LinkedIn Insight Payal-Rtmnu-Engg.-AEE-Sum18-Sol - Grad Plus


Q.1(a) Explain with neat block diagram the operation of Thermal Power plant.

Ans. Fuel Like coal,oil gas possess energy stored in them. These fuels when burnt gives out this in the form of heat. The function of a thermal generating system is to convert this heat into electrical energy.

The working of Thermal station can be desired in terms of three cycles. They are.

  1. Water Steam Cycle
  2.  Flue gas  Cycle.
  3. Coal Gas Cycle

1. Water Steam Cycle – 1. The boiler converts water steam at a temperature of 100°C. This steam is then passes to the superheater.

2. In the superheater, steam is heated to a very high temperature and pressure. The superheated steam is then passed to the steam turbine. In the turbine the steam gets expanded through its blades and its pressure decreases.

3. Thus the energy stored in the steam is imparted to the turbine rotor and steam temperature reduces to about 100°C. The steam is then passes out of the turbine into the condenser.

4. In the condenser, steam is condensed with the help of cold circulating water. The condensate thus formed passes to the steam ejector and deaerator where any steam and air left over in the condensate is removed.

5. During the flow of steam from the boiler to the deaerator through turbine, condenser etc. Some steam gets lost due to leakage. To cover up this leakage, additional water, as make-up water is taken through the raw water tank after distillation.

6. The total water is then pumped by the feed water pump to the economiser. In economiser, this water is heated by utilising the heat carried away by the flue gases.

7. Thus, the economiser act as a water preheater & this heated water is supplied back to the boiler. Due to this arrangement steam rate in the boiler increases.

B. Flue Gases Cycle

  1. In the boiler, the coal is burnt and oxygen is required to burn the coal is provided in the form of fresh air. The Flue gases leaving the boiler contain lot of heat.
  2. To utilise this heat, they are passed through superheater, economiser and air preheater. The system recovers the heat from the Fuel gas and hence the system is called heat recovery system.
  3. In superheater, steam received from the boiler is heated to a very high temperature. The economiser heats the condensate water and this hot water is fed to the boiler. The air preheater heats Fresh air to be supplied to the boiler for the purpose of combination.
  4. Gases leaving the air preheater passes out of the atmosphere through the chimney. Thus upto stage of air preheater most of the heat from flue gases is recovered.

C. Coal Ash Cycle:-

  1. Coal is brought to the unloading centers by rail or road. It is unloaded and carried to the coal storage. It is cleaned, dried and then crushed in the crushing mill. It is then passed to the pulvanizer to break it into small pieces.
  2. Magnetic seperators are placed by the sides of the conveyor belts to remove iron particles present in the coal. From bunkers coal is supplied to the boiler furnace where it is burnt to convert water into steam. Ash coming out from the boiler is collected in hoppers.
  3. The cola is then conveyed by the conveyer belts to coal bunkers which are placed at the top of the boiler.
  4. Ash handling is one of the major problems in thermal stations. Even a small boiler consumes 1 tonne of coal/hr.

Q.1.B. Draw and Explain single line diagram of electrical power system.


1. The increased demand of electricity needs more generation of electrical power. As the generation takes place at remote places, an efficient distribution is necessary.

2. The problems of AC transmission particularly in long distance transmission has led to development of DC transmission.

3. The simplified block diagram of the extra high voltage AC transmission system is shown in the fig.

4. This system can be broadly divided into two parts.

a. Transmission System

b. Distributive System

5. Each part is again sub-divided into two parts .

a. i) Primary transmission

ii) Secondary transmission

Similarly ,b.  i) Primary distribution

ii) Secondary Distribution

a. i) Primary Transmission – 1. In the figure the central section CS generator power using three phase alternator at 6.6 or 11 kv  or 13.2 or 32 KV.

2. This  voltage is then stepped up by suitable three phase transformer to 132 KV as shown.

3. Such a high voltage   of transmission required conductor of smaller cross sectional area which results in the reduction in the cost of copper or aluminium .

(Note- Reason behind step up voltage from 11 to 132 KV)

If voltage  increases current decreases  resistance  increases

If resistance increases means i) size of conductor decreases.

ii) cost of conductor decreases

iii) power loss decreases

iv) Efficiency increases

As  R\alpha \frac{\rho l}{A}

4. However cost of insulation decreases, the transmission voltage is therefore determined by economic consideration . Higher transmission voltage also reduces the line losses and improves efficiency.

