# Transformer - Losses and Efficiency

## Losses in transformer

In any electrical machine, 'loss' can be defined as the difference between input power and output power. An electrical transformer is an static device, hence mechanical losses (like windage or friction losses) are absent in it. A transformer only consists of electrical losses (iron losses and copper losses). Transformer losses are similar to losses in a DC machine, except that transformers do not have mechanical losses.
Losses in transformer are explained below -

#### (i) Core losses or Iron losses

Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used for the construction of core. Hence these losses are also known as core losses or iron losses.
• Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the transformer core. This loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density. It can be given by, Steinmetz formula:
Wh= ηBmax1.6fV (watts)
where,   η = Steinmetz hysteresis constant
V = volume of the core in m3
• Eddy current loss in transformer: In transformer, AC current is supplied to the primary winding which sets up alternating magnetizing flux. When this flux links with secondary winding, it produces induced emf in it. But some part of this flux also gets linked with other conducting parts like steel core or iron body or the transformer, which will result in induced emf in those parts, causing small circulating current in them. This current is called as eddy current. Due to these eddy currents, some energy will be dissipated in the form of heat.

#### (ii) Copper loss in transformer

Copper loss is due to ohmic resistance of the transformer windings.  Copper loss for the primary winding is I12R1 and for secondary winding is I22R2. Where, I1 and I2 are current in primary and secondary winding respectively, R1 and R2 are the resistances of primary and secondary winding respectively. It is clear that Cu loss is proportional to square of the current, and current depends on the load. Hence copper loss in transformer varies with the load.

## Efficiency of Transformer

Just like any other electrical machine, efficiency of a transformer can be defined as the output power divided by the input power. That is  efficiency = output / input .
Transformers are the most highly efficient electrical devices. Most of the transformers have full load efficiency between 95% to 98.5% . As a transformer being highly efficient, output and input are having nearly same value, and hence it is impractical to measure the efficiency of transformer by using output / input. A better method to find efficiency of a transformer is using, efficiency = (input - losses) / input = 1 - (losses / input).

#### Condition for maximum efficiency

Let,
Copper loss = I12R1
Iron loss = Wi
Hence, efficiency of a transformer will be maximum when copper loss and iron losses are equal.
That is Copper loss = Iron loss.

### All day efficiency of transformer

As we have seen above, ordinary or commercial efficiency of a transformer can be given as
But in some types of transformers, their performance can not be judged by this efficiency. For example, distribution transformers have their primaries energized all the time. But, their secondaries supply little load all no-load most of the time during day (as residential use of electricity is observed mostly during evening till midnight).
That is, when secondaries of transformer are not supplying any load (or supplying only little load), then only core losses of transformer are considerable and copper losses are absent (or very little). Copper losses are considerable only when transformers are loaded. Thus, for such transformers copper losses are relatively less important.  The performance of such transformers is compared on the basis of energy consumed in one day.
All day efficiency of a transformer is always less than ordinary efficiency of it.

1. I had no idea that figuring out how a transformer works could be so complicated! However, it is good to be able to calculate if there is a loss of power. If there are too many losses, then I am sure that means that the transformer is not efficient. When I get a new transformer for my electrical system soon, I will be sure to check on the efficiency of the transformer so that I don't have too many losses of energy. http://www.totalgenerators.com/transformers.html

2. It is really interesting how much math goes into electricity. You have got to be really good with numbers if you are working with power transformers. You wouldn't want to overload it because that could end up blowing out the power to the entire neighborhood. I have seen what those look like when they decide to blow out and it isn't very pretty. http://www.epe.com.au/products/transformers/

3. why in the hell does dWi/dI1 = 0, (which is clearly what u implied when you got the derivative of dη/dI1)...?? and the best part about this is that after assuming that dWi/dI1 = 0, you then write that Wi=R1*(I1)^2, whose derivative with respect to I1 clearly is not zero... I'm literally willing to pay 5 dollars for anyone who can explain this nonsense. It says the same thing in my university book, and I just can't get over it

1. The core losses(Wi) in a transformer are constant, unless the frequency and volume of core changes.(The formula has been given in derivation). We all know that the frequency of a power transformer doesn't change while operating.(for eg: in india standard frequency 50hz). Thus the iron loss(core loss(wi)) in a transformer is constant and hence d(Wi)/dI is Wi.

Friend please dont swear immediately when you dont understand something....chances are that you may not know some concepts.

2. sorry one correction d(Wi)/dI=0

4. प्लीज put सम लाइट व transformer on no लोड