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When a square wave is applied to primary of a transformer
then what will be output wave form of secondary ?
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Answer Posted / fabio corsi
A Fourier analysis may be applied only if a transformer is
linear, which is rarely the case; furthermore, even in a
linear transformer, the dispersed flux has to be accounted,
i.e. flux produced by the primary current which is not
chained with the secondary coil and viceversa.
In a linear transformer without dispersed flux and with zero
resistance of the coils, the waveform of the secondary
voltage is the same as that of the primary voltage, since
u1=N1*d(flux)/dt and u2=N2*d(flux/dt), the flux being the
same, i.e. u1/u2=N1/N2 no matter the waveform (provided it
can be integrated, which is also the case of a square wave).
This is no longer the case with dispersed flux, and/or with
coil resistance, and/or with non-linear core (e.g. iron), in
which case all depends on the involved parameters
(inductance of dispersed flux, hysteresis cycle of the core,
resistance of the coils and consequent time constants w.r.t
the working frequencies).
In a nutshell: "ideal" transformer(never the case actually)
==> square wave; non-ideal transformer ==> non-square
waveforms, the more distorted the less the transformer may
be approximated as ideal, possibly even spikes.
| Is This Answer Correct ? | 30 Yes | 4 No |
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