5. The three phase three wire overhead high volatge  transmission line gets terminated in the step down transformers in a substation known as receiving station .(Rs)

6. Receiving station is usually situated outside the city in order to ensure safety.

7. Here the voltage is stepped down to 33 kv from 132 kv.

a. ii) Secondary Transmission – 1. From the Receiving station, the power is then transmitted at 33 KV by underground cables ( and ocassionally by overhead lines ) to the various substation ‘SS’ located at various points in the city

2) These is known as secondary or low voltage transmission

3) At the substation (SS), this voltages is further reduced from 33kv to 3.3 kv using step down transformer.

b.1) Primary Distribution

The output of substation at 3.3 kv can be directly given to a customer whose demand exceeds 50 KVA using special feeder. This is primary distribution.

b.2) Secondary Distribution

1) Secondary distribution is alone at 440 V or 230 V

2) The reduction in the voltage level form 3.3 KV to 440 V or 230 V is done by step down transformer at the distribution substations.

3) The single phase residential load is connected between any one line and the neutral where as 3-phase, 440 V, motor load is connected across three phase line directly.

4) The standard frequency in India is 50 Hz and in USA 60 Hz

Q.2. (A) Explain with neat diagram , plate type earthing.

Ans.  Plate  type earthing


1.First of all diag 5×5 ft(1.5×1.5) pit about 20-30 f (6-9m) in the ground.[Note that ,depth & width depends on the nature and structure of the ground.

2.Bury an appropriate (usually 2’×2’×1/8” means (600×600×300mm) copper plate in that pit in vertical position.

3.Tight earth lead through nut bolts from two different places or earth pipe.

4.Use two earth lead through with each earth pipe& tight them.

5.To protect the joints from corrosion ,put grease around it.

6.Collect all the wires in a metallic pipe from the earth electrodes make sure the pipe is 1 ft (30cm) above the surface of the ground.

7.To maintain the moisture condition around the earth pipe , put a 1ft layer of powdered  charcoal(powdered wood coal) & lime mixture around the earth pipe.

8.Use  nut  bolts  to connect tightly wires to the bed plates of machine. Each machine should be earthed from two different places.

9.Earth continuity conductor which is connected to the body & metallic parts of all installation should be tightly connected to earth lead.

10.At last, test the overall earthing system through earth tester .If everything is going about the planning ,then fill the pit with soil. The maximum allowable resistance for earthing is 1Ω. If it is more than 1Ω than more the size of earth lead & earth continuity conductors .Keep the external ends of the pipe open & put the water time to time to maintain the moisture condition around the earth electrode which is important for the better earthing system.

Q.2.(B). What is Fuse? Draw and explain HRC Fuse.

Ans. Fuse is a simplest device, which breaks the circuit under abnormal conditions i.e. it is only a current interrupting under abnormal device  under fault condition and it is not able to make or break the circuit under normal condition.

2. It is used for overload or short circuit protection in medium voltage upto 33 KV and low voltage upto 400 V installation.

3. Fuses are widely used in power, heating and lighting purpose.

HRC fuse ( High Rupturing Capacity Fuse )

Diagram –

  • Fig. Shows the essential part of typical HRC.
  • HRC fuse mainly consists of heat resisting ceramic body both the ends of ceramic  body consists  of metal end caps
  • A silver current carrying element is welded to these metal end caps.
  • The current carrying element is completely surrounded by the filling powder.
  • The filling material may be plaster of parries, chalk, quartz or marble dust.
  • Filling material act as an arc quenching and cooling medium when fuse element blows off due to excessive  heat generated under abnormal condition.
  • Under abnormal conditions, the fuse element is at a temperature below its melting point. Therefore it carries the normal current without overheating.
  • When a fault occurs, the current increases and heat produced is sufficient to melt fuse element. Fuse element6 melt before the fault current reaches its first peak value.
  • Vaporisation of metal silver elements chemically reacts with filling powder and results in the formation of high resistance substance and helps in quenching the are.

Advantages ;

  • Speed of operation is very high.
  • Maintainance cost is practically zero.
  • They are capable of clearing high as well as low fault current.
  • They do not deteriorate the age.
  • They provide reliable operation.
  • They are cheaper than other circuit interrupting devices of equal breaking capacity.


1) Heat produced by the arc may affect the associated switches.
2) they have to be replace after each operator.

Q.3.(a)  Derive EMF equation of DC generator.



