Some of the frequently asked interview questions on electrical engineering are as follows:
Q. 1. What is ohm’s law?
Ans. Temperature remaining constant the current flowing between two points of a conductor is directly proportional to the potential difference between those two points of the conductor.
Thus, if a current of I-ampere flows through a conductor having a potential difference of V-volt between its terminals, then at constant temperature,
V ∝
I,
Or V = R I,
where R is a constant, called the Resistance of the conductor. R opposes the flow of current through the conductor, and hence it is called resistance. The unit of resistance is Ohm.
The resistance of a conductor depends on its length, cross-section, material and temperature.
Q. 2. What is the effect of temperature on Resistance?
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Ans. (i) Pure metals increase in resistance with the rise in temperature.
(ii) The resistance of most alloys increases very slightly with the rise in temperature. The resistance alloys used in electrical work have practically constant resistance at all temperatures.
(iii) Corbon, Insulators and Electrolytes decrease in resistance with the rise in temperature. (An electrolyte is a solution which conducts electricity, e.g. water containing sulphuric acid).
The resistance of an insulator decreases considerably, i.e., it becomes a worse insulator with the rise in temperature.
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Q. 3. What is insulation resistance?
Ans. The resistive power which prevents leaking of current through the insulation is known as Insulation Resistance. The greater the insulation resistance, the better is the condition of the circuit. However, no insulator can prevent leaking of current through it absolutely. Moreover, its insulating property deteriorates due to wear off and due to effects of various weather conditions. Therefore leakage of current through insulation goes on increasing with time.
Insulation resistance is measured in meg-ohm (1 meg-ohm= 1,000,000 ohms). When the leakage current is negligible, the insulation resistance of a circuit is ‘infinity’. This resistance is measured in kilo-ohm or even in ohm as the leakage of current gradually increases. Ultimately when the circuit reaches short-circuit condition, the insulation resistance becomes zero.
The insulation resistance of a cable depends on:
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(i) Length of the cable.
(ii) Thickness of insulation,
(iii) Material of insulation, and
(iv) Temperature of insulation material.
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Here thickness of insulation corresponds to length of conductor, i.e., greater the thickness, more is the resistance of insulation. Also length of cable corresponds to cross-section of conductor. A longer cable has a greater cross-section of insulation and therefore a smaller Insulation resistance.
Q. 4. What is force? State the relationship between different units of force?
Ans. A force is a push or pull which produces or tends to produce, stops or tends to stop motion in a body.
Force = mass x acceleration.
The unit of force depends upon the units of mass and acceleration. The unit of force is dyne, poundal and newton in C.G.S., F.P.S. and M.K.S. (S.I) systems respectively.
In GG.S system the unit of mass is 1 gm and that of acceleration is 1 cm/sec2. Therefore, one dyne is that force which produces an acceleration of 1 cm/sec2 when acting on a mass of 1 gm.
In F.P.S. system that unit of mass is 1 lb and that of acceleration is 1 ft/sec2. Hence, one poundal is that force which produces an acceleration of 1 ft/sec2 in a mass of 1 lb.
In S.I. or M.K.S. system the unit of mass is 1 Kg. and that of acceleration is 1 m/sec2. Hence, one newton is that force which produces an acceleration of 1 m/sec2 when acting on a mass of 1 Kg.
Relationship:
The relationship between different units of force is shown below:
(i) Newton and dyne:
1 newton = 1 kg x lm/sec2
= 1000 gm x 100cm/sec2
= 105 dynes.
(ii) Newton and poundal:
1 newton = 105 dynes
=105/13825 poundals
( ... 1 poundal = 13825 dynes)
= 7.23 poundals.
(iii) Poundal and dyne:
1 poundal = 1 lb x 1 ft/sec2
= 453.6 gm x 12x 2.54 cm/sec2
= 13825 dynes
Q. 5. What is weight? State the relationship between different units of weight?
Ans. The force with which a body is attracted towards the centre of earth is called weight of the body.
Weight = mass of body x acceleration due to gravity
= m x g,
where m is the mass of the body and g is the acceleration due to gravity. The value of g is 9.81 m/sec2.
Gram weight, Kilogram weight and pound weight are called the gravitational units of weight in C.G.S., S.I. (or M.K.S.) and F.P.S systems respectively.
Gram weight is the weight of 1 gm of a body. Likewise, Kg. weight is the weight of 1 kg of a body and lb. weight is the weight of 1 lb of a body.
Relationship:
The relationship between different units of weight and force used in various systems is shown below:
(i) Kg. weight and newton:
1 Kg. weight = 1 Kg x g
= 1Kg x 9.8 lm/sec2
= 981 newtons
(ii) Gram weight and dyne:
1 gm. weight = 1 gm x g
= 1 gm x 981 cm/sec2
= 9.81
= 1 gm x 96 = 9.81 dynes.
(iii) Pound weight and poundal:
1 lb weight = 1 lb x g
= 1 lb x 32.2 ft/sec2
= 32.2 poundals.
