# The meaning of KVA in single phase transformers

### What is KVA rating of transformers?

In the textbook Electric Machines (4th ed.) by Kothari and Nagrath, pg.79, it is written that:

The rating of the transformer is specified in units of VA/KVA/MVA depending upon its size. $$\text{KVA(rated)}=\frac{V\text{(rated)}\times I\text{(full-load)}}{1000}$$ where V and I are referred to one particular side.

Are V and I primary voltage (V1) and current (I1) or secondary voltage (V2) and current (I2)? And what does the term "full load" mean here? Further this page mentions that:

It is advised that the KVA of the transformer should be equal to or greater than the rating of the load to handle present requirements. The KVA rating will also be affected by the application of the load, for example motors may need an increased kVA rating.

So what actually is the KVA rating of a transformer? Please add proper references. Thank you.

• The primary VA should be slightly greater than the secondary VA for an efficient transformer. The Full Load is the maximum load the transformer is designed to handle. Jul 19, 2023 at 6:32
• Machines are generally rated by output. Jul 19, 2023 at 7:17

Are V and I primary voltage (V1) and current (I1) or secondary voltage (V2) and current (I2)?

It can be any of it. Computing V*I for primary or secondary shall result in the same value of power rating, since when we go from the primary to the secondary, the V decreases in the same proportion that I increases.

Theoreticaly, one can draw any load from a transformer, however, the transformer will eventually heat up and fail somehow. So, electrical transformer designers use this "full load" (or "power rating") as a reference to compute the temperatures and ensure that for this load the temperatures inside the transformers will not go beyond the insulation temperature limits.

kVA literally means

• k - kilo (1000)
• VA = volt-amps, a unit of power similar to watts used for provisioning power

It is a unit used in the electrical industry (utilities and electricians doing mains power installations) for the purpose of provisioning (sizing equipment to deliver power).

This also correlates to how design standards and certification agencies (think UL/CSA/BSI/TUV) rate things and give them nameplate data. That nameplate data is the basis for Load Calculation.

Typical usage is that a Load Calculation will be done on the loads to be served, e.g. in North America per NEC article 220, 220.82 for a dwelling, or Article 430 for a motor. Here are a few examples of how it's used.

• A dwelling calculates out to 21.5 kVA of service needed at 240V... that is about 89.5 amps, so the builder puts in a request for the next size up, 100 amp electrical service.
• An RV park wants to install 21 RV spaces and power each at 50A/240V, which is 12,000 VA or 12 kVA. That totals 252 kVA, but due to NEC 551.73's allowance for diversity of loads, only 45% need to be provisioned, or 113.4 kVA. The park plans to accept 480V 3-phase and create 120/240V via three transformers; this means each of the three transformers must be at least 38 kVA.
• The authority pushes back and says each transformer is actually its own "mini RV park" of 7 RV stands of 84 kVA, which only gets a 55% diversity derate giving 46.2 kVA.

The kVA figures arrived from these calculations apply to the transformer's rating as assigned by the standards/agency (so you don't need to worry about whether it pertains to input V and A versus output V and A; the difference is small anyway). So for instance the RV park will need three 46.2 kVA transformers (probably 50 kVA), and will ask the utility for 114 kVA electrical service (and probably get the next size up, 125 kVA).

Notice the part where we can have this conversation without thinking about voltages and phases. :)

* What is the difference between watts and VA (or to be more precise: the ratio)? Take an LED Christmas light string made of 37 diodes in series with a protective diode. It lights during half the power cycle, taking 50mA of 120V half the time. Multiply volts x amps and you get 6 VA, which is correct. However it's only on half the time, so it's 3 watts. So it has a Power Factor of 50%. You pay for 3 watts but the supply equipment must be sized to deliver 6 VA.

The kVA rating of a transformer is the product of its secondary voltage and its secondary full load current divided by thousand.

The kVA rating is an indicator of the transformer's size and the maximum load that it is designed to handle.

This rating enables selection of the proper transformer size for the equipment it is to cater to.

It follows that the input to the transformer will be higher, considering its efficiency.

• And the full load current will be the maximum current printed on the rating plate on the transformer, or in the manufacturer's data sheet. Jul 19, 2023 at 10:45

### Apparent Power (S)

In electrical engineering, kVA is formally called Apparent Power (S).

The Apparent (S) Power is the total power consumed by an electrical system, and it's defined as the product of Voltage (V) times Current (I):

$$S = V_{rms} * I_{rms} \space\text{(in VA)}$$

For a transformer, the kVA rating usually pertains to the maximum apparent power that it can deliver to a full-load, so you usually refer to the V2 and I2 at the secondary windings because these are connected to the load. A full-load is the expected load impedance (Z) that would draw the maximum apparent power output from the transformer:

$$Z_{full-load} = \frac{V_{rms}(rated)}{I_{rms}(full-load)}$$

Since the (rated) voltage is usually fixed in a transformer, a given full-load impedance (Z) will determine what is your full-load current (I), and this will tell you what kVA rating do you need for that particular full-load.

### More Context: Power Triangle and Impedance Triangle

The Apparent Power (S) is the total power consumed by the Impedance Load (Z). However, this Apparent Power is further composed of two kinds of power, Real (P) and Reactive (Q) Power:

Likewise, the Impedance Load (Z) follows the same principle and is composed of the Resistance (R) and Reactance (X) to form the Impedance Triangle:

The Real Power (P) is the fraction of the Apparent Power (S) that is consumed by the Resistive Load (R). $$P = V_{rms} * I_{rms} * \frac{R}{Z} \space\text{(in watt)}$$

The Reactive Power (Q) is the fraction of the Apparent Power (S) that is consumed by the Reactive Load (X). $$Q = V_{rms} * I_{rms} * \frac{X}{Z} \space\text{(in var)}$$