rpm for sometime which speed control will be desirable

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 35

The leakage flux in a transformer depends upon

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 36

In a transformer if peak voltage is fed to primary

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 36
Explanation:
When peak voltage is fed to the primary of a transformer, the following effects can be observed:
1. Iron Loss: The iron loss in a transformer is primarily due to hysteresis and eddy current losses. When peak voltage is fed to the primary, the magnetic flux in the core will be at its maximum value. This means that the core will experience a higher rate of change of magnetic flux, resulting in increased hysteresis and eddy current losses. Therefore, the iron losses will be more.
2. Copper Loss: The copper loss in a transformer is caused by the resistance of the windings. When peak voltage is applied to the primary, the current flowing through the windings will be higher due to the increased voltage. This increased current will result in higher copper losses. Therefore, the copper losses will be more.
3. Windage Loss: Windage loss in a transformer is caused by the friction and turbulence generated by the movement of air around the windings and core. The windage loss is not directly affected by the voltage applied to the primary. Therefore, the windage losses will not be significantly affected by feeding peak voltage to the primary.
In conclusion, when peak voltage is fed to the primary of a transformer, the iron losses and copper losses will be more, while the windage losses will remain relatively unaffected.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 37

The phase difference between the primary & secondary voltage of a transformer is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 37

There is 90 degree phase shift between voltage and magnetic field. So primary voltage causes 90 degree phase shifted magnetic field, which induces again 90 degree phase shifted secondary voltage. So it adds up to 180 degree. If you change winding direction, secondary voltage will now cancel that phase shift and you will get 0 degree phase shift.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 38

All of the following methods/techniques can be used for the measurement of high ac voltages EXCEPT

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 39

Consider the circuit shown in the given figure for maximum power transfer to the load, the primary to secondary turn ratio must be

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 40

Maximum electric power output of a synchronous generator is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 40

The power developed in cylindrical synchronous machine is Maximum power developed Pmax = EV/Xs Where E = Excitation voltage V = Terminal voltage Xs = Synchronous reactance δ = load angle Therefore, maximum power developed in cylindrical synchronous machine is at a load angle of 90°.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 41

Two alternators are running in parallel. If the field of one of the alternator is adjusted, it will

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 41

If the field excitation (voltage adjustment control)of one of a pair of paralleled alternators is turned down, two things will happen: first, the generator in question will develop a less lagging power factor, while the other generator will develop a more lagging power factor, with the overall power factor of the plant and load remaining the same; second, bus voltage, and by extension grid voltage, will drop slightly, unless the other generator’s voltage adjustment (field excitation) is increased proportionally, which will also result in the second generator develop an even more lagging power factor as the first generator’s power factor becomes less lagging.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 42

An unexcited single phase synchronous motor is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 42
An unexcited single phase synchronous motor is:



A: Reluctance motor

B: Repulsion motor

C: Universal motor

D: AC series motor

An unexcited single phase synchronous motor refers to a type of motor that does not require any external excitation to operate. It is self-starting and operates synchronously with the AC power supply.
Here is a detailed explanation of each option:
A: Reluctance motor:
- A reluctance motor is a type of synchronous motor that operates based on the principle of reluctance torque.
- It requires excitation to create a magnetic field and is not unexcited, so it is not the correct answer for an unexcited single phase synchronous motor.
B: Repulsion motor:
- A repulsion motor is an AC motor that uses brushes and a commutator to create a rotating magnetic field.
- It requires excitation through brushes and is not unexcited, so it is not the correct answer for an unexcited single phase synchronous motor.
C: Universal motor:
- A universal motor is a type of motor that can operate on both AC and DC power supplies.
- It requires excitation and is not unexcited, so it is not the correct answer for an unexcited single phase synchronous motor.
D: AC series motor:
- An AC series motor is a type of motor where the stator and rotor windings are connected in series.
- It requires excitation and is not unexcited, so it is not the correct answer for an unexcited single phase synchronous motor.
Therefore, the correct answer is A: Reluctance motor
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 43