Q.6. What is work? State the relationship between different units of work?
Ans. Work is said to be done when a force acting on a body causes the body to move.
Work done = force x distance moved by point of application of force. The unit of work depends upon the units of force and distance. The unit of work is erg in C.G.S. system, foot-poundal in F.P.S. system and newton-metre (or joule) in S.I. (or M.K.S.) system.
In C.G.S. system the unit of force is 1 dyne and that of distance is 1 cm. Hence, work done on a body is 1 erg if a force of 1 dyne moves it through a distance of 1 cm.
In F.P.S. system the unit of force is 1 poundal and that of distance is 1 foot. Therefore, work done on a body is 1 foot-poundal if a force of one poundal moves it through a distance of 1 foot.
In S.I. (or M.K.S.) system the unit of force is 1 newton and that of distance is 1 metre. Hence, work done on a body is 1 newton-metre (or joule) if a force of 1 newton moves it through a distance of 1 metre.
Relationship:
(i) Joule and erg:
1 joule = 1 newton x 1 metre
= 105 dynes x 100 cm
= 107 ergs.
(ii) Foot lb. and joule:
1 ft. lb = 1 lb. weight x 1 foot
= 4.45 newton x 0.3048 metre
= 1.356 joules.
Q. 7. What is power? How can we derive electrical power absorbed by a circuit?
Ans. Power is the rate of doing work.
Power = Work done/time
Obviously, the unit of power depends upon the units of work and time. Erg/Sec is the unit of power in C.G.S. system, foot-poundal/sec in F.P.S. system and newton-metre /sec or joule/sec in S.I. (or M.K.S.) system.
The electrical unit of power is Watt. The watt is the power in a circuit when a p.d. of 1 volt causes 1 ampere to flow through the circuit. It may also be defined as the amount of power in a circuit when work is being done at the rate of 1 joule per second.
If a p.d. of V-volt is applied across a circuit having a resistance of R-ohm and carrying a current of I-ampere, the electrical power absorbed by the circuit is given by:
P = VI watt = I2R watt (... V = IR volt )
V2/R watt (... I = V/R ampere )
A larger unit of power is kilowatt (KW).
1 kilowatt = 1000 watt.
Another unit of power is also in use. This is known as Horse Power.
Horse Power (British):
It is the British practical unit of power and is defined as the rate of doing 550 ft-lbs of work per second.
1 H.P. (British) = 550 ft. lbs. /sec.
= 746 joules/sec.
= 746 watts
1 kilowatt = 1.34 H.P. (British)
Horse Power (Metric):
It is the practical unit of power in M.K.S. system and is defined as the rate of doing 75 Kilogram-metre of work per second.
1 H.P. (Metric) = 75 Kg. m./sec.
= 735.5 joules/sec.
= 735.5 watts.
1 kilowatt = 1.36 H.P. (Metric)
Q. 8. What is meant by leading and lagging vector quantities?
Ans. The quantities having both magnitude as well as direction are called vector quantities, while those having only magnitude but no direction are called scalar quantities.
A vector is usually represented by a straight line. This is called vector diagram. The direction along which the line is drawn indicates the direction of the vector, while the length of the line is a measure of its magnitude. Electric current, electric pressure and electric power are all vector quantities.
Therefore they are represented by a vector diagram. Since electric current and electric pressure are sinusoidal, their relative positions may be shown in a wave diagram also. These relative positions of the vectors can be readily and precisely understood if both the vector and the wave diagrams for an a.c. circuit are drawn side by side.
Vectors are supposed to rotate along anti-clockwise direction which is the positive direction of rotation. Clockwise direction is considered to be the negative direction of rotation. When one vector is ahead of another along anti-clockwise direction, the former is said to be a leading vector and the latter is said to be a lagging vector.
In figure 22 (a) two vectors OA and OB are shown. OA is ahead of OB by an angle θ along anti-clockwise direction. Therefore OA is a leading vector and OB is a lagging vector. OA is said to lead OB by an angle θ or OB is said to lag behind OA by an angle θ. Here θ is the angle of phase difference or phase angle between OA and OB. The wave diagram of the two vectors is shown in figure 22 (b). From this diagram the relative positions of the two vectors and their instantaneous values at every instant can be readily determined.
Q. 9. State the bad effects of low power factor.
Ans. The power absorbed by an a.c. circuit is given by the product of three factors—supply pressure across the circuit, current flowing through the circuit and the power factor of the circuit. Supply pressure is usually maintained constant. Hence, for a given amount of power, as the power factor decreases, the current drawn by the circuit increases in the same proportion.
This has the following bad effects on the machines and supply system:
(i) To obtain certain power at a low power factor, one has to increase the total current. This results in increased resistive losses or I2R losses so that the efficiency obviously decreases.
(ii) Low power factor limits the output of generators and transformers because of the high currents drawn from these machines which results in a rise in temperature.
