# Transformer

A transformer is a static piece of equipment used either for raising or lowering the voltage of an A.C supply with a corresponding decrease or increase in current. It essentially consists of two winding’s, the primary and secondary , wound on a common laminated magnetic core. The winding connected to A.C source is called ** primary winding** and the one connected to load is called

**. Depending upon the number of turns of the primary (N**

*secondary winding*_{1}) and secondary (N

_{2}), an alternating e.m.f E

_{2}is induced in the secondary. This induced e.m.f E

_{2}in the secondary causes a secondary current I

_{2}. Consequently, terminal voltage V

_{2}will appear across the load. If V

_{2}>V

_{1}, it is called a

**. On the other hand, if V**

*step up transformer*_{2}<V

_{1}, it is called a

*.*

**step down transformer**## Working

When an alternating voltage V_{1} is applied to the primary, an alternating flux, Φ is set up in the core. This alternating flux links both the winding’s and induces e.m.f E_{1} and E_{2} in them according to * Faraday’s law of electromagnetic induction*. The e.m.f E

_{1}is termed as primary e.m.f and e.m.f E

_{2}is termed as secondary e.m.f.

E_{1} = -N_{1}dΦ/dt

E_{2} = -N_{2}dΦ/dt

E_{2}/E_{1} = N_{2}/N_{1}

The magnitude of E_{2} and E_{1} depends upon the number of turns of the secondary and primary respectively. If N_{2}>N_{1}, then E_{2}>E_{1} and we get a step up transformer. On the other hand, if N_{2}<N_{1}, then E_{2}>E_{1} and we get a step down transformer. If load is connected across the secondary winding, the secondary e.m.f E_{2} will cause a current I_{2} to flow through the load. Thus, a transformer enables us to transfer A.C power from one circuit to another circuit with a change in voltage level.

The following points may be noted carefully:

- The transformer action is based on the laws of electromagnetic induction.
- There is no electrical connection between the primary and secondary. The A.C power is transferred from primary to secondary through magnetic flux.
- There is no change in frequency.
- The losses that occur in a transformer are:

- core losses – eddy current and hysteresis losses
- copper losses – in the resistance of the windings

Generally, these losses are very small so that output power is nearly equal to the input primary power.

## Ideal Transformer

An ideal transformer is one that has

- no winding resistance
- no leakage flux
- no iron losses

## E.M.F Equation Of A Transformer

Consider that an alternating voltage V_{1} of frequency *f* is applied to the primary. The sinusoidal flux Φ produced by the primary can be represented as:

Φ = Φ_{m}sin ωt

The instantaneous e.m.f e_{1} induced in the primary is

e_{1} = – N_{1}dΦ/dt = – N_{1}dΦ/dt (Φ_{m}sin ωt)

= – ωN_{1}Φ_{m} cos ωt = -2π*f*N_{1}Φ_{m}cos ωt

= -2π*f*N1Φ_{m}sin(ωt-90)

It is clear from the above equation that maximum value of induced e.m.f in the primary is

E_{m1} = 2π*f*N_{1}Φ_{m}

The R.M.S value E_{1} of the primary e.m.f is

E_{1} = Em1/√2 = 2πfN1Φ_{m/√2}

E_{1} = 4.44*f*N_{1}Φ_{m}

E_{2} = 4.44*f*N_{2}Φ_{m}

In and ideal transformer, E_{1} = V_{1} and E_{2} = V_{2}

## Voltage Transformer Ratio (k)

As we know that, equation for the induced e.m.f is

E_{2}/E_{1} = N_{2}/N_{1} = K

The constant K is called * voltage transformation ratio*. Thus if K = 5, then E

_{2}= 5E

_{1}

### For an ideal transformer

E_{1} = V_{1} and E_{2} = V_{2} as there is no voltage drop in the winding.

E_{2}/E_{1} = V_{2}/V_{1} = N_{2}/N_{1} = k

There are no losses. Therefore, volt-ampere input to the primary are equal to the output volt-ampere.

V_{1}I_{1} = V_{2}I_{2}

I_{2}/I_{1} = V_{1}/V_{2} = 1/k

Hence, current are in the inverse ratio of voltage transformation ratio. This means that if we raise the voltage, there is a corresponding decrease of current.