A synchronous motor is switched on to supply with its field winding shorted on themselves it will

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 44

The power factor of pure resistive circuit is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 44
Power Factor of a Pure Resistive Circuit:
In a pure resistive circuit, the power factor is unity (D). Let's understand why:
Definition of Power Factor:
The power factor is the ratio of the real power (P) to the apparent power (S) in an AC circuit. It represents the efficiency of the circuit in converting electrical power to useful work.
Characteristics of a Pure Resistive Circuit:
In a pure resistive circuit, the load consists of a resistor connected to an AC power source. The characteristics of a pure resistive circuit are as follows:
- The voltage and current are in phase with each other.
- There is no reactive component (inductive or capacitive) in the circuit.
- The power consumed by the load is purely real power.
Explanation:
In a pure resistive circuit, since there is no reactive component, the voltage and current waveforms are in phase with each other. As a result:
- The cosine of the phase angle between voltage and current is 1.
- The power factor (PF) is equal to 1 (unity).
Conclusion:
Therefore, the power factor of a pure resistive circuit is unity (D). This means that the real power consumed by the circuit is equal to the apparent power.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 45

Under which method of starting an induction motor is expected to take largest starting current?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 45

In Direct on line starting, an induction motor is expected to take largest starting current. 

In Direct on line starting (DOL) , the contactor will be controlled by separate start and stop buttons, and an auxiliary contact on the contactor is used, across the start button, as a hold in contact. i.e. the contactor is electrically latched closed while the motor is operating.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 46

A dc motors develops a torque of 200 N-m at 25 rps. At 20 rps, it will develop a torque of____ N-m

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 46

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 47

An eight pole wound rotor induction motor operating on 60 Hz supply is driven at 1800rpm, by a prime mover in the opposite direction of revolving magnetic field. The frequency ofrotor current is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 47

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 48

A 6-pole, 50 Hz, 3-f induction motor is running at 950 rpm and has rotor Cu loss of 5kw. Its rotor input is ____ kw

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 48

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 49

A 6 pole, 50 hz-, 3-f induction motor has a full load speed of 950 rpm. At half load, its speed would be ____ rpm

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 49

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 50

A 60 w lamp given a luminous flux of 1500 lumen. Its efficiency is

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 51

Which of the following is a cold cathode lamps?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 51
Answer:
Cold cathode lamps are a type of gas discharge lamp that operate at a low pressure and do not require a heated filament to emit light. Among the options provided, the correct answer is D: Neon lamp.
Here is a detailed explanation for each option:
- Sodium vapour lamp: Sodium vapour lamps are a type of high-intensity discharge lamp that use an electric arc through sodium vapor to produce light. They do not fall under the category of cold cathode lamps.
- High pressure mercury vapour lamp: High pressure mercury vapour lamps also fall under the category of high-intensity discharge lamps. They use an electric arc through mercury vapor to produce light and require a heated filament. They are not cold cathode lamps.
- Low pressure mercury vapour lamp: Low pressure mercury vapour lamps are a type of fluorescent lamp that operate at a low pressure and require a heated filament to emit electrons. They are not cold cathode lamps.
- Neon lamp: Neon lamps are a type of cold cathode lamp that use neon gas to produce light. They operate at a low pressure and do not require a heated filament. Neon lamps are commonly used for decorative and indicator purposes.
In conclusion, among the given options, the only cold cathode lamp is the Neon lamp.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 52

Which of the following lamp has least capacity to sustain voltage fluctuations?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 52

Effect of Voltage variation in incandescent lamp

Within the normal range of supply voltage variations the light output will change by about 3.5% for a 1% change of voltage. The effect of voltage on life is much more pronounced; 5% over-voltage will roughly halve the lamp life, whereas 5% under-voltage will approximately double it.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 53