(iii) Low power factor increases the voltage drop in the transmission and distribution lines resulting in poor voltage regulation of the supply system.
Causes of Low Power Factor:
The cause of low power factor is the addition of more and more inductive loads in the circuits. Loads in the a.c. circuits are mostly inductive having high inductance and comparatively low resistance. Hence, with the addition of more inductive load, the phase angle between applied voltage and the circuit current increases. As a result the power factor which is the cosine of the phase angle decreases.
Q. 10. What is meant by inductance?
Ans. When electric current flows through a conductor or a coil, a circular magnetic field is set up in and around that coil, the magnitude and direction of the field depending on current. If the current is alternating, the flux will be pulsating. This pulsating flux, when links with the coil, induces an e.m.f. in the coil. This induced e.m.f. is called e.m.f. of self-induction or self-induced e.m.f. According to Lenz’s Law e.m.f. of self-induction is found to oppose the current flowing through the coil.
The magnitude of the self-induced e.m.f. depends on two factors:
(i) Rate of change of current in ampere per second,
(ii) Inductance of the coil in henry.
Inductance of the coil may be defined as the number of weber-turns per ampere linking with the coil. Its unit is henry. Thus, if a coil of inductance L henry carries a current of i ampere for t second, and if the magnetic field produced in and around the coil has a flux of Φ weber, then,
where N is the number of turns of the coil.
Rate of Change of Current = i/t ampere per second, and e.m.f. of self-induction = – (Li/t) volt.
The (-) negative sign indicates that it is an opposing e.m.f.
Q. 11. Differentiate between the use of capacitor in D.C. and A.C. circuits.
Ans. When a Capacitor is connected across a d.c. supply, there cannot be a continuous flow of current because the circuit is broken by the dielectric. Initially there is a charging current, but this current rapidly decreases and finally ceases when the voltage across the capacitor becomes equal and opposite to the applied voltage.
But the effect of connecting a capacitor is quite different in an a.c. circuit. When a capacitor is connected across an a.c. supply, no such steady condition exists. The capacitor is charged and discharged during alternate quarter cycles, and these charge and discharge currents give rise to an alternating current in the circuit.
Q. 12. What is meant by reactance?
Ans. When electric current flows through a conductor or a coil, not only the resistance of the coil opposes the flow of current through it. Its inductance also, in conjunction with supply frequency, offers an opposition to the flow of current. This opposition is known as Inductive reactance of the coil. It is usually denoted by XL and its unit is ohm. Thus, if the inductance of a coil be L henry and if the frequency of supply be f hertz, the inductive reactance of the coil
XL = 2πfL ohm.
Likewise, if a capacitor be connected in the circuit, its capacitance, in conjunction with supply frequency, offers an opposition to the flow of current through the circuit. This opposition is called Capacitive Reactance of the capacitor. It is usually denoted by XC and its unit is also ohm.
Thus, if a capacitor of capacitance C farad is connected in a circuit where the supply frequency is f hertz, the capacitive reactance of the capacitor:
XC= 1/2πfC ohm.
Inductive reactance and capacitive reactance not only oppose the flow of current through an a.c. circuit, but they also cause an angle of phase difference between the applied voltage and the circuit current. Inductive reactance causes the current to lag behind the applied voltage by 90°, while the current tends to lead the applied voltage by 90° due to capacitive reactance.
Thus, it is evident that the inductive reactance and capacitive reactance act in opposition to each other. Therefore, when a coil and a capacitor are connected in series, the resultant reactance acting in the circuit is the difference between inductive reactance and capacitive reactance.
If the resultant reactance be denoted by X, then:
X = XL – XC ohm
(i.e., when XL is greater than XC, X = XL – XC ohm; when XC is greater than XL, X = XC – XL ohm).
Q. 13. What is meant by impedance?
Ans. Impedance of an alternating current circuit is the opposition to the flow of current offered jointly by the resistance and reactance of the circuit. It is usually denoted by ‘Z’ and its unit is ohm.
If an a.c. circuit has a resistance of R ohm and a reactance of X ohm, the impedance of the circuit
The current flowing through the circuit is given by the voltage applied across the circuit divided by the impedance of the circuit. If a voltage of V volt is applied across an a.c. circuit carrying a current of I ampere, then
I = V/Z ampere.
This shows that for a given applied voltage as the impedance increases, the current flowing through the circuit decreases.
Q. 14. What is meant by skin effect?
Ans. Skin effect is the tendency of an alternating current to flow along the outer layers of a conductor rather than through the inner core. A conductor may be regarded as consisting of a number of filaments, each carrying part of the total current and each producing around itself a magnetic field which links with other filaments.
The flux linking the central filaments is considerably greater than that linking the filaments near the surface. As a result a conductor has much greater self-inductance at the centre. This causes the current concentrating in the outer layers.
