What is reactive power? Is it necessary for power system operation?

Reactive power is the virtual oscillating power in the power system that enables active power to do the work in the system.So it is necessary for the power system.

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Define the following: Average demand, Maximum demand, Demand factor, Load factor.

• Average Demand: the average power requirement during some specified period of time of considerable duration is called the average demand of installation.
• Maximum Demand: The maximum demand of an installation is defined as the greatest of all the demand, which have occurred during a given period. It is measured accordingly to specifications, over a prescribed time interval during a certain period.
• Demand Factor: It is defined as the ratio of actual maximum demand made by the load to the rating of the connected load.
• Load Factor: It is defined as the ratio of the average power to the maximum demand.

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State the factors, for the choice of electrical system for an aero turbine.

The choice of electrical system for an aero turbine is guided by three factors:
• Type of electrical output: dc, variable- frequency ac, and constant- frequency ac.
• Aero turbine rotational speed: constant speed with variable blade pitch, nearly constant speed with simpler pitch- changing mechanism or variable speed with fixed pitch blades.
• Utilization of electrical energy output: in conjunction with battery or other form of storage, or interconnection with power grid.

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State the methods of improving power factor?

Methods of improving power factor:
• By connecting static capacitors in parallel with the load operating at lagging power factor. 
• A synchronous motor takes a leading current when over excited and therefore behaves like a capacitor.
• By using phase advancers to improve the power factor of induction motors. It provides exciting ampere turns to the rotor circuit of the motor. By providing more ampere-turns than required, the induction motor can be made to operate on leading power factor like an overexcited synchronous motor.

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Mention the disadvantages of low power factor? How can it be improved?

Disadvantages of low power factor:
• Line losses are 1.57 times unity power factor.
• Larger generators and transformers are required.
• Low lagging power factor causes a large voltage drop, hence extra regulation equipment is required to keep voltage drop within prescribed limits.
• Greater conductor size: To transmit or distribute a fixed amount of power at fixed voltage, the conductors will have to carry more current at low power factor. This requires a large conductor size.

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Define the following terms:- • Reliability, • Maximum demand, • Reserve-generating capacity, • Availability (operational).

Reliability: It is the capacity of the power system to serve all power demands without failure over long periods.
Maximum Demand: It is maximum load demand required in a power station during a given period.
Reserve generating capacity: Extra generation capacity installed to meet the need of scheduled downtimes for preventive maintenance is called reserve-generating capacity.
Availability: As the percentage of the time a unit is available to produce power whether needed by the system or not.

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Comment on the working principle of operation of a single-phase transformer

Working principle of operation of a single-phase transformer can be explained as
An AC supply passes through the primary winding, a current will start flowing in the primary winding. As a result, the flux is set. This flux is linked with primary and secondary windings. Hence, voltage is induced in both the windings. Now, when the load is connected to the secondary side, the current will start flowing in the load in the secondary winding, resulting in the flow of additional current in the secondary winding. Hence, according to Faraday’s laws of electromagnetic induction, emf will be induced in both the windings. The voltage induced in the primary winding is due to its self inductance and known as self induced emf and according to Lenze’s law it will oppose the cause i.e. supply voltage hence called as back emf. The voltage induced in secondary coil is known as mutually induced voltage. Hence, transformer works on the principle of electromagnetic induction.

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Explain the process of commutation in a dc machine. Explain what are inter-poles and why they are required in a dc machine.

Commutation: It is phenomenon when an armature coil moves under the influence of one pole- pair; it carries constant current in one direction. As the coil moves into the influence of the next pole- pair, the current in it must reverse. This reversal of current in a coil is called commutation. Several coils undergo commutation simultaneously. The reversal of current is opposed by the static coil emf and therefore must be aided in some fashion for smooth current reversal, which otherwise would result in sparking at the brushes. The aiding emf is dynamically induced into the coils undergoing commutation by means of compoles or interpoles, which are series excited by the armature current. These are located in the interpolar region of the main poles and therefore influence the armature coils only when these undergo commutation.