Unbalanced forces are maximum in case of

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 54

Which motor is used in tramways

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 54

Mostly DC series motors are used in this system. For trolley buses and tramways, DC compound motors are used where regenerative braking is required. The various operating voltages of DC traction system include 600V, 750 V, 1500V and 3000V.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 55

In arc welding, the voltage on ac supply system is in the range

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 55
Explanation:
The voltage on the AC supply system in arc welding is typically in the range of 70-1000 V. Here is a detailed explanation of why this range is chosen:
1. Safety considerations:
- The voltage should be high enough to create an arc between the electrode and the workpiece, allowing for efficient welding.
- At the same time, it should not be too high to avoid electrical hazards and potential injuries to the welder.
2. Welding process requirements:
- The voltage range is determined by the specific welding process being used.
- Different materials and thicknesses require different voltage levels to achieve proper penetration and weld quality.
3. Electrode characteristics:
- The type and size of the electrode also play a role in determining the voltage range.
- Different electrodes have different voltage requirements for optimal performance.
4. Power supply capabilities:
- The AC power supply system should be capable of providing the required voltage range for arc welding.
- The voltage is typically adjusted using a welding power source or transformer.
In conclusion, the voltage on the AC supply system in arc welding is in the range of 70-1000 V to ensure both efficient welding and safety.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 56

In gas welding the gases used are

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 56

Gases used in welding and cutting processes include: shielding gases such as carbon dioxide, argon, helium, etc. fuel gases such as acetylene, propane, butane, etc. oxygen, used with fuel gases and also in small amounts in some shielding gas mixtures.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 57

For welding duty, the rectifier commonly used are

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 57

The rectifier types are commonly called "Welding Rectifiers" and produce DC or, AC and DC welding current. They may utilize either single phase or three phase input power. They contain a transformer, but rectify the AC or DC by the use of selenium rectifiers, silicon diodes or silicon controlled rectifiers.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 58

In hygrometers the principle of measurement is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 58

 Hygrometer uses for measuring the humidity present in the surrounding environment. The term humidity means the amount of water vapour present in the gas. The physical properties of the material changes by the effect of the humidity and this principle use in hygrometer for measurement.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 59

A signal of 10 mv at 75 MHz is to be measured which of the following instrument can be used?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 59

At high frequencies, cathode ray oscilloscope  is used

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 60

Holes are filled on the opposite side of the disc of an induction type energy meter to

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 61

Galvanometer type recorders use

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 61

In a galvanometer-type recorder, the pointer of the D'Arsonval movement is fitted with a pen-ink (stylus) mechanism. The pointer deflects when current flows through the moving coil.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 62

The curve representing Ohm’s law is 

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 62

From Ohm’s law 
V/I = K = R 
V = IR 
= Linear

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 63

A slide wire potentiometer is used to measure voltage across a dc circuit. The voltage is 1.2V as read by the potentiometer. Across the same points 20000 W/V voltmeter reads 0.6V on its 5V range. Input resistance between the two point is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 63

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 64

In a gravity controlled instrument, the deflection angle is proportional to

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 64
Explanation:
The deflection angle in a gravity-controlled instrument is proportional to the sine-inverse of the measured quantity. Here's why:
1. Gravity Control: In a gravity-controlled instrument, the deflection is determined by the force exerted by gravity on a mass or pendulum. The instrument is designed to measure the deflection caused by this force.
2. Proportional Relationship: The deflection angle is directly proportional to the measured quantity. This means that as the measured quantity increases or decreases, the deflection angle will also increase or decrease proportionally.
3. Sine-Inverse Relationship: The relationship between the deflection angle and the measured quantity is given by the sine-inverse function. The sine-inverse (also known as arcsine) of a value is the angle whose sine is equal to that value. In this case, the deflection angle is directly proportional to the sine-inverse of the measured quantity.
4. Mathematical Representation: Mathematically, we can represent this relationship as follows:
- Deflection angle = k * sin^(-1)(measured quantity)
- Where k is a constant of proportionality.
5. Example: Let's consider an example where the measured quantity is the length of a pendulum. As the length of the pendulum increases, the deflection angle will also increase, but not in a linear manner. The relationship follows the sine-inverse function, which means that the deflection angle will increase more slowly as the length of the pendulum increases.
In conclusion, the deflection angle in a gravity-controlled instrument is proportional to the sine-inverse of the measured quantity. This relationship allows the instrument to accurately measure and display the value of the measured quantity based on the deflection angle.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 65

