1 What qualities do successful engineers have? Strong in mathematics and science. Highly analytical and detail oriented. Imaginative and creative. Good communication skills. Enjoy working in teams. Enjoy building or improving the way things work. TECHNIQUE TO CLEAR ELECTRICAL TECHNOLOGY SUBJECT IMPORTANT CHAPTERS IN ELECTRICAL TECHNOLOGY WITH HIGHER WAITAGE: ( HIGHER TO LOWER PRIORITY TO STUDY) 1) SINGLE PHASE A.C.CIRCUITS 2) SINGLE PHASE TRANSFORMERS 3) D.C.MOTORS 4) D.C.CIRCUITS 5) THREE PHASE A.C. CIRCUITS 6) A.C. FUNDAMENTALS 7) INSTALLATION, EARTHING AND TROUBLESHOOTING 8) FRACTIONAL HORSE POWER MOTORS Vijay Raskar Electrical Notes….. “Vijay B. Raskar
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1
EELDAD
What qualities do successful engineers have? Strong in mathematics and science. Highly analytical and detail oriented. Imaginative and creative. Good communication skills. Enjoy working in teams.
Enjoy building or improving the way things work.
TECHNIQUE TO CLEAR ELECTRICAL TECHNOLOGY SUBJECT IMPORTANT CHAPTERS IN ELECTRICAL TECHNOLOGY WITH HIGHER WAITAGE: ( HIGHER TO LOWER PRIORITY TO STUDY)
1) SINGLE PHASE A.C.CIRCUITS 2) SINGLE PHASE TRANSFORMERS 3) D.C.MOTORS 4) D.C.CIRCUITS 5) THREE PHASE A.C. CIRCUITS 6) A.C. FUNDAMENTALS 7) INSTALLATION, EARTHING AND TROUBLESHOOTING 8) FRACTIONAL HORSE POWER MOTORS Vijay Raskar
Electrical Notes….. “Vijay B. Raskar
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Problems are very important with theory. Subject is very easy if practice. Use technique to solve the problems. More practice = More marks. Use calculator and it is very essential to clear “ELECTRICAL SUBJECT”.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
:. Sin curve of frequency double that of voltage and current waves. Therefore, Average power is zero.
CONCLUSION: The average demand for power in a purely inductive circuit over the whole period is always zero. Q.3 Derive the relationship between voltage and current in pure capacitive circuits. ANS:
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
3. SINGLE PHASE TRANSFORMERS Q.1 State and explain the principle of transformer? ANS: STATEMENT: The operation of a transformer is based on the principle of mutual induction between two circuits linked by a common magnetic field. EXPLAINATION:
See the elementary transformer.
There are two windings PRIMARY and SECONDARY.
They are separated and wound on common laminated steel core.
Vertical portion of core where windings are placed is called as LIMB.
Top and bottom portions are called as YOKES.
Primary winding is connected across mains and output is receives from secondary windings.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Electrical energy is transfer from one circuit to another.
Some energy is always lost.
All the losses in the transformer are ultimately converted into heat.
Therefore, transformer is provided with cooling system to protect the insulation of the windings.
There are two types of losses in the transformer:
1) COPPER LOSS:
TOTAL COPPER LOSS = I₁² R₁ + I₂²R₂ Where, I₁ & R₁ : For primary I₂ & R₂ : For secondary
Copper conductor is used for winding.
Loss of power caused by resistance of the winding.
Power loss is proportional to the square of the current flowing through the windings.
It is ultimately utilized in heating the windings.
Copper loss is proportional to the current and square of the KVA output. 2) CORE OR IRON LOSS:
a) Hysteresis loss:
Hysteresis loss = Ph = Kh X Bm¹·⁶ X f X v volts Where, Kh = constant Bm = Maximum flux density in tesla f = frequency v = volume of the magnetic material
This loss is takes place in the transformer core.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
It is continuously subjected to rapid reversals of magnetization by the alternating flux.
b) Eddy current loss:
Eddy current loss= Pe = Ke X Bm² X f² X v X t² watts Where, Ke = constant Bm = Maximum flux density in tesla f = frequency v = volume of the magnetic material t = thickness of the laminations
Core losses together are nearly constant and independent of the magnitude of the current.
They can be reduced by choosing silicon steel having small values of Kh and Ke.