∅ = Flux per pole in wb

P = Number of poles

Z = Total no. of conductors an armature

N = Speed of rotation of armature in rpm

A = Number of parallel paths in armature

Eg = EMF generated / path in armature.

Now, emf generated by faraday low,

E=No of conductor\times \frac{d\phi }{dt}volts

∴ Total EMF generated = \frac{zd\phi }{dt}volts

As the armature is rotating in magnetic field, conductor will cut the flux present in the air gap.

∴ Flux cut by one conductor in one rotation will be the total flux

= flux/pole × No. of poles

dΦ= Φ P webers ………………(1)

  • Speed in rpm =N rpm

speed in rps = \frac{N}{60} rps

  • Time required for one rotation will be dt = \frac{1}{\frac{N}{60}} =\frac{60}{N}Second

∴ dt = \frac{60}{N}Second second ………………… (2)

\frac{d\phi }{dt}=\frac{\phi P}{\frac{60}{N}}=\frac{\phi PN}{60}

∴ EMF generated for one conductor = 1 × \frac{d\phi }{dt}=\frac{\phi PN}{60} volts

Now total Z conductors are present If armature winding has a parallel paths, then conductors per path = \frac{Z}{A} conductor/ path

∴EMF Generated /path = Conductor/path × EMF/ Conductor

Eg. = \frac{Z}{A\times }\frac{\phi PN}{60}

= \frac{\phi PNZ}{60A}

1) For Lap wdg. A = P

∴ Eg = \frac{\phi NZ}{60}

2) For wave wdg A= 2

Eg = \frac{\phi PNZ}{60Z}

Q.3.(B) An 8 pole armature has 96 slots with 8 conductors per slot. It is driven at 600 rpm. The useful flux per pole is 10 mwb. Calculate the induced emf in armature  winding when it is

  1. Lap connected. 2. Wave connected

Ans. P=4

slots= 96


N=600 rpm

Φ= 10 × 10-3 Wb

Eg. = ?

Z = slots × conductor / slots

= 96× 8

= 768

N= 600 rpm

Φ = 10 mwb = 10 ×  10-3wb

Eg = ? For Lap

Eg = ? For wave

1) For lap winding


Eg = \frac{\phi NZP}{60A}

= \frac{10\times 10^{-3}\times 600\times 768\times 8}{60\times 8}

Eg = 76.8 volt

2) For wave wdg


Eg = \frac{\phi NZP}{60A}

= \frac{10\times 10^{-3}\times 600\times 768\times 8}{60\times 2}

Eg = 307.2 volts

Q.4. (A) Draw and explain electrical characteristics of DC series motor, necessary equation and its application.

Ans. 1. In these  motors the field winding is connected in series with the armature winding. The field coils are consisting of a few turns of thick wire. The same current is flowing through both the winding.

2.The current and voltage equation of DC series motor is given by

Ia= IF= IL ——– current equation

Vt– Eb= IaRa + Ia Rs + B.V.D

Vt= Eb+ Ia(Ra+Rs) + B.V.D

Eb= Vt-Ia(Ra+Rs) + B.V.D——– Voltage equation

I_{a}=\frac{V_{t}-E_{b}}{R_{a}+R_{s}} ——- Eb changes Ia changes hence Φ also changes means Φ is not constant for DC series motor.

A) Speed vs. Armature current characteristics.(N/Ia)

  1. E_{b}=\frac{\phi PN}{60}\times \frac{Z}{A}= V_{t}-I_{a}(R_{a}+R_{s})

Eb ∝ ΦN ∝ Vt-Ia(Ra+Rs)

2. Therefore, at low values of Ia, the volateg drop of Ia(Ra+Rs) is negiligibly small in comparison with Vt. Because Ra and Rs is very small.

Eb ∝ ΦN ∝ Vt

N ∝ \frac{V_{t}}{\phi }

Since Vt is constant

N ∝ \frac{1}{\phi }

3. In the series motor, the flux Φ is produced by the armature current flowing in the field winding. Actually Is is flowing through field winding.

Φ ∝ Is

but Ia=Is,

Therefore, Φ ∝ Ia

Hence the series motor is a variable flux machines

Equation now becomes

N ∝ \frac{1}{I_{a}}

Thus for the series motor the speed is inversely proportional to the armature and load current. Since the speed armature current characteristics is rectangular hyperbola.

5. N ∝ \frac{1}{I_{a}} shows that when the load current is small, the speed will be very large, therefore at no load or ta light loads there is possibility of dangerously high speeds, which may damage the motor due to large centrifugal forces.

