Transformers - Variation in Current, EMF and Magnetic FLux

In summary: The voltage and current will still be 180 degrees out of phase, but the waveforms will not be perfectly sinusoidal anymore.
  • #36
i m not getting your quesn...??
 
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  • #37
elemis said:
Now according to me and my knowledge of electricity this is how I feel it should play out :

If the input current were a sine curve that varies with time then :

Input voltage = cosine curve i.e. it is ahead of input current by 90 degrees.

Magnetic Flux = sine curve with same period as input voltage but LARGER amplitude.

Output voltage = out of phase with input voltage by 180 degrees.

Output current = this is where I'm faltering and I'm having issues...

Could someone please confirm what I've said is correct and help me with the bit in bold ?
Hi elemis. Just look what you've started! http://img196.imageshack.us/img196/3295/qvk34.png

In one post you suggested that you might be considering a transformer with no load. I don't think so, because if there is no load on the secondary, there will be no secondary current. With no secondary current, there will be no primary current (for an ideal transformer).

For your purposes, you should think of the transformer as transforming resistances (or impedances). So imagine the secondary is connected to a light globe. A light globe is a resistor, so the secondary current will be in phase with the secondary voltage because that is the defining V-I characteristic for a resistance.

The transformer transforms impedances, as a consequence of transforming voltages and currents, so a light globe load in the secondary means the primary current will be in phase with the primary voltage. (The load on the secondary is a resistor, and in turn that load as seen on the primary side will also be resistive.)

You already understand that the primary voltage is N × the secondary voltage, where N = Nº of turns on primary ÷ Nº of turns on secondary,
and that the primary current is 1/N × the secondary current,
∴ the impedance seen at the primary is V/ I = N² × the secondary load impedance
https://www.physicsforums.com/images/icons/icon2.gif So a resistor of R ohms in the secondary draws power from the mains equivalent to connecting a resistor of N² × R ohms in place of the primary and not having a transformer at all.

I hope that addresses your question. :smile:

Feel free to post another. :wink:
 
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  • #38

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