Q.6 Explain Auto-Transformers. Also, explain its applications. ANS:
Auto-transformer having only one winding on a laminated magnetic core.
Winding is common for both primary and secondary circuits.
Ordinary two winding transformer, it can be used as step up and step down transformer.
V1, N1 for primary side and V2, N2 For secondary side.
If neglecting the losses, leakage reactance and magnetizing current, We have, V2 = I1 = N2 = K V1 I2 N1
APPLICATION:
For starting squirrel-cage induction motors & synchronous motors.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
For furnace transformers for getting suitable supply voltage.
As variacs for getting continuously variable supply a.c. voltage.
Q.7 What are the advantages and disadvantages of auto-transformers? ANS: ADVANTAGES OF AUTO-TRANSFORMER:
Its cheaper.
Copper require very less for windings.
Transformation ratio is approaches unity.
Its regulation is better than two winding transformer.
Efficiency of transformer is high.
I²R Losses are less.
The resistance and leakage reactance is less.
DISADVANTAGES OF AUTO-TRANSFORMER:
There is always risk of electrical shock.
When it is used in high-voltage circuit, possibilities of electrical shocks.
Therefore, Low voltage and high voltage sides are not electrically separated.
Q.8 Explain isolation transformers. ANS: Definition: The transformer which are specially designed to provide electrical isolation between the primary and secondary circuits without changing voltage and current levels. EXPLAINATION:
The isolation transformer is frequently used to isolate one portion of an electrical system from another.
These systems are physically separated.
They are magnetically couple to each others.
Electrical isolation provided between primary and secondary circuits.
Isolation transformer is a 1:1 transformer and is used only for the purpose of electrical isolation.
FUNCTIONS OF ISOLATION TRANSFORMERS:
1) Isolation transformers isolate and protect the sensitive and expensive equipments from electrical system grounds.
2) Isolation transformers protect the delicate and expensive equipments against voltage spikes.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
4. D.C.MOTORS Q.1 Explain the working principle of operation of a d.c. generator. ANS: Generator: Which converts mechanical energy into electrical energy either ac or dc. Principle of Generator: “When a conductor is moved in magnetic field or magnetic field moved with respect to the conductor. According to Faraday’s law of electromagnetic induction, electromotive force is set up in the conductor”
There is relative motion between a conductor and a magnetic field.
Voltage will always be generated in the conductor.
A.C. Generators are normally known as alternators.
CONSTRUCTION:
It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.
Coil / conductor is placed in between two poles (S & N POLE).
Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.
The coil is also known as armature of the alternator.
The ends of the armature coil are connected to rings called slip rings. i.e. R1 & R2
Two carbon brushes pressed against the slip rings.
The slip rings collect the current and carry it to the external resistor (R).
OPERATION: Current supplied by alternator is necessarily alternating.
Unidirectional or direct current is desired.
Before external circuit, alternating current must be rectified or commutated.
Rectification can be done by replacing the slip-rings R1 and R2 of the alternator.
Where, M is shown in figure “COMMUTATOR”.
Commutator is made by conducting material which is cuts in two equal halves or segments.
Segments are insulated by thin sheet of mica or some other insulating material.
The ends of the armature coil AB are joined to these segments.
The external load resistor R is connected to the split ring M through brushes B1 and B2.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
When coil is undergoing for first half cycle, current flowing from load(R) (C to D).
For next half cycle, its reverses with position of a and b terminals.
Again for next cycle, current is flowing from C to D and so on.
Direction of current through the external circuit is same.
In the commercial d.c. generator, by using number of coils evenly distributed around the armature core and connecting these coils to commutator having a corresponding number of segments, a more steady direct current is obtained in the external circuit.
Q.2 Explain the working principle of operation of a d.c. Motor. ANS: CONSTRUCTION:
It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.
Coil / conductor is placed in between two poles (S & N POLE).
Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.
Two carbon brushes pressed against the slip rings.
Armature coil is supplied with direct current through the brushes B1 and B2.
WORKING / OPERATION:
Unidirectional torque is produced in the d.c. motor.
There are three positions to understand it.
Torque developed at different position of the coil.
By Flemming’s left hand rule, conductor moves upwards as well as downwards.
Conductor is placed between the two poles.