In a Q-meter measurement to determine the self capacitance of a coil, the first resonance occurred at F1 with C1 = 30 PF. The second resonance occurred at F2 = 2F1 with C2 = 60PF. The self capacitance of coil works out to be

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 65
Q-meter Measurement to Determine Self Capacitance of a Coil:
To determine the self capacitance of a coil using a Q-meter measurement, we can use the following steps:
Step 1: Identify the resonant frequencies and corresponding capacitances:
- The first resonance occurred at frequency F1 with capacitance C1 = 30 PF.
- The second resonance occurred at frequency F2 = 2F1 with capacitance C2 = 60 PF.
Step 2: Use the resonant frequencies and capacitances to calculate the self capacitance of the coil:
- The self capacitance of the coil can be determined using the formula: Cs = (4C1 C2 - C12) / (C2 - 4C1)
- Substituting the given values, we get: Cs = (4 * 30 * 60 - 302) / (60 - 4 * 30) = 20 PF.
Therefore, the self capacitance of the coil works out to be 20 PF.
Answer: D. 20 PF
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 66

The deflection of hot wire instrument depends on

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 66

The deflection of hot wire instrument depends on RMS value of alternating current.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 67

Which one of the following types of instruments can be used to determine the rms value of a.c.voltage of high magnitude (10 KV) and of any wave shape?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 67

Electrostatic Instruments are used for voltage measurements only

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 68

Eddy current damping is not used on type instruments because

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 69

Least expensive instrument for dc measurement is

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 70

A Meggar is basically a

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 71

The economic size of a conductor is determined by

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 71

"The Kelvin's law states that the most economical size of a conductor is that for which annual interest and depreciation on the capital cost of the conductor is equal to the annual cost of energy loss."

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 72

For a good voltage profile under load condition a long line needs

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 72

Shunt capacitive compensation. This method is used to improve the power factor. Whenever an inductive load is connected to the transmission line, power factor lags because of lagging load current. To compensate, a shunt capacitor is connected which draws current leading the source voltage.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 73

An industrial consumer has a daily load pattern of 2000 kw, 0.8 p.f. for 12 Hrs and 1000 kw UPF 12 Hrs. The load factor is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 73

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 74

The positive sequence reactance will be equal to  the negative sequence reactance.. the equipment is

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 75

In overhead transmission lines the effect of capacitance can be neglected when the length of line is less than

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 75

Introduction:
In overhead transmission lines, the effect of capacitance can be neglected when the length of the line is less than a certain threshold. The capacitance of the line contributes to reactive power flow and voltage drop, and neglecting it simplifies the analysis of the transmission system.
Determining the threshold length:
To determine the length at which the effect of capacitance can be neglected, we need to consider the characteristics of the transmission line and the impact of capacitance.
Capacitance in transmission lines:
When transmission lines are energized, a capacitance is formed between the conductors and the ground. This capacitance allows for the flow of displacement current, which contributes to the reactive power flowing through the line.
Effects of capacitance:
The presence of capacitance in a transmission line results in:
1. Voltage drop: The reactive power flowing through the capacitance causes a voltage drop along the line.
2. Power loss: The reactive power flow also leads to power loss due to the charging and discharging of the capacitance.
Threshold length:
To neglect the effect of capacitance, the length of the transmission line must be relatively short. This is because the impact of capacitance becomes more significant with increasing line length.
Given options:
Based on the given options, the length of the line at which the effect of capacitance can be neglected is less than 80 km.
Conclusion:
Therefore, the correct answer is option D: 80 km.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 76

