Transformers - Variation in Current, EMF and Magnetic FLux

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jim hardy

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Then your secondary current is always ninty degrees behind your output voltage
if 'output voltage' means 'secondary voltage' then i don't buy it either.
 
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psparky

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if 'output voltage' means 'secondary voltage' then i don't buy it either.
Obviously this is wrong. Delivering only Vars to everyone would not work out so well.
 

FOIWATER

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well then the output depends on the load connected?

I would say the information you have written for your primary variables is correct, and I guess I sort of got tangled up trying to relate them to the secondary, but the secondary current depends on load connected..

although, what i wrote previously about polarity is still correct.

no load - no current
resistive load - the voltage current are in phase
capacitive load - the output current will lead the output voltage, but with reference to the primary, it depends on your polarity markings.
inductive load - the output current will lag the output voltage, but with reference to the primary, it depends again on polarity markings.

http://www.allaboutcircuits.com/vol_2/chpt_9/4.html <--- this page cites the two voltages as in phase, unless you reverse your output coil. It states that the coil polarity for practical scenario's is ambigious... and also explains the polarity markings I made reference to.

http://www.allaboutcircuits.com/vol_2/chpt_9/1.html <--- i agree with gerbi as well, saying to the OP that the magnetic flux is not in phase with the primary voltage but in phase with the injected current, but yes it is of a larger magnitude.

any corrections welcomed.
 
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gerbi said the conditions were a perfect transformer with no load.
In a perfect transformer with no load.
There is no primary current and no secondary current.
The secondary voltage in just the turns ratio times the primary voltage.
The magnetic flux lags the primary voltage by 90 degrees. (Since there is no primary current, it doesn't make sense to say that the flux leads or lags the current)

If there is a load, the primary current is the secondary current/the turns ratio.

E1/E2=N1/N2
I1/I2=N2/N1

E1=primary voltage
E2=secondary voltage
N1=primary turns
N2=secondary turns
I1=primary current
I2=secondary current

gerbi, a good check of your calculations for a perfect transformer is that the input watts should equal the output watts.
 
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Opps made a mistake
the primary current is the secondary current X the turns ratio.
 

jim hardy

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Nice, Carl.

Gerbi's example in #20 of the bifilar winding is a good reference point to keep thinking straight.

I was taught ~1961 about a phase reversal but for the life of me can't recall why.
Unloaded, Flux will be integral of primary voltage and secondary voltage the derivative of Flux.

Does taking the derivative of an integral not simply get you back where you started?
Volts = sin
∫(sin) = -cos
d(-cos) = +sin

I wonder if the confusion stems simply from drawing a transformer with windings shown on opposite sides of the window as in this hyperphysics link.
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/transf.html

transf.gif


Observe flux leaves the dotted end of primary but enters dotted end of secondary. [EDIT oops no dots on that drawing but wrap your right-hand fingers around core in same direction as current through winding and thumb points in direction of MMF]

In your mind's eye flip that secondary over so it's just above the primary and observe it's wound in opposite direction.
From the TOP view primary is THEN wound CCW and secondary is wound CW.

old jim
 

psparky

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Jim,

Those were my thoughts as well. There clearly is a phase shift on the primary coil....but there is also a phase shift on the secondary coil which must put it back. Don't know that math....but it must be true because sure enough the voltage is in phase with the current on any oscilliscope thru a purely resistive load. Starting with a power factor of 1
is most convenient.
 
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The polarity of a transformer can be determined from which direction the wire enters the core.
In the picture Jim place above, Ip could be replaced with a polarity dot.
Is could be replaced with a polarity dot.
The polarity dots would be correct.

In the picture above, if the secondary conductor entered from the back of the core and exited from the front of the core (opposite to that shown), the secondary polarity mark would be near Ns.

Assuming a perfect transformer and a resitive load, could the confusion be because the transformer primary is a load, so the current and voltage have to be in phase.
In the other case the transformer is a power source so the current in the secondary has to be 180 degrees out of phase with the voltage in the secondary.
 
Could someone also confirm if voltage = rate of change of current with time ?
no, voltage can not be equal to the rate of change of current with time.as it will violate the ohm's law. according to which the current flowing in a conductor is directly proportional to the voltage across it.
 
This question about the variation of the aforementioned quantities in a transformer.

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 ?
but what do you want to ask in it???
 
i m not getting your quesn....??
 

NascentOxygen

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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 [Broken]

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 [Broken] 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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