Due to skin effect since practically no current flows along the centre of a conductor, the effective cross-section of a conductor is reduced. Hence, for the same conductor effective resistance is greater to a.c. than to d.c.
Q. 15. Explain how a coil can be non-inductive coil.
Ans. When a conductor is coiled with a number of turns, it will have inductance. Alternating current flowing through a coil produces a pulsating magnetic field in and around the coil. This magnetic field is responsible for producing self-inductance of the coil. If the magnetic field disappears, the coil will be non-inductive.
The magnetic field can be neutralized in the following way:
A conductor is plaited at the middle and the two halves are coiled together as shown in figure 23. Supply is given at the open terminals, and the current flows through one half of the conductor in one direction and through the other half in the opposite direction.
As a result no magnetic field is set up in and around the coil, since the flux produced by the current flowing through one half neutralizes the flux produced by the current flowing through the other half of the conductor. Hence, the coil is non-inductive in this case.
Q. 16. Explain magnetising force or magnetic field intensify.
Ans. Magnetising force or magnetic field intensity may be defined as the m.m.f. per unit length along the path of the magnetic flux. It is analogous to fall of potential (i.e. voltage drop) per unit length along the path of electric current in an electric circuit.
When m.m.f. is measured in ampere-tums and the path of the magnetic flux in metre, field intensity is measured in ampere- turns per metre. Thus, for a magnetising coil of N turns carrying a current of I ampere, field intensity,
H = NI/l ampere-turns per metre (AT/m),
where l is the length of the magnetic circuit in metre.
Q. 17. What is reluctance?
Ans. Reluctance is the opposition offered by the magnetic circuit to the flow of magnetic flux through it. It can be compared to resistance which is the opposition to the flow of electric current in an electric circuit.
Reluctance depends upon:
(i) Material used for the magnetic core,
(ii) Cross-sectional area of the material through which flux is passing,
(iii) Length of the magnetic path.
Thus, if l be the length of the magnetic circuit in metre, A be the cross-sectional area of the material in square metre through which the flux is passing, μr be the relative permeability of the material and μ0 be the permeability of free space, then reluctance
S = l/μ0μrA
Reluctance may also be defined as the ratio of the drop in m.m.f. to the flux produced in the magnetic circuit, just as resistance of any conductor in an electric circuit is the ratio of the potential drop to the current flowing through the conductor. If m.m.f. is measured in ampere- turns and flux in weber, reluctance is measured in ampere-turns per weber.
Let a coil of N turns carrying a current of I ampere produces a flux of Φ weber.
Then reluctance of the magnetic circuit is given by:
S = NI/Φ ampere-turns per weber (AT/wb).
Q. 18. What do you mean by permeance of a magnetic circuit?
Ans. Permeance of a magnetic circuit is the flux produced by unit m.m.f. of that circuit. It is the reciprocal of reluctance and analogous to conductance in an electric circuit. If flux is expressed in weber and m.m.f. in ampere-turn, permeance is expressed in weber per ampere-turn. Thus, permeance,
P = flux/ m.m.f. = Φ/NI weber per ampere-turn (wb/AT).
Q. 19. What is magnetic flux?
Ans. Magnetic flux may be defined as the total number of lines of force in a magnetic field. It is the stress in any medium due to a magnet.
Magnetic flux is assumed to be consisting of closed magnetic lines of force which proceed from North Pole towards South Pole in the magnetic field outside the magnet, and from South Pole towards North Pole within the magnet itself. It is analogous to electric current in an electric circuit.
The unit of magnetic flux is weber, and 1 weber = 108 maxwells.
Thus, magnetic flux,
Q. 20. What is flux density?
Ans. The flux density is the number of lines of force per unit area of the magnetic field, the area being taken at right angles to the direction of flux. It is usually denoted by B and is expressed in weber per square metre or tesla.
If the flux in a magnetic field be Φ weber and the cross-sectional area of the field be A square metre, then,
Q. 21. What is magnetic leakage?
Ans. In most of the magnetic circuits flux is to pass through an air-gap for certain distance in order that it can be utilised for different purposes. Whenever magnetic flux crosses into an air-gap, particularly where the gap is long, a certain number of lines of force is always lost due to leakage, i.e. these lines of force cannot be utilised as they take the path outside the air-gap. The flux that does not follow the intended path in a magnetic circuit is called a leakage flux. Thus,
total flux = flux in the iron path
= flux in the air-gap + leakage flux.
Flux in the air-gap is called useful flux, since it can be utilised for useful purposes.
The ratio of the total flux to. useful flux is called Leakage Factor or Co-efficient of Magnetic Leakage. It is usually denoted by λ and it varies between 1.1 to 1.5.
... λ = total flux/useful flux or flux in the air-gap
Q. 18. State the factors on which the strength of an electro-magnet depends.
Ans. The strength of an electro-magnet Is determined by the number of lines of force produced by it. The greater the number of line of force, the stronger is the electro-magnet. Hence, the factors on which the number of lines of force depends also determine the strength of an electro-magnet.
