Question

what will be the output if we apply a square wave signal in transformer input

Answer 1

This is an interesting question and is the basis of the operation of an induction coil in a motor car.
You have a switch (the rotor arm) which connects a 12 V supply to the primary of the coil and then disconnects the 12 V supply.

A relatively small voltage pulse is produced at the secondary terminals when when 12 V supply is connected but an enormous voltage pulse if produced at the switching off phase. This is the voltage which produces the spark at the spark plug.

Faraday's law of electromagnetic induction states that the induced emf is equal to minus the rate of change of flux linkage.

So when the 12 V supply is connected the current in the primary increases and eventually reaches the maximum (steady) value. So the B-field produced by the primary coil mirrors this current change and so does the flux through the secondary coil. As long as the primary current is changing there is an emf at the output of the secondary coil.
When the current is steady there is no longer an emf at the secondary terminal.

A similar thing happens when the current is switched off except that the induced emf will be in the opposite sense to that produced by the switching on phase.

To explain the difference in the sizes of the the emf when turning the current on and then turning it off you have to realise that the current in the primary does not change instantaneously. There is what is called a time constant involved. In the case of an inductor the time constant is equal to the inductance divided by the resistance of the circuit. In symbols L/R. This time constant gives an indication of how quickly the current will change. A large time constant means that the current will change over a long period of time - ie slowly.

Now the value of the inductance of the coil is the same for both phases but the resistance is not.

When the current is switched on the resistance is the resistance of the supply the leads and the primary of the transformer. A relatively low resistance means a relative large time constant means a relatively slow change of current means a relatively small emf at the secondary.

Switch off however means that the resistance is enormous - you have broken the cicuit - it is effectively open circuit so the time constant is very small so the current collapses to zero in a very short period of time which in turn means that the rate of change of flux and the secondary is enormous as is the emf at the secondary terminals.

Answer 2

Sine wave. Inductive reactance opposes rapid change of voltage.

Answer 3

As any square wave can't be ideal it have to have some definite level transition time, in this case output will be spikes or impulses.If ideal square wave is employed no output.

Answer 4

@abhinavec08
".If ideal square wave is employed no output."
Only because the rate of change of magnetic flux would be infinite and so the output would be a spike of infinite height and of infinitely small width.
So in real world this cannot happen.

Answer 5

@Farcher If the square wave is perfectly ideal so frequency at the transition edges will be infinite as time for level change is zero. Inductor can't allow sudden change so nothing will be appearing at output.

Answer 6

@ElectronicsGeezer
If you are thinking that the square wave is made up of the sum of odd harmonics when it is expanded as a Fourier series then the output could tend to look like a sinusoidal variation after the rising edge and a similar situation but in the opposite direction at the falling edge.
What actually happens depends on the relative sizes of the rise and fall time of the square wave and the time constant of the circuit.
In the practical example I described the rise and fall times will be such that you will get pulses at the output.

Answer 7

@abhinavec08
You are describing a situation where at one instant there is no current in the inductor and then in the next instant there is a current.
In someways there is no point about arguing about such a theoretical abstraction..
The situation you describe contravenes a number of laws of Physics.
Given that inductance is to do with the inertia (mass) of the charge carriers you are suggesting that from not drifting at all the charge carries instantaneously have a drift velocity.
I think that Mr Newton will have a concern about the infinite force that you applied to the charge carriers and Mr Einstein would worry about the instantaneous transmission of information that all the charge carriers as they were all told to get moving at the same time?

Answer 8

what i think there should be impulse train at out put.you know v=Ldi/dt. square wave can be represent by summation of unit step function x(t)=u(t)-2u(t-t1)+2u(t-t3) and so on.if you differentiate this function you will get a series of impulses.but according to Fourier series this square wave is x(t)=d.c term +sinusoidal harmonics frequency component now if you apply input at primary, secondary will give you sum of sinusoid containing infinite frequency component.so the frequency behavior of transformer is important.if it act as a low pass or high pass filter,band pass it will produce a distorted square wave.if it is a all pass filter it should produce an impulse train....