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# Basic Electronics - Transformer Efficiency

When the Primary of a transformer has some voltage induced, then the magnetic flux created in the primary is induced into the secondary due to mutual induction, which produces some voltage into the secondary. The strength of this magnetic field builds up as the current rises from zero to maximum value which is given by $\mathbf{\frac{d\varphi}{dt}}$.

The magnetic lines of flux pass through the secondary winding. The number of turns in the secondary winding determines the voltage induced. Hence the amount of voltage induced will be determined by

$$N\frac{d\varphi}{dt}$$

Where N = number of turns in the secondary winding

The frequency of this induced voltage will be same as the frequency of primary voltage. The peak amplitude of the output voltage will be affected if the magnetic losses are high.

## Induced EMF

Let us try to draw some relationship between induced EMF and number of turns in a coil.

Let us now assume that both the primary and the secondary coils has a single turn each. If one volt is applied to one turn of the primary with no losses (ideal case) the current flow and magnetic field generated induce the same one volt in the secondary. Hence voltage is same on both sides.

But the magnetic flux varies sinusoidally which means,

$$\phi\:\:=\:\:\phi_{max} \sin \omega t$$

Then the basic relationship between induced EMF and coil winding of N turns is

$$EMF\:=\:turns\:\:\times\:\:rate\:of\:change$$

$$E\:=\:N \frac{d\phi}{dt}$$

$$E\:=\:N\:\times\:\omega\:\times\: \phi_{max}\:\times\: \cos(\omega t)$$

$$E_{max}\:=\:N \omega \phi_{max}$$

$$E_{rms}\:=\:\frac{N \omega}{\sqrt{2}}\:\times\:\phi_{max}\:=\:\frac{2\pi}{\sqrt{2}}\:\times\:f\:\times\:N\:\times\:\phi_{max}$$

$$E_{rms}\:=\:4.44\:f\:N\:\phi_{max}$$

Where

f = flux frequency in Hertz = $\frac{\omega}{2\pi}$

N = number of coil windings

∅ = flux density in webers

This is known as **Transformer EMF Equation**.

As alternating flux produces current in the secondary coil, and this alternating flux is produced by alternating voltage, we can say that only an alternating current AC can help a transformer work. Hence **a transformer doesn’t work on DC**.

## Losses in Transformers

Any Device has few losses in practical applications. The main losses that occur in the transformers are Copper losses, Core losses and Flux leakage.

### Copper Losses

Copper loss is the loss of energy, due to the heat produced by the current flow through the windings of the transformers. These are also called as “**I ^{2}R losses**” or “I squared R losses” as the energy lost per second increases with the square of the current through the winding and is proportional to the electrical resistance of the winding.

This can be written in an equation as

$$I_{P} R_{P}\:+\:I_{S} R_{S}$$

Where

= Primary Current*I*_{P}= Primary Resistance*R*_{P}= Secondary Current*I*_{S}= Secondary Resistance*R*_{S}

### Core Losses

Core Losses are also called as **Iron Losses**. These losses depends upon the core material used. They are of two types namely, **Hysteresis** and **Eddy Current losses**.

**Hysteresis Loss**− The AC induced in the form of magnetic flux keeps on fluctuating (like rise and falls) and reversing the direction according to the AC voltage induced. Some energy is lost in the core due to these random fluctuations. Such loss can be termed as**Hysteresis loss**.**Eddy Current Loss**− While this whole process goes on, some currents are induced in the core which circulate continuously. These currents produce some loss called as**Eddy Current Loss**. Actually the varying magnetic field is supposed to induce current only in the secondary winding. But it induces voltages in the nearby conducting materials also, which results in this loss of energy.**Flux Leakage**− Though the flux linkages are strong enough to produce the required voltage, there will be some flux which gets leaked in practical applications and hence results in the energy loss. Though this is low, this loss is also countable when it comes to high energy applications.

## Power of a Transformer

When an ideal transformer is considered with no losses, the Power of the transformer will be constant, as the product when voltage **V** multiplied by current **I** is constant.

We can say that the power in the primary equals the power in the secondary as the transformer takes care of that. If the transformer, steps-up the voltage then the current is reduced and if the voltage is stepped-down, the current is increased so as to maintain the output power constant.

Hence the primary power equals the secondary power.

$$P_{Primary}\:=\:P_{Secondary}$$

$$V_{P}I_{P}\cos \phi_{P}\:=\:V_{S}I_{S}\cos \phi_{S}$$

Where **∅ _{P}** = Primary phase angle and

**∅**= Secondary phase angle.

_{S}## Efficiency of a transformer

The amount or the intensity of Power loss in a transformer, determines the efficiency of the transformer. The efficiency can be understood in terms of power loss between primary and secondary of a transformer.

Hence, the ratio of power output of secondary winding to the power input of primary winding can be stated as the **Efficiency of the transformer**. This can be written as

$$Efficiency\:=\:\frac{Power\:output}{Power\:input}\:\times\:100 \%$$

Efficiency is generally denoted by **η**. The above given equation is valid for an ideal transformer where there will be no losses and the whole energy in the input gets transferred to the output.

Hence, if losses are considered and if the efficiency is calculated in practical conditions, the below equation is to be considered.

$$Efficiency\:=\:\frac{Power\:output}{Power\:output\:+\:Copper\:losses\:+\:Core\:losses}\:\times\:100 \%$$

Otherwise, it can also be written as

$$Efficiency\:=\:\frac{Power\:input\:-\:Losses}{Power\:input}\:\times\:100$$

$$1\:-\:\frac{Losses}{Input\:Power}\:\times\:100$$

It is to be noted that the input, output and losses are all expressed in terms of power, i.e., in Watts.

### Example

Consider a transformer having input power of 12KW which is rated at 62.5 amps current having equivalent resistance of 0.425ohms. Calculate the efficiency of the transformer.

**Solution −**

Given data

- Input power = 12KW
- Rated current = 62.5 Amps
- Equivalent resistance = 0.425 ohms

Calculating the loss −

The copper loss at rated current is I^{2}R = (62.5)^{2} (0.425) = 1660W

We have

$$Efficiency\:=\:\frac{Power\:input\:-\:Losses}{Power\:input}\:\times\:100$$

Hence,

$$\eta\:=\:\frac{12000\:-\:1660}{12000}\:\times\:100$$

$$\eta\:=\:\frac{10340}{12000}\:\times\:100$$

$$\eta\:=\:0.861\:\times\:100\:=\:86 \%$$

Hence the efficiency of the transformer is 86%.