The flux produced by an electro-magnet is given by:
Φ = NI x (μ0μrA/l)
Therefore the factors on which the strength of an electro-magnet depends are as follows:
(i) The number of turns of the magnetising coil (N) placed around the pole core.
(ii) The exciting current (I) flowing through the magnetising coil.
(iii) The relative permeability (μr) of the material of the magnetic core.
(iv) Cross-section (A) of the magnetic core.
(v) Length (l) of the magnetic path.
Q. 22. Why electro-magnets are used in electric machines and apparatus?
Ans. In electric machines and apparatus electro-magnets are used instead of permanent magnets due to following reasons:
(i) The e.m.f. induced in the armature of a generator and the speed developed by an electric motor can be controlled by changing the flux produced by the magnets. The flux produced by a permanent magnet is fixed and cannot be altered, but the flux produced by an electro-magnet can be increased or decreased according to requirements by altering the exciting current flowing through the magnetising coil.
(ii) The strength of a permanent magnet is gradually reduced with time. This causes difficulty for proper operation of an electric machine. But the magnetic core of an electro-magnet is fully excited as soon as the current flows through the magnetising coil.
(iii) For reversing the direction of rotation of a d.c. motor, sometimes it becomes necessary to change the polarities of the magnets. The polarities of a permanent magnet cannot be reversed, but those of an electro-magnet can be readily reversed by reversing the direction of the flow of current through the magnetising coil.
(iv) In order to minimise eddy-current losses, pole-shoes of the magnets used in electric machines must be made of laminated sheet steel. Permanent magnets have usually solid cores, while the core and the shoe of an electro-magnet are always made of sheet steel laminations.
Q. 23. Why magnetic cores of electro-magnets used in electric machines are laminated?
Ans. The core of an electro-magnet is made of iron or steel which is a conductor to electricity. If the core is moved relative to a magnetic field or if the flux linking with the core pulsates, an e.m.f. is induced in the core. This induced e.m.f. sets up currents, called eddy-currents which circulate in large numbers in closed paths if the core is made of solid mass of steel. Owing to eddy-current the core of the magnet becomes quickly heated.
The heat liberated by eddy-currents usually serves no useful purpose, but actually represents a loss of electrical energy. This energy loss in termed as eddy-current loss which can be reduced to a minimum or to a fairly low value by flaking it impossible for large currents to circulate.
Eddy-current can be reduced to a minimum by laminating the core and by using materials of high specific resistance. In order to increase the electrical resistance to the flow of eddy-cur- rents the surfaces of thin sheets or laminations are coated with some insulating material such as varnish or paper. In solid core induced eddy-currents find a direct path, whereas with the introduction of insulated laminations these current paths are broken up.
Thus, for minimising the eddy-current losses in the cores and in the pole faces, the electro-magnets used in electric machines have laminated pole cores.
Q. 24. What are the uses of electro-magnets?
Ans. Electro-magnets are usually used in electric machines and apparatus such as generators, motors, transformers, ammeters, voltmeters, wattmeter, electric fans, electric bells etc. For lifting heavy materials or structures made of iron or steel, sometimes a specially designed electro-magnet of large size is used. Such a magnet is called Lifting Magnet. An electro-magnet having a pointed pole is also used by medical practitioners to take out very small iron filings dropped in human eyes.
Q. 25. Differentiate between magnetic flux and flux density.
Ans. Flux is defined as the total number of lines of force in a magnetic field. It corresponds to current in an electric circuit.
Flux density is the number of lines of force per unit area of the magnetic field, the area being taken at right angles to the direction of flux. It corresponds to current density in electric circuit.
If Φ weber be the flux in a magnetic field having an area A square metre, the flux density of the field
B = (Φ/A) wb/m2 or tesla
Q. 26. What is earth plate or earth electrode?
Ans. The metallic pipe or the metallic plate placed in the earth pit at the end of earth lead is known as Earth Electrode or Earth Plate. Earth electrode or earth plate must be made of that metal with which the earth lead is made. Unless otherwise specified, an earth electrode means a galvanized iron pipe of suitable diameter and length or a galvanized iron plate of suitable size or a copper plate of suitable size. If copper plate is used, earth lead must also be made of copper.
When plate earth electrode is used, it should be made of galvanized iron plate of size 60 cm x 60 cm x 6.35 mm (2 ft x 2 ft x 1/4 in) and the earth lead must also be galvanized iron wire. If copper earth plate is used, its size should be 60 cm x 60 cm x 3.18 mm (2 ft x 2 ft x 1/8 in). In this case copper wire is to be used as earth lead. In both cases the plate is vertically placed in the earth pit at least 3 metres (10 ft) below the ground level.
Where pipe earth electrode is used, the size of the pipe should be at least 3.81 cm (1 1/2 inches) in diameter and 2 metres (6 ft) in length of galvanized iron pipe. For dry and rocky places this pipe may be as long as 2.75 metres (9 ft).