The computer plays very important role in the operation of the d.c.motor.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Q. 1 What is ohm’s law? Or what fundamental relation does Ohm’s law express? Give the three forms in which the law may be expressed? ANS: OHM’S LAW: “The current flowing through a solid conductor is directly proportional to the difference of potential across the conductor and inversely proportional to its resistance, temperature remains constant”. EXPRESSIONS:
I V/ R I = (V/ R)*CONSTANT : . V = IR …………..Ohm’s law equation Where, V= Voltage, Unit: Volt (V) I= Current, Unit: Ampere (A) R= Resistance, Unit: Ohm ( )
Q.2 Define the terms resistance, specific resistance, conductance and specific conductance? ANS: (1) Definition of resistance: “It is defined as the property of a substance by virtue of which it opposes the flow of current through it. Unit of resistance is Ohm ( ) Pure metals means less resistance and vice versa. (2) Definition of specific resistance: “It is the resistance of a specimen piece of a material having unit length and unit cross-sectional area”. :. It is also called as resistivity. :. R = . L / A
Where, R = Resistance = Specific resistance
l = Length A = Area Unit: Ohm-meter
(3) Definition of conductance (G): “It is the reciprocal of resistance” Where resistance is measure of opposition offered by the conductor to the flow of current, the conductance is measure which current may flow through a conductor. Unit of conductance is Mho ( )¯¹ (4) Definition of specific conductance: “It is the reciprocal of the specific resistance” G = 1 / R = .A / l
Where, R = Resistance = Specific resistance
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l = Length A = Area Unit of specific conductance or conductivity: siemens/meter. Q.3 what are the factors on which the resistance of a material depends? ANS: The resistance of a conductor depends upon the following factors:
(i) Type and nature of material: The resistance varies from material to material. Based on capacity to carry free electrons. Example: silver is the best resistance and iron offers high resistance.
(ii) Purity and hardness also affects the resistance. Pure metal less resistance and vice versa.
(iii) Length of the conductor: The resistance of a conductor is directly proportional to the length. Resistance of the longer wire is greater.
(iv) Cross sectional area: The resistance of a conductor is inversely proportional to the area.
(v) Temperature: The resistance is directly proportional to the temperature. In most materials, resistance increases with temperature.
NOTE: CURRENT: The rate of transfer of electric charge per unit time.
CHARGE = CURRENT X TIME
VOLTAGE OR P.D.: The difference between the electric potentials or pressure at any two given points in the electric circuit.
e.m.f.(ELECTROMOTIVE FORCE): A p.d. generated by a source of electrical energy across its terminals which tends to produce an electric current in a circuit.
P.d = work done per charge. Q.4 Derive the expression for equivalent resistance of a circuit having several resistance connected (i) in series and (ii) parallel. ANS: (i) SERIES CIRCUITS:
DERIVATION:
Consider the three resistances R1, R2 and R3 connected in series as shown in circuit diagram.
They are connected in series.
Current flowing through the circuit is “I” and voltage across each resistance is V1, V2 and V3 across R1, R2 and R3 respectively.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
From circuit, V = V1 + V2 + V3 …………….(1) By Ohm’s law, V = IR (Substitute in equation 1) :. IR = IR1 + IR2 + IR3 IR = I ( R1 + R2 + R3 ) Hence, R = R1 + R2 +R3 In general, R = R1 + R2 + R3 +…..Rn SUMMARY
The same current flows through each resistance in turn.
The supply voltage is sum of the individual voltage drops across the resistances.
The total or equivalent resistance is sum of all the connected resistances. (ii) PARALLEL CIRCUITS:
DERIVATION:
Consider the three resistances R1, R2 and R3 connected in series as shown in circuit diagram.
They are connected in series.
Current flowing through the circuit is “I” and voltage across each resistance is V1, V2 and V3 across R1, R2 and R3 respectively.
I = I1 + I2 + I3 …….(1) By ohm’s law, V = IR I = V / R (Substitute in equation 1)
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
The same potential difference is applied across all of the resistances.
The total current drawn from the supply is equal to the sum of the currents flowing through the individual resistances.
The total or equivalent resistance is sum of all individual reciprocals of resistances.
Q.5 Classify the electrical networks. ANS: In an electrical networks or circuit, there are sources of energy and parameters like resistors, capacitors and inductors etc. The behavior of entire network depends on these elements. Accordingly, they can be classified as,
(i) Linear Circuits (ii) Non-linear Circuits (iii) Bilateral Circuits (iv) Unilateral Circuits and (v) Active Networks (vi) Passive networks
(i) Linear Circuits: Circuits whose parameters are always constant at any condition like
variation in voltage and currents etc. Current is directly proportional to voltage.