6. Hence, a series motor must never run unloaded. It should be coupled to a mechanical load either directly or through gearing.It should never be coupled by belt, which may break at any time.

7. With the increase in armature current (which is also the field current) , the flux also increases and therefore the speed is reduced.

B. Torque -armature characteristics

The torque equation is given by

T= \frac{1}{2\Pi }\left ( \frac{PZ}{A} \right )\phi I_{a}

T ∝ ΦI

For DC series motor, Φ ∝ I

Therefore, T ∝ Ia .I

T∝ Ia2

Therefore , This equation shows that T vs Ia curve will be parabolic. This characteristics is initially parabolic and finally becomes linear when Ia larges.

c. Speed- Torque characteristics

The speed torque characteristics of DC series motor can be derived from its speed -armature current (N vs Is) and torque-armature current(T vs Ia) characteristics. Hence speed torque characteristics is shown in the fig.

Application- DC series motor is used where high starting torque is required

Example Cranes, hoists , conveyor belts, trains, electric locomotions etc.

Q.4.(B) Numerical.

A 250V, DC shunt motor runs at 1000rpm at no load and takes 8A. The total armature and shunt field resistance are 0.2 Ωand 250 Ωres. Calculate the speed when loaded and taking 50 A. Assume flux to be constant.

Ans. = Given

Vt = 250V

At No load

N1 = 1000 rpm

IL1 = 8A

Ra = 0.2Ω

Rsh = 250Ω

At load

N2 = ?

IL2 = 50A

Φ = constant

Step 1) Ish= \frac{V_{t}}{R_{sh}} = \frac{250}{250} = 1A

Step 2) Ia1 =IL1-Ish = 8-1 = 7A

Ia2 = IL2-Ish = 50-1 = 49A

Step 3) Eb1 = Vt-Ia1Ra =  250-7(0.2) = 248.6 volts.

Eb2 = Vt – Ia1Ra = 250-49(0.2) = 240.2 voles.

Step 4) N ∝ \frac{Eb}{\phi } Φ = constant


N_{2}=\frac{Eb_{2}}{Eb_{1}}\times N_{1}

N_{2}=\frac{240.2}{248.6}\times 1000

∴ N=966.2 rpm

Q.5(A) Define tariff. What are types of tariff and explain simple rate tariff.

Ans. Tariff- i. Tariff is defined as the rate of selling the electrical energy to the consumers. e.g. 3.00/- per brush.

ii) The tariff rates for different types of consumers such as domestic types of consumers such as domestic, industrial, farmers etc. will be different.

iii) The tariff is dependent on many factor such as capacity of the consumer to pay, cost of production, fixed and running expenses.

There are various types of tariff and are as follows.

  1. Flat rate tariff
  2. Block rate tariff
  3. Simple rate tariff
  4. Two part tariff
  5. Maximum Demand tariff
  6. Power factor tariff

Simple rate tariff –

  • If the rate per unit of consumed electricity is fixed, then it is called as simple rate tariff
  • This tariff is also called as uniform rate tariff
  • The cost per unit remain always constant . And independent or increase or decrease in number of units consumed.
  • the energy consumption is measured by means of an energy meter installed at the consumers place.
  • This is the simplest type of tariff and easy for consumes to understand
  • Disadvantages
    • Every consumer is charged equally. There is no discrimination.
    • This does not encourage the consumers to use more electricity
    • The cost of energy per unit is high.

Q.5(B) Numerical.

 A consumer has following load schedule for a day

Ans. Step 1. Energy Consumption = (10×60×4)+(1×1000×2)+(2×90×6)

= 5480 Whr = 5.48 Kwhr

Energy Consumption for a day = 5.48 units

Step 2. Energy Consumption for 31 days = 169.88 units × 31 days = 169.88 units

Step 3. Bill for month of 31 days = (3× 20) +(5×30)+[(169.88-20-30)×7.5]

= 60 +150+(119.880×7.5)

= 60+150+899.1

        Bill       = 1109.1 Rs. 

Q.6(A) Define.

1. Luminous Flux-  1. It is defined as that radiant energy from hot body which produces the visual sensation upon the human eye.

2. It is denoted by F and expressed in lumens.

3. The whole of electrical power supplied to a lamp is not converted, Luminous Flux but some of the power is lost by heat conduction.