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Explain different types of D.C motors? Give their applications

Different type of DC motors and their applications are as follows:-
• Shunt motors: It has a constant speed though its starting torque is not very high. Therefore, it is suitable for constant speed drive, where high starting torque is not required such as pumps, blowers, fan, lathe machines, tools, belt or chain conveyor etc.
• Service motors: It has high starting torque & its speed is inversely proportional to the loading conditions i.e. when lightly loaded, the speed is high and when heavily loaded, it is low. Therefore, motor is used in lifts, cranes, traction work, coal loader and coal cutter in coalmines etc.
• Compound motors: It also has high starting torque and variable speed. Its advantage is, it can run at NIL loads without any danger. This motor will therefore find its application in loads having high inertia load or requiring high intermittent torque such as elevators, conveyor, rolling mill, planes, presses, shears and punches, coal cutter and winding machines etc.

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Explain different losses in a transformer.

There are two types of losses occurring in transformer:
• Constant losses or Iron losses: The losses that occur in the core are known as core losses or iron losses. Two types of iron losses are:
o eddy current loss 
o Hysteresis loss. 
These losses depend upon the supply voltage, frequency, core material and its construction. As long as supply voltage and frequency is constant, these losses remain the same whether the transformer is loaded or not. These are also known as constant losses.
• Variable losses or copper losses: when the transformer is loaded, current flows in primary and secondary windings, there is loss of electrical energy due to the resistance of the primary winding, and secondary winding and they are called variable losses. These losses depend upon the loading conditions of the transformers. Therefore, these losses are also called as variable losses.

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State Maximum power transfer theorem

The Maximum power transfer theorem explains about the load that a resistance will extract from the network. This includes the maximum power from the network and in this case the load resistance is being is equal to the resistance of the network and it also allows the resistance to be equal to the resistance of the network. This resistance can be viewed by the output terminals and the energy sources can be removed by leaving the internal resistance behind.

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State Norton’s Theorem

The Norton’s theorem explains the fact that there are two terminals and they are as follows:
• One is terminal active network containing voltage sources 
• Another is the resistance that is viewed from the output terminals. The output terminals are equivalent to the constant source of current and it allows giving the parallel resistance. 
The Norton’s theorem also explains about the constant current that is equal to the current of the short circuit placed across the terminals. The parallel resistance of the network can be viewed from the open circuit terminals when all the voltage and current sources are removed and replaced by the internal resistance.

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State Thevenin’s Theorem:

According to thevenin’s theorem, the current flowing through a load resistance 
Connected across any two terminals of a linear active bilateral network is the ratio open circuit voltage (i.e. the voltage across the two terminals when RL is removed) and sum of load resistance and internal resistance of the network. It is given by Voc / (Ri + RL).

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Name the types of motors used in vacuum cleaners, phonographic appliances, vending machines, refrigerators, rolling mills, lathes, power factor improvement and cranes.

Following motors are used: -
• Vacuum cleaners- Universal motor.
• Phonographic appliances – Hysteresis motor.
• Vending machines – Shaded pole motor.
• Refrigerators – Capacitor split phase motors.
• Rolling mills – Cumulative motors.
• Lathes – DC shunt motors.
• Power factor improvement – Synchronous motors.

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What are the different methods for the starting of a synchronous motor.

Starting methods: Synchronous motor can be started by the following two methods:
• By means of an auxiliary motor: The rotor of a synchronous motor is rotated by auxiliary motor. Then rotor poles are excited due to which the rotor field is locked with the stator-revolving field and continuous rotation is obtained.
• By providing damper winding: Here, bar conductors are embedded in the outer periphery of the rotor poles and are short-circuited with the short-circuiting rings at both sides. The machine is started as a squirrel cage induction motor first. When it picks up speed, excitation is given to the rotor and the rotor starts rotating continuously as the rotor field is locked with stator revolving field.

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Explain the application of storage batteries.

