How does current increase on the secondary side of the transformer?

[foolios]

Can someone explain what happens with a transformer as far as an example like the picture I've posted below?

I read that the less turns on the secondary side will result in less voltage but more amperage. How is it that current increases? Where do the extra electrons come from? I can't picture what's actually happening inside that wire. Can someone elaborate further, please?

::"image removed due to hosts image bandwidth constraints"

 

[GreekIslandLover]

The formula Watts = Amps * Volts is the answer. Primary power will always equal secondary power.

If you have 100v and 10 amps on the primary side, you will have 1000w in the primary. If the transformer is a step down ratio of 5:1 you will have 20v on the secondary side. Again, using the W= A*V forumla, transposing it to find Amps, you get A= W/V. We know that primary power = secondary power, so 1000w/20v = will give you 50 amps on the secondary.

 

[a1]

http://www.electricianforum.co.uk/electricalformulas.html

There are some electrical formulas that I have put together.

I will do more when I get some time.

 

[springcrocus]

How is it that current increases? Where do the extra electrons come from? I can't picture what's actually happening inside that wire. Can someone elaborate further, please?

Thicker wire on the secondary. For a 10:1 ratio, imagine the secondary having ten wires in parallel, each the same gauge as the primary but with only one tenth of the turns.

 

[JustStartingOut]

Foo

My understanding is that it is not that there are extra electrons giving a larger current in the secondary....

you need to remember that current can be thought of as the amount of charge (number of electrons) that pass through a point in the circuit in a certain amount of time.

current = charge / time

So if you have a charge of 10 Coulombs passing through a point in the circuit in 1 second, then the current is

I = Q / t

I = 10 / 1

I = 10A

But if the same 10 Coulombs passed through that point in only 0.5 seconds then the current is given by

I = Q / t

I = 10 / 0.5

I = 20A

So you should be able to see that it is not just the amount of charge passing through a point, but also how quickly that charge moves that gives rise to a larger current.

Hope that makes sense... I'm sure some of the others will correct me if I'm wrong.

 

[SPECIAL LOCATION]

Another way to consider it is that the Transformer doesn't generate or consume any power, (excluding negligible losses for the sake of this illustration)

The power being drawn on the load side, will have to come from the supply side. and a 100% efficient transformer would have power in = power out.

If the number of windings are identical both sides, voltage & current remain the same both sides, as in an isolating transformer e.g. shaver socket.

if the winding are different such that the voltage is lower, say 230v - 12v,

For example lets say we draw 4amp @ 12v (48watts), on the secondary,

if there had been 4 amps on the primary side you would have have 920watts!

where would this other 872watts have dissapeared too?

which as GreekIsland said...

Primary VoltAmps = Secondary VoltAmps.

 

[foolios]

Thicker wire on the secondary. For a 10:1 ratio, imagine the secondary having ten wires in parallel, each the same gauge as the primary but with only one tenth of the turns.

Ahhh, ok, that's something that helps me picture whats happening. When I picture thicker wire on the secondary side I can see how there is larger number of electrons that can move in the same given instant as compared to the primary side. So, this ten times thicker wire has more mass to move, therefore a greater amount of amps as a result.

Does this also explain why we can use less voltage? We don't need to push the amps as hard on the secondary side like we did on the primary because more Coulombs will move at a time. ??

Ok, if that seems right so far. What actually happens to the greater voltage from the primary side? Is it just not absorbed because of the less turns in the wiring of the secondary side or because thicker wire on the secondary side is more resistant?

Thanks so much for the replies.

 

[elvis]

It's all about induction.

It is not about the difference in voltage between the two coils. The voltage that exists in the secondary coil only exists as a result of the voltage in the primary coil, it is induced.

The 12V in the secondary coil will remain in the secondary coil for as long as the 230V exists in the primary. The 230V in the primary coil flows around the primary coil for as long as you have it connected to a power source, it is not absorbed.

Remember both coils are electrically separated, there is no physical connection between them. The voltage in the secondary coil is induced via the magnetic field that is generated in the primary coil by the input voltage. This (primary) magnetic field produces a magnetic field in the secondary coil and the voltage generated in the secondary coil is dependent on the strength of this magnetic field and the number of turns in the secondary coil.

So, we can control inputs by controlling the input voltage, the number of turns in the primary coil and length of the coils. We can control the output by understanding strength of the generated secondary field, the effect this has on the secondary coils and again controlling the number and length of turns we need in the secondary coil to cause the induction of the required secondary voltage.

Hope this helps...