DC source for transformer

Exactly same thing will happen that happen to An AC motor last night when another guest try to operate on DC.

such a great answer

This is not possible, in order for a transformer to work you need the Alternating Magnetic field that AC power provides. DC power doesn't provide this.

For example take an isolation transformer, when you apply the AC current to the primary winding of the transformer it causes a magnetic field to form around the copper conductor. As the field expands and contracts on the primary side the flux lines of the magnetic field cut threw the winding of the secondary side of the transformer and induces an AC current into the secondary side of the transformer, this is referred to as mutual induction. The output voltage you get on the secondary side is depends on the turns ratio of the winding.

If your looking to change your DC voltage you should look into to getting a DC-DC converter.

dont we use inductors in dc circuits!

as per my knowladge while applying dc supply to the inductor winding it will not get dammage until we apply sufficient potential to over come the winding resistance. in the same way in case of 1 phase transformers and induction motors winding willnot gooff until we apply sufficient potential to winding for to come winding resistance.

in dc motors we are applying dc supply to the field winding, it is not burning because of its winding resistance is in few 100s of ohms but in ac motors or transformers the winding resistance is in few 10s of ohms

in case of transformers if we apply dc to primary change of flux linkages will not take place so we cannot get secondary voltage status of winding dedpend up on at what potential we apply the voltage to winding

No one said that we don't use inductors in DC circuits :)

"what will be the output of the tranformer"

Smoke, maybe molten metal.

hii all,

when u give a dc supply, the supply does not go to its steady state value directly but has a infinitesimal transient from zero to its final value. by applying final value theorem, we can arrive at the conclusion that the output will be a very steep impulse very near to time zero. this high value may or may not damage the transformer depending on its ratings. correct me if i am wrong.

regards,

cv

"this high value may or may not damage the transformer depending on its ratings. correct me if i am wrong."

You are not incorrect for the issue of "time zero", but you fail to continue on. Immediately thereafter, where is the impedance going to come from in the transformer to provide any counter EMF that will limit the current flow? Without impedance, all you have left is resistance of the winding wires, which if you look at the "peak inrush" data of a transformer, can allow currents in excess of 2000% of rated. With AC, that may only happen for a fraction of a second because impedance is established in the transformer. Without that, it continues to destruction.

thank you for your valuable insight into my reply. what i meant by my above reply is that the dc source gets drained completely in that impulse at time zero. the magnitude of the impulse depends on the dc source which we apply. if the source itself gets drained, how can it continue to provide current after time zero even though the resistance of the windings is small.

another doubt i have is that when we test for DC winding resistance, then why doesnt anything happen? the test readings come out as good as ever.

regards,

cv

DC voltage has frequency 0, so there will be no mutual induction between the primary and secondary,so supplying dc will generate heat and burn the coils because of this no load impedance

while looking in AC, according to Faraday's law, voltage induced is directly proportional to the product of number of turns and "change in rate of flux"(only in AC)....

<...what will be the output of the tranformer?...>

With a bit of luck and a following wind, exactly nothing, as the circuit protective device(s) should operate before it melts/catches fire, etc.