Storage batteries are used for various purposes, some of the applications are mentioned below:

• For the operation of protective devices and for emergency lighting at generating stations and substations.
• For starting, ignition and lighting of automobiles, aircrafts etc.
• For lighting on steam and diesel railways trains.
• As a supply power source in telephone exchange, laboratories and broad casting stations.
• For emergency lighting at hospitals, banks, rural areas where electricity supplies are not possible.

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What is slip in an induction motor?

lip can be defined as the difference between the flux speed (Ns) and the rotor speed (N). Speed of the rotor of an induction motor is always less than its synchronous speed. It is usually expressed as a percentage of synchronous speed (Ns) and represented by the symbol ‘S’.

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Why back emf used for a dc motor? highlight its significance.

The induced emf developed when the rotating conductors of the armature between the poles of magnet, in a DC motor, cut the magnetic flux, opposes the current flowing through the conductor, when the armature rotates, is called back emf. Its value depends upon the speed of rotation of the armature conductors. In starting, the value of back emf is zero.

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What are the various kind of cables used for transmission?

Cables, which are used for transmitting power, can be categorized in three forms:
• Low-tension cables, which can transmit voltage upto 1000 volts.
• High-tension cables can transmit voltage upto 23000 volts.
• Super tension cables can transmit voltage 66 kV to 132 kV.

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How can you relate power engineering with electrical engineering?

Power engineering is a sub division of electrical engineering. It deals with generation, transmission and distribution of energy in electrical form. Design of all power equipments also comes under power engineering. Power engineers may work on the design and maintenance of the power grid i.e. called on grid systems and they might work on off grid systems that are not connected to the system.

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What is reverse power relay?

Reverse power relay are used for the protection of the generating stations. The generating stations feed power to the grid and in case when the generating stations are off, there is no generation in the plants and it pulls off power from the grid. To stop this power flow from the grid to the generating stations, reverse power relays are used.

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Why is negative feedback more preferable in control system?

The role of feedback system in control system is to check deviation from the desired result, that is, it takes the sample output back to the input and compares the output signal with the input signal for any kinds of error. Negative feedback results in a better stability for the system. It rejects any kind of disturbance signal and is less sensitive to any kind of parameter variations.

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Why AC systems are more used or preferred over DC systems?

The reasons why AC systems are more widely used and more preferable than DC systems are as follows:

1. It is comparably easy to maintain and modify the voltage of AC electricity for transmission and distribution than it is for DC electricity.
2. The entire plant cost, ie, circuit breakers, isolators, transformers, etc is much lower in comparison to that of DC transmission.
3. All the power stations produce AC electricity, and hence it is advantageous and also easy to use AC rather than converting it and using DC.
4. AC is practically a sine wave current, and hence in case of a large fault in the network, the sine wave will naturally tend to zero at some point and hence it is very easy to interrupt it in an AC system rather than in a DC system.

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What are the differences between an alternator and a generator?

Alternator and generator, both devices have the same basic functionality. They both convert mechanical energy into electrical energy and they both have the same principal of electromagnetic induction. The main and only difference between these two devices lies in the construction. Alternator has a stationary armature and a rotating magnetic field for high voltages, while generator uses stationary magnetic field and rotating conductor which rolls on the armature with slip rings.

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Why is star delta starter preferred with induction motor?

Star delta starter is preferred with induction motor because it reduces the starting current about 3-4 times of the direct current which causes the voltage to drop and hence it causes less damages and losses. Also, it increases the starting torque and hence prevents from the damaging of the motor winding.

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Why a Auto Transformer is used in a grid?

auto-transformers, where they can be used (depends on the transformation ratio), are cheaper than "regular" transformers. 

Other than that, they operate exactly like a conventional transformer. Transformers do not care about the direction of the power flow, they can always operate with MW flow from HV to LV or vice-versa. 

Tap changers are used to help control voltage, particularly if you have on-load tap changers. These devices are usually installed at the HV side of the transformer (so they operate at lower current levels). 