Sparking occurs when a load is switched off because the circuit has

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 76

Wires have inductance associated with it. When load is disconnected current through the wire changes instantaneously. Now inductors can not allow a sudden change in current through them. As a result high voltage grows at the open end of the wire causes breakdown in air. For this there is sparking. Spark plug operates under same principle. Although value of inductance is higher.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 77

Which relay is used for feeders?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 77
Feeders and Relays:
Feeders are electrical circuits that transmit power from the source to the load. In order to protect these feeders from faults and overloads, relays are used. Relays are devices that sense abnormalities in the circuit and initiate actions to isolate the faulty section.
Types of Relays:
There are different types of relays used for various purposes in electrical systems. Some common types of relays include:
1. Overcurrent Relay: These relays sense the current flowing through the circuit and operate when the current exceeds a predetermined value.
2. Differential Relay: These relays compare the current entering and leaving a particular section of the circuit, and if there is a difference, it indicates a fault.
3. Directional Relay: These relays are used to detect the direction of current flow and operate when the current flows in the opposite direction.
4. Distance Relay: These relays measure the impedance or distance to a fault in the circuit and operate when the impedance exceeds a certain value.
Relays for Feeders:
Among the various types of relays, the most commonly used relay for feeders is the Translay relay. The Translay relay is a combination of an overcurrent relay and a directional relay. It provides both overcurrent protection and directional fault sensing for feeders.
The Translay relay is ideal for protecting distribution feeders as it can detect faults and initiate actions to isolate the faulty section. It operates based on the direction of current flow and the magnitude of the current, ensuring effective protection for the feeders.
Conclusion:
In conclusion, the Translay relay is the relay commonly used for feeders. It combines the features of an overcurrent relay and a directional relay, providing effective protection for distribution feeders.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 78

Which of the following is a conducting medium for electric current?

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 79

Dielectric strength of SF6 is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 79

Correct Answer :- b

Explanation : The dielectric strength which is 30 % less than that of oil at atmospheric pressure increases rapidly with increase of pressure.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 80

Thermal relays are often used in

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 80
Thermal Relays
Thermal relays are protective devices used in electrical systems to monitor and control the temperature of equipment. They are specifically designed to protect motors from overheating by monitoring the motor's current and temperature. When the temperature exceeds a certain limit, the thermal relay trips and interrupts the power supply to the motor, preventing any damage.
Applications of Thermal Relays
Thermal relays find their applications in various electrical systems, including:
1. Motor Starters: Thermal relays are commonly used in motor starters to protect the motors from overheating. They help prevent damage due to prolonged overloading or excessive current flow.
2. Generator Protection: Thermal relays are also used in generator protection systems. They monitor the generator's current and temperature and provide overload protection by tripping if the conditions exceed the specified limits.
3. Transformer Protection: Another application of thermal relays is in transformer protection. They ensure that the transformers operate within safe temperature limits and trip if there is an abnormal temperature rise that could lead to transformer failure.
4. Other Applications: While motor starters, generator protection, and transformer protection are the most common applications of thermal relays, they can also be used in other electrical systems that require temperature monitoring and protection.
Conclusion
In conclusion, thermal relays are versatile protective devices used in various electrical systems. They play a crucial role in safeguarding motors, generators, and transformers from overheating and potential damage. Therefore, option C, "motor starters," is the correct answer.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 81

Differential protection principle is used in the protection of

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 81

Differential relay is very commonly used relay for protecting transformers and generators from localised faults. Differential relays are very sensitive to the faults occurred within the zone of protection but they are least sensitive to the faults that occur outside the protected zone.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 82

