In Simple words,

There are two type of losses in a transformer;

Copper losses ( I²R)depends on Current which passing through transformer winding while Iron Losses or Core Losses or Insulation Losses depends on Voltage.

Thats why the rating of Transformer in kVA,Not in kW.

Also read here about :

Transformer

How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)?

TRANSFORMER NAMEPLATE (GENERAL REQUIREMENTS).

There are two type of losses in a transformer;

**1. Copper Losses****2. Iron Losses or Core Losses or Insulation Losses**Copper losses ( I²R)depends on Current which passing through transformer winding while Iron Losses or Core Losses or Insulation Losses depends on Voltage.

Thats why the rating of Transformer in kVA,Not in kW.

Also read here about :

Transformer

How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)?

TRANSFORMER NAMEPLATE (GENERAL REQUIREMENTS).

Great post, now im knowing why tranformer rating in KVA..

ReplyDeleteHi Great and help me to why we calculate KVA for Generator and UPS.

DeleteAgain I am little bit confuse about the unit of Transformer.

ReplyDeleteWe write, P=VxI and whatever product of VxI comes out to be, we just denote it with unit "WATT". Why not "VA" as in Tranformers?

then in case of Transformer due to CU loss (I) and Iron loss (V) we write its unit "VA". Why not "WATT" as in case of P=VI?

Dear Mukesh Khatri.

DeleteAs P = V x I = Watts ..But this is Real Power...As Also wee write P = V x I = VA..This is Apparent Power...

Its recommended that read this article...I am sure that it will clear your confusion..

Active, Reactive, Apparent and Complex Power. Simple explanation with formulas

Dear Mukesh Khatri

DeleteWhen we write the equation as P=V*I, then it unit is always VA, that is apparent power.

But when the power not only depends on voltage & current but also depends on their phase relation ship then we generally use the formula P=V*I*cos(THETA).... Then the unit becomes Watt...

this is the question I am asking that why the unit of Apparent power (S)=VI is VA? Why not to write Watts?

ReplyDeleteSo what if it is apparent power and we are writing it as VA and in case of active power we write Watts. Why? Couldn't get the logic behind.

I read your post it is really good but unit is still not clear to me logically.

Dear Mukesh Khatri@

DeleteSee.... Watt is a unit of real power or actual power...i.e. the power which we utilize in our home [ Electrical Appliances operate on Real Power]

in other hand... Apparent Power [ S= VxI = VA and its formula

Apparent Power = √ (True power2 + Reactive Power2)] is not real power..its complex power [ ... I.e we can't use Reactive power but only real power in watts but we pay for both real and reactive power. As reactive power provide only Electric or magmatic field.

Therefore we take real power in Watts....and Apparent power in VA.

nice ans

Delete\

hi mukesh,i see you are confused about kva and kw.just keep in mind that where ever there is a generation point of electricity its not loaded with anything so its calculated in kva,but when we put on the load on to it then we calculate it on kw becuase of the power at the consuming end,if you are not consuming the electricity then there is no kw.

DeleteReal Power=Voltage*Current*Cos(Pi).Watts

DeleteReactive Power=Voltage*Current*Sin(Pi)

Apparent Power=Resultant of Real and reactive power so cos(pi) and Sin(pi) will become one.hence we will get apparent power Volt-Amps only

it is the totoal power for real and reactive power so it is to the v-i

Deletemy question in simple words is how you come to know that P=VxI= VA?

ReplyDeleteIf you have the proof of denoting VxI with unit VA then what was the necessity of using unit the VA inspite of Watts as a unit for apparent power?

What is the problem with the unit WATT that we don't mention it as a unit of apparent power when we clearly see it is product of VxI ..

Dear Mukesh Khatri@

DeleteSee.... Watt is a unit of real power or actual power...i.e. the power which we utilize in our home [ Electrical Appliances operate on Real Power]

in other hand... Apparent Power [ S= VxI = VA and its formula

Apparent Power = √ (True power2 + Reactive Power2)] is not real power..its complex power [ ... I.e we can't use Reactive power but only real power in watts but we pay for both real and reactive power. As reactive power provide only Electric or magmatic field.

Therefore we take real power in Watts....and Apparent power in VA.

I know, Apparent power(s) is sum of route of Active Power(wattful) and Reactive Power(wattless).

DeleteI am afraid how does this justify that the unit of Apparent Power will be "VA"?

You mean to say it is complex power and we utilize real power not reactive power therefore Apparent power's unit should be "VA".

Does common sense accept this justification to denote Apparent Power as " VA " not " watts"?