The direction of the power flow depends on the system conditions. I don't know if this particular generator is connecting a 220 kV "system" network to a somewhat radial ("sub-transmission") 132 kV serving loads, in which case MW will flow from 220 kV to 132 kV. You could also have lots of generation connected to the 132 kV system, in which case MW could flow from the 132 kV to the 220 kV. But I can tell this: the transformer doesn't care about it, it will work properly (as long as within its capability, obviously) either way.

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Advantages of Power factor improvement and Correction

 Advantages of Power factor improvement and Correction: 

Following are the merits and benefits of improved Power factor;

1. Increase in efficiency of system and devices
2. Low Voltage Drop
3. Reduction in size of a conductor and cable which reduces cost of the Cooper
4. An Increase in available power
5. Line Losses (Copper Losses) I2R is reduced
6. Appropriate Size of Electrical Machines (Transformer, Generators etc)
7. Eliminate the penalty of low power factor from the Electric Supply Company
8. Low kWh (Kilo Watt per hour)
9. Saving in the power bill
10. Better usage of power system, lines and generators etc
11. Saving in energy as well as rating and the cost of the electrical devices and equipment is reduced 

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Why We Need to Install a Starter with a Motor?

Motor below 1 Hp is directly connect without starter because their armature resistance is very high and they have ability to afford the high current due to high resistance. So the Armature winding safe from the high starting current.

But a large size of motors has a very low armature resistance. if connect  this type of motor direct to Supply (3-phase Supply) then the large current will destroy the armature wading due to low resistance because motor is not running in this time. Why motor is not running in this time when we connect motor to supply? Obviously, because their is no Back E.M.F in the motor. the back E.M.F of the motor is reach at full rate when motor is running at full speed.

       So this is the answer that why we connect a starter with motor in series.  Starter in series with motor ( I.e. Resistance) is reduce the high starting current and armature takes a low current and motor will be start. But this is not end of our story. After starting the  motor at low current, the starter resistance reduce by turning a starter handle ( not in each case, in other system or case, this can be automatically) so the armature will take high current and motor armature will be rotate at full speed ( in other words, the speed of the motor will be increase).

For more Explanation see the example.

We know that the armature current can be finding by this formula,

I_a = V-Eb/R_{a       ,}             ( I=V/R, Ohm Law)

        Where,      

I_a =Armature current

V= Supply voltage

Eb= Back E.M.M                                           

R_a = Armature resistance

Suppose

A 5 Hp (3.73killowatt) motor with 440 volts having armature resistance 0.25 ohm resistance.

And the normal full load current is 50 amperes.

if we connect to direct to supply without starter the result will be.

So putting the values in equation

I_a= 440-0/0.25 = 1760 A

ahh! This high current will destroy armature winding because its 35.2 times high with respect to normal full load current.

1760/50 = 35.2

So that's why we need to install a starter with a motor.

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Why alternator rated in kVA. Not in kW?

The power √3 V_L I_L Cos φ delivered by the alternator for the same value of current, depends upon p.f. (Power Factor=Cos φ) of the load. But the alternator conductors are calculated for a definite current and the insulation at magnetic system are designed for a definite voltage independent of p.f. (Cos φ) of the load. For this reason apparent power measured in kVA is regarded as the rated power of the alternator.

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What are the fundamental differences between E.M.F (Electro motive force) and P.d (Potential difference)?

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Under what condition is D.C supply applied safely to the primary of a transformer?

 As i mentioned ,This is a simple electrical interview question. So if some one ask this question in interview or somewhere..then the right answer will be this....I think u got the main Idea 
Also
As we know that Transformer works only on A.C, in case of D.C Supply, the primary of transformer may start to smoke and burn....But this is the only one condition where we can operate a Transformer on D.C (although the Circuit is Useless)...For more information....in the above blog...U will see a post where we talked about the topic.....1 Transformer primary with A.C supply, 2. Transformer primary with D.C Supply.