(ii) Non-linear Circuits: Circuits whose parameters changes with variation in current and voltages.
(iii) Bilateral Circuits: Circuit whose characteristic is same as direction of operation like transmission line. It has same relationship between current and voltage for current flowing in either direction.
(iv) Unilateral Circuits: Circuits whose characteristics are dependent on the direction of its operation is called unilateral circuits. E.g. vacuum diodes which allow the current at one direction.
(v) Active Circuits: Having source of energy in the circuit. It may be either voltage or current.
(vi) Passive networks: A networks contains no source of energy.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
The load on a 3-phase system may be star or delta connected.
If all phases impedances of the three phases load are exactly identical in respect of magnitude and their nature. It is called as balanced load.
The lighting and heating loads are resistive in nature.
The lighting and heating loads are connected between line and neutral.
The electricity is distributed on three phase of equal loads.
In such condition, all single phase loads forms a balanced star-connected load. Q.3 Write a short notes on “balanced system”. ANS: A 3-phase system is balanced when it possesses following characteristics:
1) The three phase e.m.f.s are equal and displaced by 120 electrical degrees. 2) Impedance in any one phase should be identical as compare with other two phases. 3) The resulting currents in different phases with same phase angles. 4) Power and reactive power flow should be equal in each phase.
Q.4 Derive the voltage, current and power relations in star connections in a star connection. ANS:
EXPLAINATION:
Star connection is shown in figure (a).
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7. A.C. FUNDAMENTAL Q.1 What is an alternating current? Compare it with a direct current. ANS: Alternating current is one which periodically passes through a definite cycle of changes in respect of magnitude as well as direction. Or Alternating current is current which flows periodically through a definite cycle of changes in respect of magnitude as well as direction. Direct current: d.c. is that current which flows continuously in one direction and has constant magnitude with respect to time.
Comparison between a.c. and d.c. or advantages of a.c.
1. The ac voltage can be raised or lowered very easily by using transformer. In dc, not so easy and economical.
2. Construction and operating costs per kilowatt are low. 3. We can build up high voltage, high speed a.c. generators with large capacities. It is not
possible in dc. 4. Due to adoption of high voltages, a.c. transmission is always efficient and economical.
Because, higher the voltages lesser the current through transmission. 5. ac motors are simple in construction, cheaper and more efficient than dc. 6. ac can be easily converted to dc for the application like traction, search lights,
projections etc. Q.2 What is the principle of Generator? Also, Explain generation of alternating current and voltage. ANS: Generator: Which converts mechanical energy into electrical energy either ac or dc. Principle of Generator: “When a conductor is moved in magnetic field or magnetic field moved with respect to the conductor. According to Faraday’s law of electromagnetic induction, electromotive force is set up in the conductor”
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Conductor A cuts up and conductor B cuts down through the field.
Both the polarity of generated e.m.f. and current is reverse.
By fleming’s left hand rule, direction of current in conductor A is away and for B towards.
Current flows through D to C.
Similar to position 2, emf in the coil is maximum in position 4.
In general, the variations in the magnitude of emf of the alternator is exactly similar to those in first half revoltion.
Current supplied by alternator is necessarily alternating.
Unidirectional or direct current is desired.
Before external circuit, alternating current must be rectified or commutated.
Rectification can be done by replacing the slip-rings R1 and R2 of the alternator.
Where, M is shown in figure “COMMUTATOR”.
Commutator is made by conducting material which is cuts in two equal halves or segments.
Segments are insulated by thin sheet of mica or some other insulating material.
The ends of the armature coil AB are joined to these segments.
The external load resistor R is connected to the split ring M through brushes B1 and B2.
ACTION OF A COMMUTATOR:
When coil is undergoing for first half cycle, current flowing from load(R) (C to D).
For next half cycle, its reverses with position of a and b terminals.
Again for next cycle, current is flowing from C to D and so on.
Direction of current through the external circuit is same.
In the commercial d.c. generator, by using number of coils evenly distributed around the armature core and connecting these coils to commutator having a corresponding number of segments, a more steady direct current is obtained in the external circuit.