A cable carrying alternating current has

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 82
Explanation:
The cable carrying alternating current has different types of losses. Let's break down the losses mentioned in the answer choices:
1. Hysteresis losses: These losses occur due to the magnetic properties of the materials used in the cable. When the magnetic field changes direction with the alternating current, the material experiences hysteresis and dissipates energy in the form of heat.
2. Leakage losses: These losses occur due to the leakage of current from the cable. When the cable is not perfectly insulated, some current may escape and be lost as heat.
3. Copper losses: These losses occur due to the resistance of the copper conductor in the cable. As current flows through the cable, some energy is lost as heat due to the resistance of the copper.
4. Friction losses: Friction losses are not typically associated with the transmission of electricity through a cable. This option is incorrect.
Based on the above explanations, we can conclude that the correct answer is option B: hysteresis and leakage losses only.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 83

Cables for 220 KV lines are invariably

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 83

Introduction:
In the given question, we are asked to identify the type of insulation used for 220 KV lines. Let's evaluate the options and determine the correct answer.
Evaluation:
Among the given options (A, B, C, and D), we need to choose the one that is invariably used for 220 KV lines. Let's analyze each option:
1. Paper Insulated:
- Paper insulation is commonly used for low voltage applications.
- It is not suitable for high voltage transmission lines like 220 KV.
- Hence, option A is incorrect.
2. Rubber Insulated:
- Rubber insulation is mainly used for flexible cables and low voltage applications.
- It is not suitable for high voltage transmission lines.
- Therefore, option B is incorrect.
3. Mica Insulated:
- Mica insulation is used for high voltage applications, but it is not commonly used for 220 KV lines.
- While it can be used for 220 KV lines in certain cases, it is not invariably used for all 220 KV lines.
- Hence, option C is incorrect.
4. Compressed Oil or Compressed Gas Insulated:
- Compressed oil or compressed gas insulation is commonly used for high voltage transmission lines, including 220 KV lines.
- It provides effective insulation and minimizes the risk of electrical breakdown.
- This type of insulation is commonly used for long-distance transmission lines where high voltages are involved.
- Therefore, option D is the correct answer.
Conclusion:
Based on the evaluation of the given options, we can conclude that the correct answer is option D: Compressed oil or compressed gas insulated. This type of insulation is invariably used for 220 KV lines to ensure efficient and safe transmission of high voltage electricity.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 84

In a cable, the voltage stress is maximum at

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 84
Explanation:
The voltage stress in a cable refers to the maximum voltage that the cable can withstand before it breaks down or experiences electrical failure. The location of the maximum voltage stress depends on the design and construction of the cable. In the case of a typical cable, the voltage stress is maximum at the core of the conductor.
Reasoning:
- The core of the conductor is the central part of the cable where the current flows. It is usually made of a conducting material such as copper or aluminum.
- The voltage stress is maximum at the core because this is where the electric field is concentrated. The electric field is the force that causes the flow of current in the conductor.
- The surface of the conductor, sheath, and insulator also experience voltage stress, but it is lower compared to the core.
- The surface of the conductor is in contact with the surrounding environment and may experience some voltage stress due to external factors such as humidity or temperature.
- The sheath is the protective layer that surrounds the conductor and provides insulation. It also experiences some voltage stress, but it is lower compared to the core.
- The insulator is the material that separates the conductor from other conductive parts or the ground. It also experiences some voltage stress, but it is lower compared to the core.
Overall, the voltage stress is highest at the core of the conductor in a cable.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 85

When diameter of the core and cable is doubled the value of capacitance

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 86

Which of the following method or technique can be used for the measurement of high dc voltages?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 86
Measurement of High DC Voltages
There are several methods and techniques that can be used for the measurement of high DC voltages. One of these methods is the generating voltmeter. The generating voltmeter works by generating a known voltage that is proportional to the voltage being measured. This known voltage is then compared to the voltage being measured to determine its value.
Another method that can be used is the electrostatic voltmeter. The electrostatic voltmeter measures voltage by using the force between two charged plates. As the voltage being measured increases, the force between the plates also increases, allowing for the measurement of high DC voltages.
The peak voltmeter is also a method that can be used for the measurement of high DC voltages. The peak voltmeter measures the peak value of the voltage being measured. This method is commonly used for measuring high voltage pulses.
In summary, the measurement of high DC voltages can be done using various methods and techniques. These include the generating voltmeter, electrostatic voltmeter, and peak voltmeter. Each method has its advantages and is suitable for different applications.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 87

