Answer / madhu balaji

whenever there is increase in current,inductor stores
mechanical energy,and whenever there is decrease in
current, it will dissipate the stored energy. Even when
square input is provided, there will be change in flux
linking with primary and secondary coil. normally pulse
transformer are provided with rectangular electric pulses.
definitely it will operate....

Answer / surendhar

To answer this question, we will need the frequency of the
square wave input. Lets discuss the two possible cases,

1. where the frequency of the input square wave is high
(around KHz), then we will get an output which resembles
the input wave but not the exact shape.
Reason: Square wave can be repreented by sinusoidal waves
of all frequency ( eg sinusoidal wave of freq(f+2f+3f+.nf)=
Square wave of fundamental frequency, f ) Assuming that the
input frequency is so high that inductor (transformer
winding) doesnt saturate, we will get the same output at
the secondary winding. But you dont get the exact square
shape at the secondary! This is because, as we know that
the impedance offered by inductance increases with increase
in frequency, so the high freq sin components would be
replicated at the secondary side with lesser amplitude and
this makes the output deviate slightly in shape from the
input.

2. In Second case, the frequency is low (say 50Hz) then the
inductor gets shorted as it would get saturated. So, there
is no change in magnetic field resulting in zero output
voltage at the secondary terminal.

Answer / vivek

It will not operate. The ideal square wave has regions of constant power supply which will cause the transformer windings to overheat due to saturation of field. Also it is against the principle of operation of a transformer, Induction. For induction to happen, constant change of flux should be there which can be provided only by a sine wave.

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