GRAPHICAL REPRESENTATION OF E.M.F. OR CURRENT:
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Q.3 Derive the equation of the alternating voltages and currents? ANS:
Generator: Which converts mechanical energy into electrical energy either ac or dc. Principle of Generator: “When a conductor is moved in magnetic field or magnetic field moved with respect to the conductor. According to Faraday’s law of electromagnetic induction, electromotive force is set up in the conductor”
There is relative motion between a conductor and a magnetic field.
Voltage will always be generated in the conductor.
A.C. Generators are normally known as alternators.
CONSTRUCTION:
It consists of a single turn rectangular conducting coil (AB) like copper or aluminum.
Coil / conductor is placed in between two poles (S & N POLE).
Coil is placed in such a way that to rotate at constant speed in uniform magnetic field.
The coil is also known as armature of the alternator.
The ends of the armature coil are connected to rings called slip rings. i.e. R1 & R2
Two carbon brushes pressed against the slip rings.
The slip rings collect the current and carry it to the external resistor (R).
DERIVATION: Let, B= Flux density of the magnetic field (teslas) l = Length of the conductor (metres) r = Radius of circular path (metres) ῳ = Angular velocity of the coil (radians / second) v = Linear velocity of the coil sides ( metres / second) when the coil lies in YOY plane and zero e.m.f. then, angle ϴ = ῳt ……………………………………………………………..(1)
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After some time coil rotates at angle ϴ. It resolved two components (1) horizontal (v.cosϴ) and (2) vertical (v.sinϴ). Therefore, E.M.F. generated in each coil side: B.l.v.sinϴ volts ……………….(2) Total e.m.f. generated in the coil = e :. e = 2 B.l.v.sinϴ volts ……………………………………….(3) When coil will be horizontal plane then ϴ = 90 and conductor cuts maximum flux (Em) :. Em = 2 B.l.v.sinϴ = 2 B.l.v.sin90 = 2 B.l.v.1 ( sin90 = 1) = 2 B.l.v volts Therefore, the standard equation of the voltage or e.m.f, e = Em.sinϴ e= Em.sinῳt e = Em.sin(2π.f.t) e = Em.sin(2π.t / T) Similarly, we can write for current, i = Im.sinϴ i = Im.sinῳt i = Im.sin(2π.f.t) i = Im.sin(2π.t / T) Q.4 Define peak factor, form factor, waveform, cycle, periodic time, frequency and amplitude. ANS: (i) PEAK FACTOR: “ It is the ratio of maximum value to the R.M.S. value of an alternating quantity”. It is also called as crest or amplitude factor. Peak factor = maximum value R.M.S. value Where, R.M.S. value = 0.707 X Maximum value Maximum value = √2 X R.M.S. value Substitute in peak factor, Peak factor = maximum value 0.707 X Maximum value Peak factor = 1.4144 (ii) Form factor: “ It is the ratio of R.M.S. value to the average value of an alternating quantity”. Form factor = R.M.S. value Average value Where,
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Average value = 0.637 X maximum value R.M.S. value = 0.707 X Maximum value Substitute in form factor, Form factor = 0.707 X Maximum value 0.637 X maximum value = 1.11 (iii) waveform: “ It is the pictorial representation of an alternating quantity in a graphical form”. (iv) cycle: “Each repetition of a complete set of changes undergone by the alternating quantity is called cycle”. (v) Periodic time: “ It is the time required by an alternating quantity to complete its one cycle”. It is denoted as “T” in seconds. (vi) Frequency: “ The number of cycle completed per second by an alternating quantity is known as frequency”. It is denoted as “f” in hurtz. f = 1 / T (vii) Amplitude:
The maximum value attained by an alternating quantity during it’s positive or negative half cycle”.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Q.1) Briefly describe the different types of wires used in the electrical wiring work. How these wires normally specified? ANS: There are nearly (commonly) seven types of wires. They are as follows:
Q.4) Explain conduit wiring? Or What are the various types of conduit used in conduit wiring? ANS: CONDUIT WIRING: V.I.R or P.V.C. wires are run through black enameled or galvansed metallic tubing called conduit. There are two types of conduits:
Since the wire is exposed to air, it is subjected to deterioration due to oxidation caused by heating. This decreases the diameter of wire.
Slow speed for the current interruption.
Risk of fire hazards due to external flash on blowing.
Q.6 What are the advantages and disadvantages of H.R.C. Cartridge fuse? ANS: ADVANTAGES:
High speed operation.
Ability to clear high values of fault current.
