Working Principle of a Transformer
““Any changes in the coil of wires’ magnetic-fields, will cause an induction of emf. The magnitude of the induced potential is identical to the flux’s changing rate.
It can be written as –
E = – N * dϕ/dt
E is the induced emf, & N, ϕ is the number of turns and the magnetic flux produced, respectively.
The negative sign represents that the change in the magnetic field’s direction is opposite to induced emf. It is also known as Lenz’s law.
Now, we know that transformers have two windings. The alternating power is applied to the primary windings. The flow of current causes generation of a magnetic field around it. This property is known as mutual inductance. Now the current flows according to Faraday’s Law. The maximum strength of the magnetic field will be equal to d ‘phi’/dt. Magnetic lines of force now expand outside from the coil. The soft iron core concentrates the field lines and forms a path. The magnetic fluxes connect the primary windings as well as the secondary windings.
Now, as the flux also passes through the secondary windings, a voltage generates there. The induced emf’s magnitude will be given according to Faraday’s law. It will be = N * dϕ /dt.
The frequency and power of the supplied voltage never change in the whole process.
The induced voltage depends on the turn’s ratio.
Transformer EMF Equation
Let us assume magnitude flux as phi.
We know that magnetic flux varies sinusoidally.
So, ϕ = ϕm * sin (2 * π * f * t)
f is the frequency of the flux and N is the number of turns
Now, E = N * dϕ/dt
or, E = N * d (ϕm sin (2 * π * f * t)) /dt
or, E = N * 2 * π * f * ϕm * cos (2 * π * f * t)
For E =Emax, cos (2 * π * f * t) = 1
Emax = N * 2 * π * f * ϕ m
Now, Erms = Emax / √ 2
Erms = N * 2 * π * f * ϕm / √2
Erms = 4.44 f * N * ϕm
This is known as Transformer EMF Equation.
Losses of a Transformer
Loss in an electrical device or circuit means loss of power. A real transformer has different types of losses, but an ideal transformers never suffers a loss. There are several types of loss inside a transformer. Some of them are –
- A. Core Loss / Iron Loss
- B. Copper loss or Ohmic Loss
- C. Stray Loss
- D. Dielectric Loss
- A. Core loss / Iron Losses: The loss occurs due to alternating flux, inside the iron core is known as Core Loss or Iron loss. This type of losses are known as No Load Losses.
There are two categories of core loss. They are –
- i) Hysteresis Loss
- ii) Eddy Current Loss
i) Hysteresis Loss –An alternative magnetic force generates in the core of the transformer. That magnetizing force causes a hysteresis loop and that causes hysteresis loss.
Ph = η * Bmax * n * f * V
Ph = Hysteresis Loss
η = Steinmetz hysteresis coefficient
Bmax = Maximum flux density
n = Steinmetz exponenet
f represents the magnetic reversal per second
V = volume of magnetic material
Hysteresis loss contributes 50% of no load loss.
ii) Eddy Current Loss – Faraday’s Laws are behind the cause of Eddy Current Loss. The magnetic-fluxes cause a potential in the core. Now, due to this emf, current flows. This current is termed as Eddy Current and it is an undesired current. Loss due to this current is Eddy Current Loss.
The eddy current loss is expressed as –
Pe = Ke * Bmax2 * f * V * t2
Pe = Eddy Current loss
Ke = Eddy current constant
Bmax refers to the maximum flux density and f is the frequency of the magnetic reversal per second.
V = volume of magnetic material
t = magnetic thickness
- B. Copper loss or Ohomic Loss: This type of loss occurs due to the windings’ wire resistance. If Ip, Rp is current and resistance of primary winding and Is, Rs is current and resistance of secondary windings, then the loss will be given by the equation –
Po = Ip2Rp + Is2Rs
As the wires are of coppers’, the loss is termed as Copper loss. This type of loss is also known as Load Losses because this loss occurs only when load is connected with the secondary windings.
- C. Stray Loss: The reason behind such losses is the leakage field. It is a negligible loss.