For medium pressure installation two separate and distinct earth electrodes are placed in two separate earth pits, and from each electrode a separate earth lead is drawn out and connected to earth bar or to the outer covering of the main switch. According to Indian Code of Practice the distance between two earth pits must not be less than 5 metres.
Q. 27. How do we determine the distance of earth pit from a building?
Ans. The distance of an earth pit from the nearest building shall not be less than 1.5 metres (5 ft).
Each and every earth wire shall be connected to an earth electrode. If the phase wire or live wire of a consumer’s circuit comes in contact with a bare conductor, the earthing arrangement shall be such that, the circuit current will rise up to 3 1/2 times the rated current of the main fuse or times the setting of the over load earth leakage circuit breaker.
As a result the fuse will melt or the circuit breaker will trip. After the completion of earthing arrangement, test shall be carried out to see that the arrangement satisfies this condition; otherwise an earth leakage circuit breaker is to be installed there. Besides, so long the leakage of current to earth continues, the potential difference between bare conductor and consumer’s earth terminal shall not exceed 40 volts (R.M.S.)
Q. 28. Explain earth continuity conductor.
Ans. The conductor by which an electric appliance or instrument or metal sheath of conductor of a wiring system remains connected to earthing lead is known as Earth Continuity Conductor. With the help of this wire earth continuity of a complete wiring system and earthing of appliances are maintained.
In case of conduit wiring the metal conduit and in case of metal sheathed wiring the Outer sheath may be used as earth continuity conductor. Similarly, armoring of an armored cable may also be used for the same purpose. But in case of cleat, casing, C.T.S. or P.V.C. wiring a separate earth wire is to be drawn everywhere along with the wiring line.
The following articles of a wiring system are to be connected to earth continuity conductor:
The outer metallic covers of electric supply line and appliances are to be earthed. In all junction boxes, switch boards and similar other places where a wiring line terminates, special arrangement must be provided to maintain electrical continuity of earth continuity conductor. Such conductor is sometimes kept within the same sheath along with current-carrying conductors, sometimes again within-the cover of flexible cord.
Q. 29. Explain the resistance of an earth continuity conductor including earth lead.
Ans. The combined resistance of metal conduit, metal sheath of wires, earth continuity conductor and main earth lead shall not exceed 1 ohm. . But excluding any added resistance or earth leakage circuit breaker measured from the connection with the earth electrode to any point in the earth continuity conductor in the completed installation shall not exceed one ohm.
The resistance of an earth electrode placed in an earth pit shall not exceed 5 ohms under normal condition. But in case of rocky soil resistance as high as 8 ohms may be allowed. There is no clear indication in Indian electricity Rules about what should be the maximum earthing resistance
The two formulae are as follows:
i. In case of d.c. supply:
Earthing Resistance = [(1/2 x Voltage to earth) / (2.5 x current-carrying capacity of the largest size of fuse wire or circuit breaker of the circuit)] ohm
ii. In case of a.c. supply:
Earthing Impedance Z = [Voltage to earth/ (minimum fusing current of fuse x factor of safety)] ohm
Here Z = allowable earth impedance. Its unit is ohm.
Note:
In order to determine earthing impedance, the value of the factor of safety will be chosen by the planner of the wiring circuit.
Q. 30. What are the articles not to be earthed?
Ans. The following articles need not be earthed:
(i) Small pieces of metal conduits used to protect small pieces of C.T.S. or P.V.C. wires;
(ii) Metal conduit pipes used as wail tubes for the protection of cleat wiring passing through partition wall;
(iii) Metal box covered with non-conducting substance and used in connection with C.T.S. or P.V.C. wiring;
(iv) Lamp caps;
(v) Name plate separated by insulating substance, screws and other small substances;
(vi) Metal chain (with which pendant lamp fitting is suspended) not touched with flexible cord of twisted type;
(vii) lamp fitting in a room the floor-of which is made of non-conducting substance (the fitting must remain at a height out of reach and away from any earthed body).
Besides Electrical Appliances the Articles to be Earthed:
When it is not possible to maintain an electrical wiring system isolated from water pipe, gas pipe, telephone wire etc., these must be properly connected to earth by means of bonding with earth continuity conductor.
The articles to be connected to earth through bonding in such cases are as follows:
(i) Bath tub, bare metallic pipe, water tank etc. of bath room;
[Note: It is essential to earth all metal appliances in bath room, kitchen, scullery, laundry, milking shed etc.]
(ii) If possible all iron or steel frame, joist, girder etc. coming within reach;
(iii) Electric crane, lift etc. moving on iron frame.
Q. 31. What is meant by final sub-circuit?
Ans. The circuit which is connected to a single ‘way’ of a sub-distribution board or fuse board for supplying current to one or more load points, is known as Final Sub-circuit. Supplier’s line or cable comes to energy meter through supplier’s sealed cut-out, and from the meter it goes to consumer’s main switch. This line is called Supply Main or Main Line.