For the same voltage boost, the reactive power capacity is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 87

Reactive Power Capacity for Shunt Capacitor:
- Shunt capacitors are connected in parallel with the load.
- The purpose of a shunt capacitor is to provide reactive power to the load and improve the power factor.
- When a shunt capacitor is connected, it draws reactive power from the system and supplies it to the load.
- Shunt capacitors have a higher reactive power capacity because they directly supply reactive power to the load.
Reactive Power Capacity for Series Capacitor:
- Series capacitors are connected in series with the load.
- The purpose of a series capacitor is to provide voltage boost and improve the power factor.
- When a series capacitor is connected, it increases the voltage across the load, resulting in a voltage boost.
- However, series capacitors do not directly supply reactive power to the load.
- The reactive power capacity of series capacitors depends on the voltage boost they provide.
Comparison:
- Shunt capacitors have a higher reactive power capacity because they directly supply reactive power to the load.
- Series capacitors do not directly supply reactive power to the load, so their reactive power capacity is lower compared to shunt capacitors.
Conclusion:
For the same voltage boost, the reactive power capacity is more for shunt capacitors compared to series capacitors. Therefore, the correct answer is option A: more for shunt capacitor.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 88

With the help of a reactive compensator it is possible to have

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 88
Reactive Compensator:
A reactive compensator is a device used in electrical systems to improve power factor and voltage regulation. It is designed to compensate for the reactive power, which is the power that oscillates between the source and load due to the inductive or capacitive nature of the load.
Constant Voltage Operation:
Constant voltage operation means maintaining a stable voltage level regardless of the changes in the load or system conditions. It is important to have a constant voltage to ensure proper functioning of electrical equipment and to prevent damage.
Unity Power Factor Operation:
Unity power factor operation means achieving a power factor of 1, which indicates that the reactive power is fully compensated, and the power flow is purely active power. A unity power factor is desirable as it minimizes power losses and improves the overall efficiency of the electrical system.
Benefits of a Reactive Compensator:
A reactive compensator provides several benefits, including:
1. Improved Voltage Regulation: By compensating for the reactive power, a reactive compensator helps in maintaining a stable voltage level even under varying load conditions.
2. Power Factor Correction: A reactive compensator helps in improving the power factor of the system by supplying or absorbing reactive power as required. This results in reduced power losses and improved energy efficiency.
3. Flexibility: A reactive compensator can be adjusted to provide the necessary reactive power compensation based on the system requirements. This allows for better control and optimization of the power factor and voltage regulation.
4. Stability and Reliability: With a reactive compensator, the voltage and power factor of the system can be maintained within acceptable limits, ensuring stable and reliable operation of electrical equipment.
Conclusion:
In conclusion, a reactive compensator provides both constant voltage and unity power factor operation. It helps in maintaining a stable voltage level and improving the power factor of the system. By compensating for the reactive power, it ensures efficient and reliable operation of electrical equipment. Therefore, the correct answer is option D: either constant voltage or unity p.f.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 89

A shunt reactor at 100 MVAr is operated at 98% of its rated voltage and at 96% of it srated frequency. The reactive power absorbed by the reactor is

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 89

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 90

The junction capacitance of a p-n junction depends on

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 91

In p+n junction diode under reverse bias, the magnitude of electric field is maximum at

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 91

Electric field is always maximum at the junction

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 92

How can the channel width in a junction field effect transistor be controlled?