Its operation is silent and without flame, gas or smoke.
It is safe from fire hazards.
Performance and protection is reliable.
Being totally closed, there is no deterioration of the fuse element. DISADVANTAGES:
Costly
We have to replace after its operation.
Overheating is possible at the contacts.
Q.7 Explain, why H.R.C. Cartridge fuses are preferred over the rewirable type, particularly high values of current. ANS: There are many advantages of H.R.C. Cartridge fuses over rewirable fuses. Advantages of H.R.C. Cartridge fuse over rewirable fuse:
High speed operation.
Ability to clear high values of fault current.
Its operation is silent and without flame, gas or smoke.
It is safe from fire hazards.
Performance and protection is reliable.
Being totally closed, there is no deterioration of the fuse element.
Q.8 Why fuses should always be provided in the live (phase) wire? ANS: Because, electrons flows in the live or phase wire. Neutral is used for return path. If we have to protects the circuit from excess current and install the fuse at the live part only. If we install it at neutral path then circuit will not be protected and can be damage. Hence, fuses should always be provided in the live (phase) wire. Q.9 Define current rating of fusing elements? ANS: It is the value of continuous current which the using element can carry safely undue heating, melting and deterioration. Q.10 Define minimum fusing current?
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
ANS: it is defined as the minimum value of current which causes melting of the fuse. Q.11 Define fusing factor? ANS: It is the ratio of Minimum fusing current to the Current rating of the fusing element.
Its value is always more than one.
Q.9 Explain MCCB. ANS: EXPLAINATION:
MCCB is a Moulded Case Circuit Breaker.
It is compact type air circuit breaker.
It is enclosed in a moulded case.
Moulded means insulating body.
It uses air for arc quenching medium.
The resistance of the arc is increased by cooling, lengthening and splitting the arc with the help of arc chute.
It protects against the overload and short circuits.
Overload protection is given by bimetal strips.
Short-circuit protection is given by magnetic attraction.
ADVANTAGES OF MCCB:
Practically, no maintenance is required.
It avoids single phasing.
No recurring cost.
Gives indication in case it trips on fault.
RATINGS / SPECIFICATION OF MCCB:
It is available at 16A to 1600.
Its breaking capacity as high as 85Ka.
Breaking capacity ratings are same for 3P as well as 4P. (P = POLES)
APPLICATIONS OF MCCB:
It is suitable for both a.c. as well as d.c. application.
Distribution feeders.
Distribution transformer.
Diesel generating sets.
L.T. capacitors.
Rectifier panels.
U.P.S.
Motors, furnaces and other electronics equipments.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
The casing becomes live with respect to earth at equal potential.
See the Fig: Fault with non-earthed appliance. When fault is occurred then also appliances are under working condition. But, if any human or living things touch the appliances they will get a
shock. Very dangerous to living things. During fault, human being or living things get shocks.
See the Fig: Fault with earthed appliance.
If the fault is occurred then there is no harmful effect due to earthing.
Fault current or leakage current is flowing through the earthing conductor at the earth.
Circuit and living things to be protected.
Appliances under working without any damages.
Q.12 Discuss in brief, the commonly used methods of earthing. ANS: EXPLAINATION:
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Convertor converts a.c. to d.c. and charges the battery.
Battery provides a.c. power to the computer with the help of invertor.
On-line UPS provides protection against loss of data in the case of power failure.
On-line UPS is costly.
Q.14 What are the essential tests to be made on a wiring installation before it is put into service? ANS: There are five tests. They are as follows:
1) Insulation resistance to earth. 2) Insulation resistance between conductors. 3) Polarity test for Single-pole Switches. 4) Continuity test for earth wire. 5) Test for earth Resistance Measurement.
EXPLAINATION:
1) Insulation resistance to earth:-
We can test the insulation of the conductor by using Megger.
We can measure the resistance between conductor and earth by 500V megger.
Rotating part is their which gives readings of the megger.
Reading is noted down which shows the insulation level.
Insulation level is more than 1 megaohm for the entire installation.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
9. FRACTIONAL HORSE POWER MOTOR Q.1 Explain the principal of working of a single-phase induction motor with the help of double revolving field theory. ANS:
EXPLANATION:
Construction of 1-phase motor is similar to the polyphase motor.