But on account of heavy load in big factories or in large buildings, very often when the line is drawn from the main switch up to bus-bar chamber and from there up to different sub-bus-bar chambers or power distribution boards, the line from consumer’s main switch up to bus-bar chamber is also included within the main line. For this reason the bus-bar chamber that rises from the ground floor up to the top most floors in a multi-storeyed building is known as Rising Main.
If a distribution fuse board is installed next in sequence to consumer’s main switch, the line from the main switch up to distribution board is called Sub-Main Line. In case there is a bus-bar chamber after the main switch and the lines are drawn from bus-bar chamber to different sub-busbar chambers or power distribution boards, the line from bus-bar chamber to a sub-busbar chamber is considered to be a sub-main line.
Lines drawn from bus-bar chamber or sub-busbar chamber or power distribution board up to distribution fuse boards may be considered as Circuits. Every line which runs from distribution fuse board towards load points is looked upon as a Final Sub-circuit. Sometimes a line may go to a load point direct from a power distribution board a sub-busbar chamber (even from a bus-bar chamber) etc. In that case every line is regarded as a final sub-circuit.
The rule about final sub-circuit is as follows:
Every final sub-circuit must come out of a separate way of a distribution board. Where there is only one sub-circuit, it may be connected directly to the main switch board.
Wiring of every final sub-circuit will be completely separated from that of another final sub-circuit. To follow this rule, one neutral conductor is required for every circuit which can be on or off with a single-pole switch. Care must be taken to see that every pair of live and neutral wires are kept together properly in order in the distribution board tor the convenience of testing or disconnecting. Current flowing towards load points must not exceed the current-carrying capacity of wires used for the sub-circuit.
Q. 32. What is the use of plug point with lamp circuit?
Ans. In a house wiring usually lamp, wall-plug etc. are connected to the same circuit. Under this condition it should be noted how much is the actual limit of the current that the cables used in the wiring can safely carry. Considering the final sub-circuit which includes discharge lamps, the sum-total of currents taken by all the discharge lamps together must not exceed the current- carrying capacity of that final sub-circuit. If the lamps are lighted by means of only inductor, the normal current of the final sub-circuit should be 1 ¼ times the total current of all the lamps together.
If in a final sub-circuit both incandescent lamps and inductor-lighted discharge lamps are used,
[(power taken by inductor-lighted discharge lamps x 2) + (power taken by incandescent lamps x 1)] / line voltage
must not exceed the current of the sub-circuit.
Q. 33. What is the exception in case of temporary wiring?
Ans. In case of temporary load points where bayonet-holders for lamps have been used, total power demand of load must not exceed 1,000 watts per final sub-circuit.
Q. 34. How to determine the number of points per ‘way’?
Ans. Usually the number of ways is so estimated that either the total load connected does not exceed 800 watts or the number of points connected does not exceed 10 in each single way. In fact, the number of points that will be connected to every single way depends on the size of cable and the ampere rating of cut-out used in it. However, it should also be noted that the voltage drop in a final sub-circuit does not exceed 1.5 volts.
Q. 35. How to ensure protection of circuits?
Ans. The fuse for a final sub-circuit should be suitable for as much current as the current-carrying capacity of cables used for its wiring. As for example, if a cable of size 1/.044″ or 3/.029″ which can safely carry 5-ampere current is used for the wiring of a final sub-circuit, fuse for such a circuit should also be rated 5 amperes. But if a flexible cord of size 14/.0076″ is used anywhere in this circuit, for the protection of that cord a fuse of 2-ampere current rating should be used for that sub-circuit (as flexible cord of size 14/.0076″ can carry only 2-ampere current). A 5-ampere fuse will not do in this case.
Similarly, if in a heater circuit a 15-ampere wall-plug is connected while the flexible cord of size 40/.0076″ is used for heater connection, a 10-ampere fuse should be used in place of 15-ampere one simply because the cord can safely carry a current of 10 amperes only.
Q. 36. State the size of consumer’s main fuse.
Ans. The size of fuse of the consumer’s main, board will depend on the size of supplier’s fuse and the total load actually connected in different circuits of the house. Suppose, in a house, there is 25-ampere fuse in the supplier’s meter board. But if the total load actually connected in the house draws only 15 amperes, consumer’s main fuse should have a current rating of 15-ampere only. Fuse size should never be estimated on the basis of size of cable used for the supply main.
Current-rating of main switch is always higher than that of main fuse.
Q. 37. What are the uses of one sub-distribution boards?
Ans. When outlets from a sub-distribution board or a fuse board are divided into ‘ways’ and each final sub-circuit is connected to a separate way, the advantage is that in the event of a short- circuit in any one sub-circuit, the other sub-circuits remain unaffected and continue to function normally. But if a fault occurs in a distribution board, all the sub-circuits coming out of it are affected.