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 92
Channel Width Control in a Junction Field Effect Transistor
There are several methods for controlling the channel width in a junction field effect transistor (JFET). However, the correct answer to the given question is option A: by two back biased p-n junctions. Let's explore this in more detail:
1. Introduction to JFET:
- A JFET is a three-terminal semiconductor device that utilizes the electric field control of a junction formed between a doped semiconductor material (p or n-type) and a non-doped semiconductor material (intrinsic).
- The JFET operates in depletion mode, meaning that the channel is formed without any external voltage applied.
2. Channel Width:
- The channel width in a JFET refers to the distance between the source and drain terminals through which the current flows.
- The channel width directly affects the conductivity and therefore the performance of the JFET.
3. Control of Channel Width:
- The channel width in a JFET can be controlled by varying the width of the depletion region formed by the two back biased p-n junctions.
- When a voltage is applied across the gate and source terminals, the p-n junctions become reverse-biased.
- The reverse biasing widens the depletion region, reducing the channel width and thus controlling the current flow.
4. Back Biasing:
- Back biasing refers to the application of a voltage across the gate and source terminals to control the channel width in a JFET.
- By adjusting the back bias voltage, the width of the depletion region and hence the channel width can be controlled.
- A higher back bias voltage increases the width of the depletion region, narrowing the channel width and reducing the current flow.
In conclusion, the channel width in a junction field effect transistor (JFET) can be controlled by two back biased p-n junctions. By adjusting the back bias voltage, the width of the depletion region and thus the channel width can be varied, allowing for control over the current flow through the device.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 93

A Zener diode works on the principle of

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 93

Due to zener effect in reverse bias under high electric field strength, electron quantum tunneling occurs. It’s a mechanical effect in which a tunneling current occurs through a barrier. They usually cannot move through that barrier.

Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 94

As compared to full wave rectifier using two diodes, the four diode bridge rectifier has the dominant advantage of

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 94
Comparison of Full Wave Rectifier and Bridge Rectifier:
1. Full Wave Rectifier:
- Uses two diodes to rectify the AC input signal.
- The diodes are connected in a center-tapped configuration.
- The output waveform has less ripple compared to a half-wave rectifier.
- Requires a transformer with a center-tapped secondary winding.
- The peak inverse voltage requirement for each diode is equal to the peak input voltage.
2. Bridge Rectifier:
- Uses four diodes to rectify the AC input signal.
- The diodes are connected in a bridge configuration.
- The output waveform has less ripple compared to a full wave rectifier.
- Does not require a center-tapped transformer, can use a regular transformer.
- The peak inverse voltage requirement for each diode is equal to half of the peak input voltage.
Advantages of Bridge Rectifier over Full Wave Rectifier:
The four diode bridge rectifier has the dominant advantage of lower peak inverse voltage requirement. This means that the diodes in the bridge rectifier can handle lower voltage ratings compared to the diodes in a full wave rectifier. This advantage is significant because lower voltage rated diodes are generally more cost-effective and readily available in the market. Additionally, lower voltage rated diodes tend to have better performance characteristics such as lower forward voltage drop and faster switching times, which can result in higher efficiency and better overall performance of the rectifier circuit.
Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 95

In a transistor, the forward bias across the base emitter junction is kept constant and the reverse bias across the collector base junction is increased. Neglecting, the leakage across the collector base junction & the depletion region generation current, the base current will

Detailed Solution for Technical Test SSC JE: Electrical Engineering (EE)- 1 - Question 95
Explanation:
When a transistor is in active mode, the base-emitter junction is forward biased and the collector-base junction is reverse biased. In this scenario, if the forward bias across the base-emitter junction is kept constant and the reverse bias across the collector-base junction is increased, the following will happen:
- The reverse bias across the collector-base junction increases the width of the depletion region in this junction.
- As the width of the depletion region increases, the electric field across it also increases.
- The increased electric field across the depletion region causes a higher potential barrier for the majority charge carriers (electrons in an NPN transistor) to cross from the collector to the base region.
- This results in a decrease in the collector current, as fewer electrons are able to cross the depletion region and reach the collector.
- According to Kirchhoff's Current Law, the sum of the currents entering the base region (base current) and the collector region (collector current) should be equal. Therefore, the decrease in the collector current leads to a decrease in the base current as we