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How To Calculate Your Electricity Bill. Easy and simple Explanation.

A consumer consumes 1000 watts load per hour daily for one month.Calculate The Total Energy bill of the consumer if per unit rate is 9 (In INR ,Rs, $, DHR, or Riyal etc) [Take 1 month = 30 Days]

Solution

1unit = 1kWh. 

So Total kWh = 1000 x 24 x 30 = 720000 watts/hour

we Want to convert it into Units, Where is 1unit = 1kWh.
So total Consumed units. 720000/1000...... (k=kilo=1000)
Total Units = 720.
Cost of per unit is 9.
So total Cost or bill= 720 x 9 = 6480. (In INR,Rs, $, DHR, or Riyal etc)

[Also note that…why we multiply with 24 although the daily rate is given.. It is not Daily Rates...It’s the rate Per Unit...Where 1 unit = 1 kWh (Also Called 1 =B.T.U = Board of Trade Unit)... It means...If you switched on a 1000 watt bulb for 1 hour...It mean you consumed 1000 watts for an hour... (1000 watts for 1 hour = 1kWh = 1 unit of Energy) So if the rate of unit is 5 Dollars, then you will pay 5 Dollars as a bill for your bulb]

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What is the purpose of ground wires in over-Head Transmission lines?

Ground wires are bare conductors supported at the top of transmission towers. They serve to shield the line and intercept lighting stroke before it hits the current carrying conductors below. Ground wires normally do not carry current. Therefore, they are often made of steel. The ground wires are solidly connected to ground at each tower.

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Why the reactance of a system under fault condition is low and faults currents may raise dangerously high value. ? (With simple example)

 Because the total Power is same, and under fault condition (Short circuit)...There is no load (Impedance (Z), or Reactance ( X_L)= Resistance, and in case of no load, there will be no reactance or resistance, so current will be high) in this condition...So current will be too high, and when power is same, and current increases, voltage will be decrease.

Example,

Suppose, (In normal condition)

 P= 10 watt, V = 5 Volts, and Current = 2 Amp.

But in Short circuit Condition, (When current is too high)

Then,

P = 10 Watts,    I= 10 A, so

 V = P/I….. 10 Watts/10A=1 V.

In case of short circuit, there will be no load (load = may be inductive (X_L) or resistive) so when X_L(We can say it resistance or opposition of current) = Zero, then Current will be too high.

So we can see that, in case of short circuit, (Faults condition) XL (inductive Reactance) =0, so Current increase, voltage decreases.

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Will a D.C Shunt Motor operate on an A.C Supply?

The Shunt winding has a large number of turns so that it has appreciable inductance. When A.C is applied to a shunt motor, the large inductive reactance of shunt winding will reduce the field current too much. Consequently, Shunt motor will not usually run on A.C Supply.

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Why A.C needs more insulation than D.C at same voltage level?

For the same working voltage, the potential stress on the insulation is less than in case of DC system than that AC system. therefore , a DC line require less insulation.
in other words A DC System has a less potential stress over AC system for same Voltage level, this is why AC needs more insulation over DC system

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What is The Difference between a VOLTAMETER and a VOLTMETER?

A VOLTA-METER is a device used to carry out electrolytes and a VOLTMETER is a high resistance device used for measuring potential difference or voltage between two points in an electrical Circuits.

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What is the difference between a battery and a capacitor?

 There are many differences..but the major one is that Electrical Energy is stored in battery or cell in the form of chemical energy, and transformed again in the form of electrical energy, while in a capacitor, electrical charge or energy stored in the form of electrostatic field. 

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What is the normal or average life expectancy of a Transformer ?

When a Transformer is operated under ANSI / IEEE basic loading conditions (ANSI C57.96), its normal life
expectancy is about 20 years. The ANSI / IEEE basic loading conditions for Transformer are:
 
i. The Transformer is continuously loaded at rated kVA (kilo Volt Ampere) and rated Voltages (Transformer must be operated at the rated Voltage and kVA)
 
ii. The average temperature of the ambient air during any 24-hour period is equal to 30°C (86 °F) and at no time exceeds 40°C (104 °F).
 
iii. The height where the transformer is installed,
does not above 3300 feet  or 1000 meters.