## Practical Transformer

The practical transformer differs from the ideal transformer in many respects. The practical transformer has

- Iron losses
- Winding Resistance
- Magnetic leakage (leakage reactance)

### Iron losses

Since the iron core is subjected to alternating flux, there occurs eddy current and hysteresis losses in it. These two losses together are known as * iron losses* or

**. The iron losses depend upon the supply frequency, maximum flux density in the core, volume of the core etc. It may be noted that magnitude of iron losses is quite small in a practical transformer.**

*core losses*### Winding Resistances

Since the winding consist of copper conductors, it immediately follows that both primary and secondary will have winding resistance. The primary resistance R_{1} and secondary resistance R_{2} act in series with the respective winding.

### Leakage Reactances

Both primary and secondary current produce flux. The flux Φ which links both the winding’s is the useful flux and is called ** mutual flux**. However, primary current would produce some flux Φ

_{1}which would not link the secondary winding. Similarly, secondary current would produce some flux Φ

_{2}that would not link the primary winding. The flux such as Φ

_{1}or Φ

_{2}which links only one winding is called

*. The leakage flux paths are mainly through the air. The effect of these leakage fluxes would be the same as through inductive reactances were connected in series with each winding of transformer that had no leakage flux. In other words, the effect of primary leakage flux Φ*

**leakage flux**_{1}is to introduce and inductive reactance X

_{1}in series with the primary winding. Similarly, the secondary leakage flux Φ

_{2}introduces an inductive reactance X

_{2}in series with the secondary winding.

## Impedance Ratio

Consider a transformer having impedance Z2 in the secondary.

Z2 = V2/I2

Z_{1} = V_{1}/I_{1}

Z_{2}/Z_{1} = (V_{2}/V_{1}) × (I_{1}/I_{2})

Z_{2}/Z_{1} = K^{2}

Impedance ratio (Z_{2}/Z_{1}) is equal to the square of voltage transformation ratio. In other words, an impedance Z_{2} in secondary becomes Z_{2}/K_{2} when transferred to primary. Likewise, an impedance Z_{1} in the primary becomes K^{2}Z_{1} when transferred to the secondary.

Similarly,

R_{2}/R_{1} = K^{2} and X_{2}/X_{1} = K^{2}

## Transformer Tests

The circuit constants, efficiency and voltage regulation of a transformer can be determined by two simple tests.

- Open Circuit test
- Short Circuit test

These tests are very convenient as they provide the required information without actually loading the transformer. Further, the power required to carry out these tests is very small as compared with full load output of the transformer.

### Open Circuit Test / No Load Test

In this test, the rated voltage is applied to the primary while the secondary is left open circuited. The applied primary voltage V_{1} is measured by the voltmeter, the no load current Io by ammeter and no load input power W_{o} by watt-meter.

As the normal rated voltage is applied to the primary, therefore, normal iron losses will occur in the transformer core. Hence watt-meter will record the iron losses and small copper loss in the primary. Since no load Io is very small, Cu losses in the primary under no load condition are negligible as compared with iron losses. Hence, watt meter reading practically gives the iron losses in the transformer. It is reminded that iron losses are the same at all loads.

Iron losses Pi = Watt meter reading = Wo

No load current = Ammeter reading = I_{o}

Applied Voltage = Voltmeter reading = V_{1}

Input Power, W_{o} = V_{1}I_{o} cos Φ_{o}

No load P.F, cos Φ_{o} = W_{o}/V_{1}I_{o}

I_{w} = Io cos Φ_{o} ; I_{m} = I_{o} sin Φ_{o}

Thus open circuit test gives P_{i}I_{o}, cosΦ_{o,} I_{w} and I_{m}

### Short Circuit Test / Impedance Test

In this test, the secondary is short circuited by a thick conductor and variable low voltage is applied to the primary. The low input voltage is gradually raised till at voltage V_{sc}, full load current I_{1} flows in the primary. Then I_{2} in the secondary also has full load value since I_{1}/I_{2} = N_{2}/N_{1}. Under such conditions, the copper loss in the winding’s is the same as that on fill load. There is no output from the transformer under short circuit conditions. Therefore, input power is all loss and this loss is almost entirely copper loss. It is because iron loss in the core is negligibly small since the voltage V_{sc} is very small. Hence, the watt meter will practically register the full load copper losses in the transformer winding’s.

Full load Cu loss, P_{c} = Watt-meter reading = W_{s}

Applied Voltage = Voltmeter reading = V_{sc}

Full load Primary Current = Ammeter reading = I_{1}

Pc = I_{1}^{2}R_{1} + I_{1}^{2}R_{2}‘ = I_{1}^{2}R_{o1}

R_{01} =P_{c}/I_{1}^{2}

Where R_{01} is the total resistance of transformer referred to primary.

