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 secondary winding. Depending upon the number of turns of the primary (N1) and secondary (N2), an alternating e.m.f E2 is induced in the secondary. This induced e.m.f E2 in the secondary causes a secondary current I2. Consequently, terminal voltage V2 will appear across the load. If V2>V1, it is called a step up transformer. On the other hand, if V2<V1, it is called a step down transformer.
When an alternating voltage V1 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 E1 and E2 in them according to Faraday’s law of electromagnetic induction. The e.m.f E1 is termed as primary e.m.f and e.m.f E2 is termed as secondary e.m.f.
The magnitude of E2 and E1 depends upon the number of turns of the secondary and primary respectively. If N2>N1, then E2>E1 and we get a step up transformer. On the other hand, if N2<N1, then E2>E1 and we get a step down transformer. If load is connected across the secondary winding, the secondary e.m.f E2 will cause a current I2 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.
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 V1 of frequency f is applied to the primary. The sinusoidal flux Φ produced by the primary can be represented as:
Voltage Transformer Ratio (k)
As we know that, equation for the induced e.m.f is
E2/E1 = N2/N1 = K
The constant K is called voltage transformation ratio. Thus if K = 5, then E2 = 5E1
For an ideal transformer
E1 = V1 and E2 = V2 as there is no voltage drop in the winding.
E2/E1 = V2/V1 = N2/N1 = k
There are no losses. Therefore, volt-ampere input to the primary are equal to the output volt-ampere.
V1I1 = V2I2
I2/I1 = V1/V2 = 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.
The practical transformer differs from the ideal transformer in many respects. The practical transformer has
- Iron losses
- Winding Resistance
- Magnetic leakage (leakage reactance)
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 core losses. 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.
Since the winding consist of copper conductors, it immediately follows that both primary and secondary will have winding resistance. The primary resistance R1 and secondary resistance R2 act in series with the respective winding.
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 Φ1 which would not link the secondary winding. Similarly, secondary current would produce some flux Φ2 that would not link the primary winding. The flux such as Φ1 or Φ2 which links only one winding is called leakage flux. 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 Φ1 is to introduce and inductive reactance X1 in series with the primary winding. Similarly, the secondary leakage flux Φ2 introduces an inductive reactance X2 in series with the secondary winding.
Consider a transformer having impedance Z2 in the secondary.
Z2 = V2/I2
Z1 = V1/I1
Z2/Z1 = (V2/V1) × (I1/I2)
Z2/Z1 = K2
Impedance ratio (Z2/Z1) is equal to the square of voltage transformation ratio. In other words, an impedance Z2 in secondary becomes Z2/K2 when transferred to primary. Likewise, an impedance Z1 in the primary becomes K2Z1 when transferred to the secondary.
R2/R1 = K2 and X2/X1 = K2
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 V1 is measured by the voltmeter, the no load current Io by ammeter and no load input power Wo 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 = Io
Applied Voltage = Voltmeter reading = V1
Input Power, Wo = V1Io cos Φo
No load P.F, cos Φo = Wo/V1Io
Iw = Io cos Φo ; Im = Io sin Φo
Thus open circuit test gives PiIo, cosΦo, Iw and Im
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 Vsc, full load current I1 flows in the primary. Then I2 in the secondary also has full load value since I1/I2 = N2/N1. 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 Vsc is very small. Hence, the watt meter will practically register the full load copper losses in the transformer winding’s.
Full load Cu loss, Pc = Watt-meter reading = Ws
Applied Voltage = Voltmeter reading = Vsc
Full load Primary Current = Ammeter reading = I1
Pc = I12R1 + I12R2‘ = I12Ro1
Where R01 is the total resistance of transformer referred to primary.
Total impedance referred to primary, Z01 = Vsc/I1
Total leakage reactance referred to primary, X01 = √(Z012 – R012)
Short-Circuit P.F, cos Φs = Pc/VscI1
Thus short circuit test gives full load Cu loss, R01 and X01.
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
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 = KhfBm1.6 watts/m3
Eddy current loss = Kef2Bm2 t2 watts/m3
Both hysteresis and eddy current losses depend upon
- Maximum flux density Bm in the core
- Supply frequency f
Since transformer are connected to constant frequency, constant voltage supply, both f and Bm are constant. Hence, core or iron losses are practically the same at all loads.
Iron or core losses, P1 = 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.
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 = I12R1 + I22R2
= I12R01 or I22R02
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 = Pi + Pc
= 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