- D. Dielectric Loss: The transformer’s insulator causes this type of loss.
There are also losses due to distorted voltage and currents.
The efficiency of a Transformer
The efficiency is the ratio of the produced power in the input to the supplied power of the output. It is represented as – η.
η = Output power /Input Power * 100%
In an ideal transformer, η comes as 1, which means Output power is equal to the input power. But in reality, a transformer suffers losses.
Loss = Input Power – Output Power
Or, Output Power = Input Power – Loss
So, Efficiency –
η = (Input Power – Loss) / Input Power * 100%
η = 1 – loss/ Input Power * 100%
Some Frequently Asked Questions
1. How is a transformer rated?
Transformers are rated in volt-amperes or kilo-volt-amperes (kVA). This rating indicates that the primary windings and the secondary windings are designed to tolerate the rated power.
2. How many types of Transformers are there?
There are many types of transformers based on different parameters. Some of them are –
- Ideal Transformers
- Real Transformers
- Step-up types
- Step down type
- Power transformer
- Single – phase types
- Three- phase types
- Centre tapped types
- Instrument types
- Pulse types
- RF types
- Audio types
3. A transformer has a turn’s ratio of 16 to 4 or 4. If the transformer secondary voltage is 220 V, determine the primary voltage.
We know that
Turns ratio =NpNs =VpVs
Here, Np = 16
Vs = 220 v
we have to find Vp
so Vp = Np*Vs/Ns = 16 * 220 / 4
Vp = 480 volt.
So the primary voltage was 480 volt.
4. What is the Reversibility of Transformer Operation?
Reversibility of Transformer Operation means using the transformer from backward. That is, giving the secondary windings an input voltage and connecting load at the primary windings.
5. Do the transformers perform in DC voltage?
No, a transformer does not perform in DC voltage. Applying Dc voltage will cause over hitting of the primary windings as the signal finds it a short-circuit.
6. What is Impedance matching?
The concept of impedance matching is that when a source voltage is connected to load, the load get the maximum power if the impedance of load is equal to the impedance of the impedance of the fixed internal source .It is one of the application of transformers.
7. A single phase transformer is with a rating of 2 kilo volt ampere has a 400v at primary windings and a 150v at secondary windings. Find out the primary and secondary full load current of the transformer.
Primary full-load current = 2kVA x 1000 / 400 V = 5 A
Secondary full-load current = 2kVA x 1000 / 150 V = 13.33 A
8. A transformer has 500 turns in the primary windings and 20 turns in the secondary windings. Find out –
a) The secondary voltage if the secondary circuit is open and the primary voltage is 100 v
b) Find out the current in primary and secondary windings when the secondary winding is connected to a resistance load of 16 ohms.
We know that turns ratio is given by
Turns ratio = Np/Ns = Vp/Vs
Np is the number of turns in primary windings.
Ns is the number of turns in secondary windings.
Vp is the voltage at primary side.
Vs is the voltage at secondary side.
Now we can write
Vs = (Ns * Vp) / Np
Or, Vs = (20*100)/500 V
Or, Vs = 4 V
Now for the second case, we know that power remains unchanged while transferring energy through a transformer.
We can write,
Pp = Ps
Where Pp is the power in primary side and Ps is the power from secondary side.
Pp = Vp * Ip
Ps = Vs * Is
Ip is the current in primary side and Is is the current in secondary side.
So, Vp *Ip = Vs * Is
Or, Ip = (Vs * Is) / Vp
Or, Ip = ((Vs*(Vs/Rs) / VpFrom ohm’s law V= IR, thus I = V/R, Here Rs is the resistance of the secondary coil.
Or, Ip = (Vs * Vs )/ (Vs * Rs)
Or, Ip = 4*4 / 100*16, Substituting the values and Rs = 16 ohm was given in the question.
So, Ip = 10 mili – ampere.
And, Is = Vs/ Rs
Is = 4/16 A = 0.25 A
To know more about transformers and electronics click here
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