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Why the Circuit Power factor (Cos θ) Decreases, when Inductance (L) or inductive reactance (XL) increases, In inductive circuit?

Explanation:  

Suppose, 

when Inductance (L) = 0.02H

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

X_L = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω

Z = √ (R²+X_L²) = √ (10² + 6.28²) = 11.8 Ω

Cos θ = R/Z = 10/11.05 = 0.85

 

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

X_L = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R²+X_L²) = √ (10² + 12.56²) = 16.05 Ω

Cos θ = R/Z = 10/16.05 = 0.75

 

Conclusion:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A, and Circuit power factor was (Cos θ) = 0.85.

But, when Circuit inductance increased from 0.02H to 0.04 H, then  Power Factor (Cos θ) decreased from 0.85 to 0.75.

Hence proved,

In inductive circuit, when inductive reactance X_L increases, the circuit power factor also Decreases.

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Why the circuit Current (I) decrease, when Inductance (L) or inductive reactance (XL) increases in inductive circuit?

Explain the statement that " In Inductive circuit, when Inductance (L) or inductive reactance (X_L) increases, the circuit Current (I) decrease" 
OR
Why the circuit Current (I) decrease, when  Inductance (L) or inductive reactance (X_L) increases in inductive circuit?

Explanation:
We know that, I = V / R, 
but in inductive circuit, I = V/X_{L
So Current in inversely proportional  to the Current ( in inductive circuit.
Let's check with an example..}  
Suppose, when Inductance (L) = 0.02H 

V=220, R= 10 Ω, L=0.02 H, f=50Hz.

X_L = 2πfL = 2 x 3.1415 x 50 x 0.02 = 6.28 Ω 

Z = √ (R²+X_L²) = √ (10² + 6.28²) = 11.8 Ω

I = V/Z = 220/11.8 = 18.64 A

Now we increases Inductance (L) form 0.02 H to 0.04 H,

V=220, R= 10 Ω, L=0.04 H, f=50Hz.

X_L = 2πfL= 2 x 3.1415 x 50 x 0.04 = 12.56 Ω

Z = √ (R²+X_L²) = √ (10² + 12.56²) = 16.05 Ω

I = V/Z = 220 / 16.05 = 13.70 A

Conclusion:

We can see that, When inductance (L) was 0.02, then circuit current were 18.64 A,

But, when Circuit inductance increased from 0.02H to 0.04 H, then current decreased from13.70 A to 18.64A.

Hence proved,

In inductive circuit, when inductive reactance X_L increases, the circuit current decreases, and Vice Virsa.

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A (50/60 Hz) Transformer. Which one will give more Output? (When operates on 50 or 60 Hz frequency)

 A Transformer is designed to be operated on both 50 & 60 Hz frequency. 

For the Same rating, which one will give more out put; when,

1.  Operates on 50 Hz
2.   Operates on 60 Hz

Obviously! It will give more out put when we operate a transformer (of same rating) on 50 Hz instead of 60 Hz.

Because in previous posts, we proved that, in inductive circuit, when frequency increases, the circuit power factor decreases. Consequently, the transformer out put decreases.

Let’s consider the following example.

Suppose,

When Transformer operates on 50 Hz Frequency

Transformer = 100kVA, R=700Ω, L=1.2 H, f= 50 Hz.

X_L = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω

Impedance Z = √ (R²+X_L²) = √ (700² + 377²) = 795 Ω

Power factor Cos θ = R/Z = 700/795 =0.88

Transformer Output (Real Power)

kVA x Cos θ

100kVA x 0.88

88000 W = 88kW

Now,

When Transformer operates on 60 Hz Frequency

Transformer =100kVA, R=700Ω, L=1.2 H, f= 60 Hz.