Only difference is stator has distributed 1-phase winding. OPERATING PRINCIPLE: When a.c. voltage is applied to the motor, current is flowing through 1-phase stator winding. Stator winding produces a resultant field which alternates along the axis of the winding.
Such a field varying sinusoidally with time along with fixed space axis.
Rotor is at rest for understanding.
When supply is connected, current is flowing through the winding.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Winding produces alternating flux along the axis of the winding.
By transformer action, field induces an e.m.f. in the rotor conductor.
Q.2 What are the special features of servomotors? State their types. ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011 Q.3 Write a short note on characteristics and application of stepper motors. ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011 Q.4 Compare different types of single-phase induction motors. ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011 Q.5 Write a short notes on D.C. Servomotors. ANS: REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011 Out of notes: All theorems which is taught in the class room. Like Thevenian’s , Norton, Maximum power, etc……………….. REFER BOOK or WAIT UPTO FRIDAY, 01-03-2011
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
IMPORTANT THEORY QUESTIONS Q.1 State the duality between d.c. series and parallel circuit. Q.2 State the kirchhoff’s law. Q.3 Define active and passive elements. Q.4 State the condition of transferring maximum power from the source to load in d.c. circuits. Q.5 Define form factor. What is its value for sinusoidal waveform? Q.6 For star connected load state the numerical relationship between V & I for phase and line. Q.7 For delta connected load state the numerical relationship between V & I for phase and line. Q.8 Define ideal and practical voltage source. Q.9 With the help of waveforms and phasor diagrams, show the phase relationship between voltage and current in pure inductive and pure capacitive circuits. Q.10 Define: a) Impedance b) current c) Power factor d) Power e) Resistance f) Form factor g) Peak factor h)True power i) Apparent power j) Reactive power k) Ideal transformer l)Isolation transformer m)Auto transformer n)Regulation of a transformer o)Symmetrical system p)Phase sequence q) Current rating r)Minimum fusing current s)Fusing factor t)Ohm’s law Q.11 What is the impedance of a.c. circuit? What is its unit? State factors on which it depends. Q.12 State the various losses in transformer. How these losses are minimized? Q.13 Draw schematic diagram of single-phase a.c. servometer and state its principle of working. Q.14 What are the factors on which the resistance of a material depends? Q.15 Derivation for series and parallel. Q.16 What is an a.c.? Compare it with a d.c. or Advantage of a.c. over d.c. Q.17Define rms value of an alternating quantity. Prove that in an alternating quantity varying sinusoidally, the maximum value is 1.414 times the rms value. Q.18 Explain choking coil. State any two properties. Q.19 Explain the difference between a single-phase and a polyphase system. Q.20 Explain with the reference to a three phase system, the terms “balanced load” and “balanced supply system”. Q.21 What is the function of a transformer in ac circuit? Q.22 Derive the emf equation of a transformer. Q.23 Explain the meaning of kVA rating of a transformer. OR Why transformer rating is in kVA? Q.24 What are the types of transformer? Explain in detail with diagrams. Q.25 Explain the principle of operation of a dc motor. Q.26 Explain types of dc motors. Q.27 Explain characteristics of dc motor (a) Shunt (b) Series (c) Compound dc motor. Q.28 Derive equation of Back e.m.f. Q.29 Derive torque equation of motor. Q.30 Explain a.c. servometer and d.c. servometer. Q.31 Different characteristics of stepper motors & its application. Q.32 Explain the principle of working of a single-phase induction motor with the help of double revolving field theory. Q.33 Explain different types of wires. Q.34 Explain MCCB, MCB AND ELCB.
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]
Q.35 Why earthing is necessary in a wiring installation? How it is achieved? Q.36 Explain types of fuses. Q.37 Explain classification of electrical network. Q.38 Explain concept of open and short circuits. Q.39 Explain the generation of ac current and voltage. Also explain generator principle. Q.40 Derive the equation of ac voltage and current. OR Derive the instantaneous value of generated emf.
PROBLEMS
LOOP METHOD: Q.1 Find current flowing through each resistance.
Solution: By loop method, Assuming current by clockwise direction for each loop.
Apply KVL equation to loop 1
4612
01046662
21
1211
II
IIII………………(1)
Apply KVL equation to loop 2
4146
0666523
21
1222
II
IIII…………………..(2)
Solving equation (1) and (2), By cramm’s rule,
Electrical Notes….. “Vijay B. Raskar” 8898443014 , [email protected]