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Why Transformer Rated In kVA, Not in KW?

Transformer Always Rated In kVA instead of KW

As the name suggest, transformer only transfer the power from one circuit to another without changing the value of power and frequency. In other words, It can only step up or step down the value of current and voltage while the power and frequency would remain same. A general date on transformer nameplate are printed for further details, such as rating in VA, single phase / three phase (power or distribution transformer), step up / step down, connection etc.

Right to the question, In simple words,

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. i.e. total losses depends on voltage (V) and current (I) which expressed in Volt ampere (VA) and not on the load power factor (p.f). That’s why the transformer rating may be expressed in VA or kVA, not in W or kW.

Let’s explain in more details to get the idea why transformer rated in VA instead of kW?

When manufactures design a transformer, they have no idea which kind of load will be connected to the transformer. The load may be resistive (R), inductive (L), capacitve (C) or mixed load (R, L and C). Its mean, there would be different power factor (p.f) at the secondary (load) side on different kind of connected loads depends on R, L and C. This way, they go for VA instead of W in case of Transformer.

Why Transformer Rated In kVA, Not in KW?

Lets clear the rating of transformer in VA instead of W  with solved example.

Losses of transformer will remain same as  long as the magnitude of current / voltage is same. No matter what power factor of the load current / voltage is.

Example:

Suppose for a single phase step up transformer

  • Transformer rating in kVA = 11kVA
  • Primary Voltages  = 110V
  • Primary Current = 100 A
  • Secondary Voltages = 220V
  • Secondary Current = 50 A.
  • Equivalent resistant on Secondary =
  • Iron losses = 30W

In first scenario, If we connect a resistive load to the secondary of the transformer at unity power factor θ = 1,

Then total losses of transformer would be copper losses + iron losses, i.e.

I²R + Iron losses

Putting the values,

(502 x 5 ) + 30W = 12.53kW

i.e. losses on primary and secondary of transfer is still same. (See below example for secondary losses as well)

The transformer output will be:

P = V x I x Cos θ

Again putting the value from secondary (Same value if we put the values from primary)

P = 220 x 50 x 1 = 11kW.

Now rating of transformer

kVA = VA / 1000

kVA = 220 x 50 / 1000 = 11kVA.

Now, In second scenario, connect a capacitive or inductive load to the secondary of the transformer at power factor θ = 0.6.

Again, total losses of transformer would be copper losses + iron losses, i.e.

I²R + Iron losses

Putting the values,

(502 x 5 ) + 30W = 12.53kW

Hence proved that losses in both of primary and secondary is same.

But The transformer output will be:

P = V x I x Cos θ

Again putting the value from secondary (Same value if we put the values from primary)

P = 220 x 50 x 0.6 = 6.6kW.

Now rating of transformer

kVA = VA / 1000

kVA = 220 x 50 / 1000 = 11kVA.

Its mean, 11kVA transformer rating means it can handle of 11kVA. It is our turn to transform and utilize the 11kVA as 11kW (we can do it by improving the power factor to 1 in case of pure resistive load) which is not predictable and even very hard to get in case of inductive and capacitive loads where power factor would have different values.

From the above example, it is clear that the rating of transformer is same (11kVA) but different output in power (11kW and 6.6kW) due to different power factor values after connecting different kind of load which is not predictable for transformer manufactures where the losses are same in both cases.

Good to know:

As like transformer, rating and capacity of Alternators/generators, Stabilizers, UPS, Transmission lines are also rated in VA instead if W While power plant capacity, AC (Air condition) and motors are rated in W rather than VA.

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68 Comments

      1. I understand your logic Mukesh Khatri. cos(phi) or sin(phi) produces a number and hence the units should not necessarily change. However, its a technicality to distinguish between real, apparent and reactive power. There are other instances where units are renamed to simplify rewriting the base units. In this case they rename the unit to distinguish between the types of power.

  1. Again I am little bit confuse about the unit of Transformer.<br /><br />We write, P=VxI and whatever product of VxI comes out to be, we just denote it with unit &quot;WATT&quot;. Why not &quot;VA&quot; as in Tranformers? <br /><br />then in case of Transformer due to CU loss (I) and Iron loss (V) we write its unit &quot;VA&quot;. Why not &quot;WATT&quot; as in case of P=VI?

    1. Dear Mukesh Khatri<br /><br />When we write the equation as P=V*I, then it unit is always VA, that is apparent power.<br /><br />But when the power not only depends on voltage &amp; 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…

  2. this is the question I am asking that why the unit of Apparent power (S)=VI is VA? Why not to write Watts? <br /><br />So what if it is apparent power and we are writing it as VA and in case of active power we write Watts. Why? Couldn&#39;t get the logic behind.<br /><br />I read your post it is really good but unit is still not clear to me logically.

