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DC Causes Transformer to Wear out??why?

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gauthamtechie

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Why does a transformer supposedly burn out when supplied with DC? I did do some searching and found explanations relating to emf as e=d(flux)/dt., i.e. change in flux for emf. Change in flux didn't happen in DC ...but what makes that the reason for transformer not to burn(or wear out) in AC, whereas wear out in DC?? Is it because all the current is drawn into primary itself in a DC and whereas part of the current component does work to link flux in secondary(in case of AC) ? - Is my understanding right?
 

Yes Gautham your are absolute right....The energy transfer in the transformer happens using alternating ( A.C.) magnetic field coupling and this magnetic coupling ....In case of DC this magnetic field coupling is stationary ..... So the whole energy will get wasted in the primary winding of the transformer and as a consequence it will burn out...

Good Luck
 
You have no reactive inductance at DC. At DC the transformer is almost a perfect short-circuit. In fact, the only impedance you have is the resistence of the copper wire (a few ohms).
 
etmabreu said:
. At DC the transformer is almost a perfect short-circuit.
Click to expand...

How does it avoid a short circuit condition in case of AC then? An explanation in Magnetizing and inductive components of current ..along with the reactance.. will be true right?? I know i'm kind of getting close to the logic but questioning things again!!!
 

Let's see:

Imagine you have a transformer with nothing connected do the secondary winding. If there is no current flowing in that winding, you could just remove it and it would work the same. Now, what is the difference between a transformer without the secondary winding and a simple coil? Answer is: none. What limits the current in a coil? The resistance and, with AC current, the inductive reactance. That reactance is due to the electromagnetic phenomena you describe.
 
ok things seem clear now! I found Mutual inductance (the absence of it in DC) a good way to sum it up. And thus it seems AC shares but DC keeps everything for itself and burns!..That along with the details as pointed by all you edaboard members has clarified my doubt... Thanks everyone!
 
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Over current is responsible for the burning of the winding.

DC supply I=V/R Current is by only the winding resistance
AC Supply I=V/Z Current is by winding resistance and Inductance Z=Sqrt. R^2+XL^2

So when applying DC supply, resistance only there to oppose the flow of electric current that is why it is burning.
 
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