There are some places such as hospital, operation theatre, cash room in a bank, engine room, workshop etc. where the entire room cannot be allowed to be dark under any circumstance. A lot of risks may have to be faced if such places suddenly become totally dark. The disadvantages are not much less even in an ordinary dwelling house under such circumstance, but risk in the above-mentioned places is more serious in nature.
Of course, leave alone the rooms where there is only a single lamp or where reduction in expenses is the purpose. But wherever special attention must be paid to avoid any inconvenience in business, every room is equipped with more than one lamp and these are invariably taken from different ways. Even sometimes these lamps are supplied from fully separate distribution board.
Suppose, the wiring of a three-storeyed building is to be done in such a way that no room of that building shall be totally dark (excepting, of course, in the event of discontinuity of supply). In that case there must be a separate sub-distribution board in each floor. But it is not that the sub-distribution board will control the load points of that floor only.
Depending on the convenience of a circuit, sub-distribution board in the lower floor will supply power to some lamps etc. of the lower floor and to some lamps etc. of the upper floor. Every room will be provided with two sets of cables,—one set will be supplied from sub-distribution board of the upper floor and the other set will be supplied from sub-distribution board of the lower floor.
With these arrangements if a fault develops in a sub-distribution board, there is no possibility of any room becoming totally dark. Although such arrangements may be considered luxury for a residential house, in places like operation theatre or where machinery and workshop are running all the time, it will certainly be considered as essential necessity. In such cases, operation theatre etc. are provided not only with connection from separate distribution boards but with alternative source of supply such as gas plant or charged battery.
Q. 38. What is a pilot lamp?
Ans. Arrangements should be made for fixing a bracket above each main board and for connecting a 20-watt lamp on it. Cables connecting this lamp will come out directly from the bus-bars of the board through a separate switch and fuse. This lamp is called a Pilot Lamp. The purpose behind this arrangement is to keep the main board always illuminated so that fuse etc. can easily be changed.
Q. 39. What are the arrangements for taking cable connections from one house to another?
Ans. If wiring is to be done to supply current from one house in which consumer’s main switch has been installed to another house, whichever of the following arrangements is suitable (for a particular case) is to be adopted for the wiring and its protection:
If the distance between the house in which the main meter board has been installed and the other house (e.g. garage, servant’s room etc.) does not exceed 3 meters (10 feet) and if there be no thoroughfare between the two houses, electric lines may be drawn from the former to the latter through a galvanised iron pipe of suitable dimensions at a height of at least 2.5 metres (8 feet) above the ground level.
Also the G.I. pipe has to be properly earthed. But in case the distance between the two houses exceeds 3 metres or if there be a thoroughfare between them, a separate main or sub-main has to be drawn from one house to another by means of weather-proof cables tied up with G.I. bearer wire (Fig. 241).
1. If current is to be taken from one house to another by means of cleat wiring, the cable used in the wiring will be weather-proof. This is also known as H.S.O.S. (House Service Overhead System) cable. Use of cable with ‘polychloroprine’ sheath or P.V.C. cable or cable with P.V.C. sheath is also approved by many. This arrangement of drawing a supply line is allowed up to a distance of 3 metres between two buildings.
2. Using cables as described above and drawing these cables over a separate catenary wire or using those cables which have in-built bearer wires (at the time of manufacture), the supply line may be drawn.
3. Other methods of drawing cables over bearer wires are also in use. One of these methods in shown in Fig. 242. In this method a piece of leather strap loops a hard rubber-sheathed cable at certain intervals for hanging it, while the upper part of the strap is fastened to the catenary wire by means of wire hook. This is also an arrangement for taking a supply cable from one building to another.
If such a cable, as has In-built bearer wire, is used, the limit of distance between two buildings will depend upon the load-bearing capacity of the bearer wire.
4. Besides these, a cable may be drawn from one house to another as shown in Fig. 244. (Approved by I.S.I.).
In this arrangement a 7/20 S.W.G. galvanised Iron wire is stretched between two buildings in taut condition. For this, a wrought iron hook is fixed to the external wall of each building at the same height above the ground level.
Now, one end of 7/20 S.W.G. galvanised iron wire is fastened to a hook and the other end is properly tied up with an eyelet at one end of a straining screw (or swivel screw) as shown in Fig. 243. The eyelet at the other end of the screw is slung into the hook of the other house and then the screw is twisted so that the screw on the right-hand side gets inside, thus making the cable taut.
Below the two hooks (as shown in Fig. 244) two leading-in-pipes are installed through the walls of the two houses. A weather-proof cable of suitable size is drawn through one pipe and then it is fixed to the galvanised iron wire by means of link clips at regular intervals of 15 cm (6 inches) up to the next building and then drawn inside this building through the other pipe.
G.I. earth wires are then clamped to hooks of both the buildings Each wire is led down the wall up to ground by fixing it to the wall with the help of galvanised iron staples and then connected to an earth electrode which is placed in an earth pit at least 1.5 metres (5 feet) away from the building.