2. Illumination- When the light falls on a surface it is illuminated. The illumination is defined as the luminous flux received by a surface per unit area. It is denoted by E and represented by lumens per square m or Lux.


3. Light Intensity-

a. Consider a point source of light O. Let df be luminous flux crossing any section of a narrow cone of the solid angle dw steradians. The apex of the cone so formed is at source.

b. Solid Angle. – It is subtended at a point in space by an area and is the angle enclosed in the volume formed by infinite number of lines lying on the surface of volume and meeting at the point. It is denoted by w and expressed in steradians.

c. Hence, Luminous intensity is defined as ratio of luminous flux df emitted by source to solid angle dw. It is denoted by and expressed in candela.

If ‘df’- luminous flux in lumens

‘dw’- solid angle in steradians

Therefore,   I=\frac{df}{dw} i.e. \frac{F}{w} lumens/steradians or candela

4. Candle power   – Candle power of the source is defined as the number of lumens emitted by that source per unit. Solid angle in a given direction. The term candela power is used interchangeably with luminous intensity and It is denoted by C.P

C.P= \frac{Lumens}{w}

Q.6(B) Explain construction and operation of mercury vapour lamp and its application.

Ans.  Mercury vapour lamp

It is an electric discharge lamp in which light is produced by electrical conduction through mercury vapour. It emits a very bright greenish blue light.

1. As shown in figure , it consists of two tubes ,one inside the other.
2.The smaller inner tube is made of hard glass(or quartz) and is surrounded by the larger glass tube or bulb.
3.The space between the two tubes is completely evacuated to prevent heat loss.
4.The outer tube also absorbs harmful ultraviolet radiations.
5.The inner tube contains argon gas and a small quantity of mercury. It is called as arc tube.
6.It contains two main electrodes A and B and auxiliary electrode C. The  main electrode are and of tungsten and are coated with barium oxide.
7.The auxiliary electrode c is placed near the main electrode A and it is connected to electrode B through resistance  R.
8.The choke is connected in series to limit the current .
9.The capacitor C, is connected across the mains supply line to improve the power factor.

1.When switching ON the supply an initial discharge is established in argon gas between the main electrode A and auxiliary electrode C.
2. After few seconds , the discharge take place between the two main electrode and spreads throughout the inner tube.
3.The heat produced by this gaseous discharge causes mercury to get vapourized .
4.The potential across the electrodes A and B rises from about 20V to 150V . It takes about 4-5 minute for mercury arc to build up to full brightness.
5.The mercury vapour lamps must be operated in its vertical
6.After the lamp is switched OFF , before it can restart it must be cooled to reduce the vapour pressure.

1.It life is of about 9000 working hours. Life is long.
2.It has high efficiency of about 50 lumens/watt.
3.Mercury lamps are available in various size upto 2 kw.

1.Mercury vapour lamp cost is high as compared to incandescent lamp.

It has widely used in shopping centers, industrial installation, railway yards ,street lighting, highway etc.

Q.7.(A) Give difference between squirrel cage and slip ring Induction motor.


Squirrel Cage rotor  Wound or slip ring rotor
1. Rotor is in  the form of bars which are shorted at the ends with the help of of end rings. 1. Rotor is in the from of 3 phase winding.
2. No slip ring or brushes. 2. Slip ring and brushes are used.
3. Simple construction 3. Construction is complicated
4. External resistance cannot be connected. 4. It  is possible to connect the external resistance to the rotor.
5. Small and moderate starting torque. 5. High starting torque can be obtained
6. Starting torque cannot be adjusted. 6. It is possible to connect the external resistance to the rotor.
7. Frequent maintainance is not required due to absence of slip ring and brushes. 7. Frequent maintainance is necessary due to the use of slip ring and brushes.
8. Rotor resistance stator cannot be used. 8. Rotor resistance starter can be used.
9. less rotor copper loss 9. High rotor copper loss
10. Higher Frequency 10. Low Efficiency
11. Application include lathes, fans, water pumps , blowers etc. 11. Application includes cranes, elevations. compressors, lifts etc.

Q.7(B) A 415 V, 50 Hz, 3Φ Induction motor has 4 poles. calculate

1. Synchronous speed

2. % slip if it runs at a420 rpm.