X_L = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω

Impedance Z = √ (R²+X_L²) = √ (700² + 452.4 ²) = 833.5 Ω

Power factor = Cos θ = R/Z = 700/833.5 =0.839

Transformer Output (Real Power)

kVA x Cos θ

100kVA x 0.839

=83900W = 83.9kW Output

Now see the difference (real power i.e., in Watts)

88kW- 83.9kW = 4100 W = 4.1kW

If we do the same (As above) for the power transformer i.e, for 500kVA Transformer, the result may be huge, as below.

(Suppose everything is same, without frequency)

Power Transformer Output (When operates on 50 Hz)

500kVA x 0.88 = 44000 = 440kW

Power Transformer Output (When operates on 60 Hz)

500kVA x 0.839 = 419500 = 419.5kW

Difference in Real power i.e. in Watts

440kW – 419.5kW = 20500 = 20kVA

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Why Power in pure Capacitive Circuit is Zero (0)?

We know that in Pure capacitive circuit, current is leading by 90 degree from voltage ( in other words, Voltage is lagging 90 Degree from current) i.e the phase difference between current and voltage is 90 degree.
So  If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]
So if you put Cos 90 = 0→Then Power will be Zero (In pure capacitive circuit)

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Why Power in pure Inductive Circuit is Zero (0).

We know that in Pure inductive circuit, current is lagging by 90 degree from voltage ( in other words, Voltage is leading 90 Degree from current) i.e the pahse difference between current and voltage is 90 degree. 
So  If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that,
We know that Power in AC Circuit
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0]
So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive circuit)

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Why Power in a circuit is Zero (0), in which Current and Voltage are 90 Degree out of phase?

If Current and Voltage are 90 Degree Out of Phase, Then The Power (P) will be zero. The reason is that, 
We know that Power in AC Circuit 
P= V I Cos φ
if angle between current and Voltage are 90 ( φ = 90) Degree. then
Power P = V I Cos ( 90) = 0
[ Note that Cos (90) = 0] 
So if you put Cos 90 = 0→Then Power will be Zero (In pure Inductive circuit)

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A Voltmeter, an ammeter (Ampere meter) and a battery cell are connected in series. It is observed that ammeter practically shows No Deflection. Why?

Due to the large resistance of the voltmeter, the circuit resistance becomes very high. As a result, very small current will flow in the circuit. this small current f on passing through the coil of voltmeter will produce some deflection. However, in case of ammeter (Ampere meter), most of this small current will flow through the shunt. Consequently the deflection of the ammeter will be practically nil.

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The Main difference between Active and passive Commonest (Very Easy Explanation with Examples)

Active Components:
Those devices or components which required external source to their operation is called Active Components. 
For Example: Diode, Transistors, SCR etc...
Explanation and Example: As we know that Diode is an Active Components. So it is required an External Source to its operation. 
Because,  If we connect a Diode in a Circuit and then connect this circuit to the Supply voltage., then Diode will not conduct the current Until the supply voltage reach to 0.3(In case of Germanium) or 0.7V(In case of Silicon). I think you got it :) 

Passive Components:
Those devices or components which do not required external source to their operation is called Passive Components. 
For Example: Resistor, Capacitor, Inductor etc...
Explanation and Example: Passive Components do not require external source to their operation. 
Like a Diode, Resistor does not require 0.3 0r 0.7 V. I.e., when we connect a resistor to the supply voltage, it starts work automatically without using a specific voltage. If you understood the above statement about active Components, then you will easily get this example. :)

In other words:
Active Components:
Those devices or components which produce energy in the form of Voltage or Current are called as Active Components
For Example: Diodes Transistors SCR etc…
Passive Components:
Those devices or components which store or maintain Energy in the form of Voltage or Current are known as Passive Components
For Example:  Resistor, Capacitor, Inductor etc...
In very Simple words;
Active Components: Energy Donor
Passive Components: Energy Acceptor
Also Passive Components are in linear and Active Components are in non linear category.

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