Candidly speaking still I couldn't understand the logic, where from I could say (YES! this is why Apparent Power has unit "VA")

Look...There are three kinds of Power. Active Power, Re-active Power and Apparent power. now use common sense,,,is it right to just use the same unit for these three different quantities. The same question rises here that we also know that Re-active power = P = VxI...But the unit is in VARS....Why??? Should was not it in Watts. Also this is not a technical answer [ as we have already given answer in the above reply]

DeleteSee @Mukesh...Apparent power is the product of voltage(V) & current(I). Now, the unit of voltage is volt & unit of current is ampere. So the unit of apparent power is volt-ampere.

DeleteNow to get back to watts...when the apparent power is further multiplied with power factor(cosine of the angle between voltage & current in an AC circuit) the result is active power which is different from apparent power. So we can't represent the unit of active power with the same unit as of apparent power and hence the unit of active power is given a name called watts.

Is that clear now?

great sir tq now 1ly iam clear about this tx a lot

Deletegreat work by your team

ReplyDeleteThanks for Appreciation.

Deletegreat explanation....hats of to u.......

ReplyDeleteThanks for appreciation

Deletedear sir;

ReplyDeleteI need a clarification regarding..

1.why we should not connect earth to the ground?

2.how to calculate the bus bar size?

Fist the understading P=VA=WATT is wrong,

ReplyDeleteP=VA Cos(phi) = watt for AC power

S=VA = Apparent Power

One can write P=VA only withe explanation that power factor Cos (phi) is unity.

For DC curcuit P=VA=Watt since there is no power factor as well as no apparenat power.

W and VA are the same, they are both power. VA unit is just used for apparent power to discriminate from the real power unit (W), if we use the unit watt(W, for both apparent and real power, how would somebody what power you are referring to? this is to avoid confusion. The same logic on big engines, they have this ratings: BHP, KW

ReplyDeleteWhy motor rating is in KW? not in VA..

ReplyDeletesimply put: its bcoz of power factor, the mptor is at the load side hence PF is taken into consideration, when u check on the name plate you will find the following rated parameters:

DeleteVoltage

Amperage

Power Factor....thus

Voltage*Amperage*PF =Kw

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Why motor rating in kw nt in kva,there are also cu and iron loss

ReplyDeleteplease help to solve this problem:

ReplyDeleteA 440V, 60hz, six-pole, 3-phase induction motor is taking 50kVA at 0.8 power factor and is running at a slip of 2.5 %. The stator copper losses are 0.5kW and rotational losses are 2.5kW. Calculate:

a) the rotor copper losses

b) the shaft power

c) the efficiency

d) the shaft torque

whats a explanation by faruq sir...you clear the concept of watt and volt-amp

ReplyDeleteIf i want to install a 250 kva generator set, how can i choose the rating of the transformer t be installed? Can anybody help me please!

ReplyDeletePls Answer this Q.... What is the current of an 237V, 328.4kW three-phase load with cosFI=0.99 ?

ReplyDeletePower in a Three Phase Circuit

DeleteP =√3xVxIxCosθ

And Current

I = P / √3xVxCosθ

Putting the values

I = 328.4kW / (√3x237x.0.99)

I = 800A

For More Details and Formulas…. Check This Post

Simple Electrical Formulas

P=VI

ReplyDeletePower=Voltage*Current

Quote "we can't use Reactive power but only real power in watts but we pay for both real and reactive power"

ReplyDeleteElectrical Technology kindly clarify, i read somewhere that our energy meters only charge us based on the Real power(Kw) consumed bt here you've stated that even the reactive power is paid for....

Can you please explain the effect of low/high impedance of transformer?

ReplyDeletewhy we need to connect neutral to earth in a Transformer??

ReplyDeletewhy in theTransformer, Neutral is Connected to Earth??

ReplyDeleteOne thing, since the earth is a good conductor, it can be used as a common line in distribution system. So the metal usage for the neutral wire can be minimized by earthing the neutral wire in many places of a distribution network.

DeleteThe other thing is safety related. The leakage current of consumer equipments and machinery can be bypassed to earth only if the neutral point of the transformer is earthed.

i like to know about multiplication factor

ReplyDeleteThe answer to the common man:

ReplyDeleteThe power indicated in Watts (W) is the power consumed by an equipment.

But the power indicated in Volt-Ampere (VA) is the capacity of an equipment to deliver the power.

You can connect a 1000W heater to a 1KVA transformer. Here the transformer is just transferring 1000W power to heater.

very nice explenation

ReplyDelete