# Electrical transformer basics

The efficiency of electrical power transmission has improved for using higher voltages. Alternating current has entirely replaced by the direct current for power transmission and distribution. For producing higher voltages ac generator is better than dc generator.

Transformer is an electrical device for transferring electrical energy one circuit to another circuit without changing frequency. It is widely using in alternating current now a days.

In another word a transformer is an electrical machine which raises or low voltage of an ac supply with a corresponding decrease or increase in current. It has two windings are the primary and secondary on a common laminated magnetic core shown in figure 1.

The primary winding is connected with a.c. source and the secondary winding is connected to a load. For applying alternating voltage V_{1} on primary side its magnitude changes due to the primary windings N_{1}. Alternating voltage V_{1} creates an e.m.f. E_{1} in primary side. An alternating e.m.f. E_{2} is induced in the secondary depending upon the number of turns of the primary (N_{1}) and secondary (N_{2}). Induced e.m.f. E_{2} creates a current I_{2} in secondary side. V_{2} voltage appears between two terminals across the load.

According to voltage performance transformer is two types are step up-transformer and step-down transformer. When secondary voltage V_{2} is greater than primary voltage V_{1} the transformer is called step up transformer. If the secondary voltage V_{2} is less than primary voltage V_{1} it is known as step down transformer. Step up transformer is used for raising voltage. On other hand step down transformer is used for lowering voltage.

## Circuit diagram of transformer

## Some pictures of practical transformer

photo by wikipedia

photo by wikipedia

photo by wikipedia

photo by wikipedia

## Working principal of electrical transformer

Transformer works according to Faraday’s law of electromagnetic induction. Physical basis of a transformer is mutual induction between two circuits linked by a common magnetic flux. When alternating voltage V_{1} is applied to primary side then an alternating flux Φ is created in the core. As a result the alternating flux links both windings which induce e.m.f. E_{1} in primary side and e.m.f. E_{2} in secondary side according to Faraday’s law of electromagnetic induction.

Primary e.m.f.

E_{1} = – N_{1 }dΦ / dt ——————— (i)

Secondary e.m.f.

E_{2} = – N_{2} dΦ / dt ———————– (ii)

From equation (i) & (ii),

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

The magnitude of e.m.f. E_{1} and e.m.f. E_{2} depend upon the number of turns on the primary and secondary. If the number of turns on secondary N_{2} is greater than the number of turns on primary N_{1} then E_{2 }> E_{1} or V_{2} > V_{1} and such transformer is step up transformer. If N_{2} < N_{1} then E_{2} < E_{1} or V_{2} < V_{1} then transformer is step down transformer. Secondary e.m.f. E_{2} causes a current I_{2} if a load is connected to secondary section. Actually it is load current. Thus an electrical transformer delivers a.c. power one circuit to another circuit with changing voltage level but frequency does not change. Ideal transformer and practical transformer have a little bit difference.

Following properties of electrical transformer may be noted

(i) Input power and output power have same frequency,

(ii) Transformer is based on the action of the laws of electromagnetic induction,

(iii) The power is transferred from primary coil to secondary coil through magnetic flux. There is no

electrical connection between primary and secondary coil.

(iv) There are two losses occur in transformer

a) Core losses (hysteresis losses and eddy current losses)

b) Copper losses (in the resistance of the windings)

## **Voltage transformation ratio of electrical transformer**

From above equations of induced e.m.f. primary and secondary coil,

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

Here K is a constant called voltage transformation ratio. If K = 4 then N_{2} / N_{1} = 4 and E_{2} / E_{1} = 4 or E_{2} = 4E_{1}.

- Since E
_{1}= V_{1}and E_{2}= V_{2}. For that there is no voltage drop in the windings

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

- As there are no losses. So the input power and the output power of the transformer are same in volt-amperes.

Input power = V_{1}I_{1}

Output power = V_{2}I_{2}

Input power = output power

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

or I_{2} / I_{1} = V_{1} / V_{2} = 1 / K

This equation proves that currents are inverse ratio of voltage transformation ratio. It means if we raise voltage correspondingly decrease of current.

Now we will see an example so it will give a clear concept how voltage raising decrease current.

**Math**

**A 1000/100V, 10kV transformer has 66 turns in secondary. Calculate primary and secondary full load current. Neglect all losses. **

**Answer: **

V_{2} / V_{1} = 100 / 1000 = 1 / 10

We know, ** **

** ** V_{1}I_{1 }= V_{2}I_{2} = 10000

or I_{2} = 10000 / V_{2}

or I_{2} = 10000 / 100 = 100 A

I_{1} = 10000 / V_{1} = 10000 / 1000 = 10 A

Alternatively, I_{1} = I_{2} K = 100 (1 / 10) = 10 A

**No load voltage of transformer**

Figure 2 shows a transformer with one coil connected to an alternating source left side which is called primary side. In secondary side is shown open circuit so there is no load. The primary coil receives energy from alternating source. It delivers the energy received to secondary coil. The current I_{0} flows from primary coil and creates flux Φ in the iron core in the direction given by flux. The flux Φ links both primary and secondary sides. The primary current I_{0} is very small and voltage drop also tiny so we can neglect it.

As we applied alternating current in primary side, depending on it the flux changes and a voltage induced E_{1} in primary coil. Induced e.m.f. E_{1} is proportional to the number turns N_{1} in the primary. We also know from Faradays law of electromagnetic induction,

V_{1} = – E_{1}

E_{1} α N_{1} (dΦ / dt) ———————- (iii)

Since the both coil are wound with same iron core the flux is changing within the primary coil, it is also changing within the secondary coil. The change of flux will induce an e.m.f. E_{2} in secondary and this e.m.f. also proportional to the number of turns N_{2} in secondary coil. Generally we assume no load transformer as an ideal transformer. Since there is no load to the secondary circuit. Induced voltage E_{2} will appear only in secondary coil.

Hence

E_{2} = V_{2}

E2 α N_{2} (dΦ / dt) ———————— (iv)

Dividing equation (iii) & (iv),

E_{1} / E_{2} = N_{1} / N_{2 } ———————– (v)

As we know E_{1} is equal to V_{1}. Since there is no load or open circuit in secondary side hence E_{2} = V_{2}. From this we can write equation (v),

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

Using above equation we can find voltage or number of turns for a transform for primary and secondary.

**Math**

**A transfer can covert the high voltage of a 6000V transmission line to use on a 60V distribution system. There are 1000 turns in high voltage side then how many turns must be the low voltage side have?**

**Answer:**

Here, V_{1} = 6000V,

V_{2} = 60V,

N_{1} = 1000,

N_{2} = ?

We know,

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

or N_{2} = (V_{2} / V_{1}) N_{1}

or N_{2} = (60 / 6000) 1000

Finally N_{2} = 10 turns

In this way we can find turn number or voltage using this equation.