3. Frequency of rotor current for above slip

4. Frequency of rotor current at standstill

5. Motor speed if slip is 3%

Ans. V= 415 V,  f=50 Hz, P= 4

Step 1. N_{s}=\frac{120f}{P}=\frac{120\times 50}{4}= 1500 rpm

    Ns=1500 rpm

Step 2. % Slip = \frac{N_{s}-N_{r}}{N_{s}}\times 100=\frac{1500-1420}{1500}=5.33%

  % slip= 5.33%

Step 3. f’= sf= 0.0533×50= 2.665 Hz.      f’=2.665 Hz

Step 4. f’ standstill = sf = 1 × 50 = 50 Hz.    f’standstill = 50 Hz

Step  5. Nr=?  Nr= (1-s) Ns = (1-0.03) 1500 = 1455 rpm.  Nr= 1455 rpm

Q.8(A) Why 1Φ  Induction motor is never self starting? How to make it self starting?

Ans. 1. Constructionally, this motor is more or less similar to polyphase induction motor, except that a its stator is provided with a single-phase winding and b) a centrifugal switch is used in some types of motor, in order to cut out a winding, used only for starting purpose.

2. It has distributed stator winding and a squirrel cage rotor.

3. When fed from a single-phase supply, its stator winding produces a flux (or field) which is only alternating. i.e. one which alternates along one space axis only. It is not a synchronously revolving (or rotating) flux, as in the case of  a two or a three phase stator winding fed from a two or three phase supply.

4. Now an alternating or pulsating flux acting on a stationary squirrel cage rotor cannot produce rotation (only a revolving flux can) . That is why single phase motor is not self starting.

Provisions to make it self starting

  1. To make a single phase induction motor self starting, the magnetic field produced by the stator winding  should be change from alternating type to rotating type.
  2. This is possible only if more than one single phase winding are present in the stator core with space displacement.
  3. Also that the current flowing in the windings should have a phase displacement (time displacement) between them.
  4. If two single phase winding are wound in the stator core than ideally the windings should be spaced  90°physically apart and the phase displacement between the current in the two windings should be as large as 90° electrical to create maximum torque on the rotor.

Q.8 (B) Write short note on  capacitor start capacitor run induction motor with neat diagram.

Ans. Diagram- 

Consider from the diagram , IA= IS

Starting capacitor as Cs

Running capacitor as Cr

Cr = Running capacitor      Im= Main wdg current

Cs = Starting capacitor       Is = Starting wdg current

1 In this type of motor, the auxiliary winding and main winding is connected with two capacitor Cs and Cr connected in parallel.

2 The auxiliary winding and capacitor remain connected in circuit all the times.

3 Since the capacitance which will give optimum running performance is not the same as the capacitor which will give best starting torque. Also capacitor used for normal running (Cr) should be of continuous duty rating an another capacitor (Cs) used for starting should be short duty rating.

4 Centrifugal switch is used to take out Cs our of auxiliary winding circuit when motor has attained per-determined speed.

5 During starting when when the switch ‘S’ is closed, capacitor Cs and Cr in parallel so that the total capacitance in auxiliary winding circuit in sum of their values.

6 When the motor reaches about 75% of full speed, the centrifugal switch ‘S’ open and cuts the starting capacitor out of auxiliary  winding circuit.

7 Only CR remains in the auxiliary winding.

8 Such a motor operates a a two phase motor from single phase supply, thereby, producing a constant torque.

9 As shown in fig. Current drawn by the main windings lags the supply voltage ‘V’ by some angle as it is inductive in nature where as Is Leads V by certain angle as it is capacitive in nature.

Main winding = Inductive in nature + Internal Resistance

L, r hence lags by some angles to V

Auxiliary winding = capacitive in nature + Internal Resistance Cr + Cs + r hence Leads by some angles to V


10 The two currents Is and Im are out of phase by near about 80°. Their resultant current I is small.

The main advantages of these motors is high starting torque that they can produce. The Ist can be as high as 300 to 400% of the full load torque.

Role of Capacitor.

1 The capacitor always remains in the circuit. Therefore it decides the performance of the motor at the time of starting as well as during the running mode.

2 The selection of the value of C is a comprise between two factors namely starting torque and better running performance.

3 Fig. Shows torque speed characteristics of capacitor start capacitor run motor.

4 Note that starting torque is less than that of capacitor start motor.

5 Advantage of this motor is its high power factor which is due to continuous presence of capacitor in the circuit .

6 Disadvantage is its higher cost.

Characteristics :-


Due to high Ist, capacitor start capa. run I.M used for following applications.

  1. Grinders
  2. Fan and air conditioners
  3. Compressors
  4. Refrigerators
  5. Conveyors


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