Total impedance referred to primary, Z_{01} = V_{sc}/I_{1}

Total leakage reactance referred to primary, X_{01} = √(Z_{01}^{2} – R_{01}^{2})

Short-Circuit P.F, cos Φ_{s} = P_{c}/V_{sc}I_{1}

Thus short circuit test gives full load Cu loss, R_{01} and X_{01}.

## Losses In A Transformer

The power losses in a transformer are of two types, namely:

- Core or Iron losses
- Copper losses

These losses appear in the form of heat and produce

- An increase in temperature
- A drop in efficiency

### Core Losses

These consist of hysteresis and eddy current losses and occur in the transformer core due to the alternating flux. These can be determined by open circuit test.

Hysteresis loss = *K*_{h}*f*B_{m}^{1.6} watts/m^{3}

Eddy current loss = K_{e}*f*^{2}B_{m}^{2} t^{2} watts/m^{3}

Both hysteresis and eddy current losses depend upon

- Maximum flux density B
_{m}in the core - Supply frequency
*f*

Since transformer are connected to constant frequency, constant voltage supply, both f and B_{m} are constant. Hence, core or iron losses are practically the same at all loads.

Iron or core losses, P_{1} = Hysteresis loss + Eddy current loss

= Constant losses

The hysteresis loss can be minimized by using steel of high silicon content whereas eddy current loss can be reduced by using core of thin laminations.

### Copper Losses

These losses occur in both the primary and secondary winding’s due to their ohmic resistance. These can be determined by short-circuit test.

Total Cu losses, Pc = I_{1}^{2}R_{1} + I_{2}^{2}R_{2}

= I_{1}^{2}R_{01} or I_{2}^{2}R_{02}

It is clear that copper losses vary as the square of load current. Thus if copper losses are 400 W at a load current of 10A, then they will be (1/2)^{2} × 400 = 1– W at a load current of 5A.

Total losses in a transformer = P_{i} + P_{c}

= Constant losses + Variable losses

It may be noted that in a transformer, copper losses account for about 90% of the total losses.

## Efficiency Of A Transformer

Like any other electrical machine, the efficiency of a transformer is defined as the ratio of output power to input power.

Efficiency = Output Power/Input Power

It may appear that efficiency can be determined by directly loading the transformer and measuring the input power and output power. However, this method has the following drawbacks:

- Since the efficiency of a transformer is very high, even 1% error in each watt meter may give ridiculous results. This test, for instance, may give efficiency higher than 100%.
- Since the test is performed with transformer on load, considerable amount of power is wasted. For large transformers, the cost of power alone would be considerable.
- It is generally difficult to have a device that is capable of absorbing all of the output power.
- The test gives no information about the proportion of various losses.

Due to these drawbacks, direct loading method is seldom used to determine the efficiency of a transformer. In practice, open circuit and short circuit tests are carried out to find the efficiency.

Efficiency = Output/Input = Output/ (Input+losses)

The losses can be determined by transformer tests.

## Three Phase Transformer

In a three phase system, the voltage is lowered or raised either by a bank of three single phase transformer of by one 3 phase transformer. In either case, the winding’s may be connected in Y-Y, Δ-Δ,Y-Δ or Δ-Y. For the same capacity, a 3 phase transformer weighs less, occupies less space and costs about 20% less than a bank of three single phase transformers, Because of these advantages, 3-phase transformers are in common use, especially for large power transformations.

The three single phase core type transformers, each with winding’s on only one leg, have secondaries may be connected in star or delta. This arrangement gives a 3 phase transformer, If the primary is energized from a 3 phase supply, the central limb carries the fluxes produced by the 3-phase primary winding. Since the sum of the three primary currents at any instant is zero, the sum of three fluxes passing through the central limb must be zero. Hence no flux exists in the central leg and it may, therefore, be eliminated. This modification gives a three-leg transformer. In this case, any two legs will acts as a return path for the flux in the third leg. All the connections of a 3 phase transformer are made inside the case so that only three primary leads and three secondary leads are brought out of the case.

### Phase transformation ratio (K)

It is the ratio of secondary phase voltage to primary phase voltage and is denoted by K.

Phase transformation ratio, k = Secondary phase voltage/Primary phase voltage