    1. Dear Mukesh Khatri@<br />See…. 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] <br />in other hand… Apparent Power [ S= VxI = VA and its formula <br />Apparent Power = √ (True power2 + Reactive Power2)] is not real power..its complex power [ … I.e we can&#39;t use Reactive power but only real power in

    2. 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.<br />

      1. dear mukesh i think ur thinking is full of confusion
        P = V X I is VA because there is no load and no changing angle between voltage and current so that we take take pf 1. but after load there will be changing angle between voltage and current and pf is not constant which depends upon load so that it is kw

        ex– V X I X PF = VA ( PF=1)
        V X I X PF = KW ( PF = LESS THAN 1 VARY WHICH DEPENDS UPON INDUCTIVE, CAPACITIVE , RESISTANCE )

    3. Real Power=Voltage*Current*Cos(Pi).Watts<br />Reactive Power=Voltage*Current*Sin(Pi)<br />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

  3. my question in simple words is how you come to know that P=VxI= VA?<br /><br />If 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? <br /><br />What is the problem with the unit WATT that we don&#39;t mention it as a unit of apparent power when we clearly see it is product of VxI ..

    1. Dear Mukesh Khatri@<br />See…. 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] <br />in other hand… Apparent Power [ S= VxI = VA and its formula <br />Apparent Power = √ (True power2 + Reactive Power2)] is not real power..its complex power [ … I.e we can&#39;t use Reactive power but only real power in

    2. I know, Apparent power(s) is sum of route of Active Power(wattful) and Reactive Power(wattless).<br /><br />I am afraid how does this justify that the unit of Apparent Power will be &quot;VA&quot;? <br /><br />You mean to say it is complex power and we utilize real power not reactive power therefore Apparent power&#39;s unit should be &quot;VA&quot;.<br /><br />Does common sense accept this

    3. 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

    4. See @Mukesh…Apparent power is the product of voltage(V) &amp; current(I). Now, the unit of voltage is volt &amp; unit of current is ampere. So the unit of apparent power is volt-ampere.<br /><br />Now to get back to watts…when the apparent power is further multiplied with power factor(cosine of the angle between voltage &amp; current in an AC circuit) the result is active power which is

  4. dear sir;<br /> I need a clarification regarding..<br /> 1.why we should not connect earth to the ground?<br /> 2.how to calculate the bus bar size?

  5. Fist the understading P=VA=WATT is wrong,<br />P=VA Cos(phi) = watt for AC power<br />S=VA = Apparent Power <br />One can write P=VA only withe explanation that power factor Cos (phi) is unity.<br /><br />For DC curcuit P=VA=Watt since there is no power factor as well as no apparenat power.<br />

  6. 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

    1. simply 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:<br />Voltage<br />Amperage<br />Power Factor….thus<br />Voltage*Amperage*PF =Kw

  7. please help to solve this problem:<br /><br />A 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: <br />a) the rotor copper losses<br />b) the shaft power<br />c) the efficiency<br />d) the shaft torque

  8. If 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!

  9. Quote &quot;we can&#39;t use Reactive power but only real power in watts but we pay for both real and reactive power&quot;<br />Electrical Technology kindly clarify, i read somewhere that our energy meters only charge us based on the Real power(Kw) consumed bt here you&#39;ve stated that even the reactive power is paid for….

    1. One 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.<br />The 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

  10. The answer to the common man:<br />The power indicated in Watts (W) is the power consumed by an equipment.<br />But the power indicated in Volt-Ampere (VA) is the capacity of an equipment to deliver the power.<br />You can connect a 1000W heater to a 1KVA transformer. Here the transformer is just transferring 1000W power to heater.

  11. Transformer rating Depends upon the power factor of the load but this is not specified at the time of manufacturing so rating of transformer and generator in KVA but in motor nameplate manufacturer always mentioned the pf thats why its rating is in kw.

  12. Rating depends on maximum current flowing through conductor without overheating and damage of conductor wire. If heavy load connected to a thin wire, when we switch on, wire get melt due to high current required for heavy load flowing through thin conductor generate heat and wire get melts.
    Same thing happens in transformer since transformer connected to load (grid) having resistance capacitance and inductance three basic loads and all are parallel in power distribution system. All three basic circuit element taking current from transformer winding but resistance only dissipate power, reactive component only store energy in form of charge storage in capacitor and magnetic flux in case of inductor so it is called reactive power. Total current drown from transformer coil is vector sum of active plus reactive current which is greater then current that is dissipated in pure resistor and rating of any conductor (transformer coil) depends on current flowing through it,since voltage of line is fixed(220,440 etc)and In KVA, V is fixed(line voltage or generating voltage) A denote the maximum current that can withdrawn by load without damage of transformer conductor.we can not say rating as KW because it only denote the product of voltage V and current that flowing through resistor only which is always less then A. But A is the actual current(reactor+resister) flowing through transformer coil, and this A define rating of transformer conductor and hence transformer. That,s why its rating is KVA.

  13. Iron losses and copper losses are common for all electrical machines (be it motor/ alternator or transformer). That is not the reason for transformer to be rated in VA. The real reason is that all AC sources (transformer or alternators) are rated in VA (or KVA or MVA) is because the power factor of current is decided by the load connected. And since the load power factor is not known (while the transformer is being manufactured), they are rated considering resistive load (PF = 1). If the load is inductive + resistive (for example, induction motor), some part of power from transformer will go into supplying the reactive power (VAR) and remaining for supplying the active power (W). But the power supplied from transformer (apparant power, VA) will still be vector sum of VAR and W.

  14. I know this is an old thread but your calculations in the examples is wrong. The total loss is 12.545kW or 12.530kW depending on the iron losses. At the top you say that the iron losses are 30W but in the calculation you use 45W.

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