Question: Why does AC current change through a transformer?

[Griffin Kowash]

Hello physics friends! I have a question about AC current I hope you can help me understand. So I know that putting AC through a transformer will raise or lower the voltage, but because power has to be kept constant, the current changes in the opposite direction. But according to Ohm's law, the current is directly proportional to the voltage, so if the voltage goes up, the current goes up. These two things seem to contradict each other.

Also, how would you calculate the output current, taking into account the resistance? I feel like I'm missing some essential component of AC theory, but I haven't been able to figure out what it is. Thank you!

(ﾉ◕ヮ◕)ﾉ*:・ﾟ✧

 

[Philip Wood]

You are misusing "if the voltage goes up, the current goes up". Ohm's Law applies to the p.d across, and current through, a given conductor. You have two different circuits, the primary circuit and the secondary circuit. Ohm's law can't 'span' both circuits! It IS possible to apply Ohm's Law separately to the resistive parts of the primary and secondary circuits, but that's another story!

 

[Baluncore]

The primary and secondary are electrically isolated. Only the shared magnetic field couples them. The primary current flows in the primary only, the secondary current flows in the secondary only. The turns ratio decides the voltage ratio and therefore the inverse current ratio.

 

[willem2]

Hello physics friends! I have a question about AC current I hope you can help me understand. So I know that putting AC through a transformer will raise or lower the voltage, but because power has to be kept constant, the current changes in the opposite direction.

You've already started on the wrong path where you say "because power has to be kept constant".

It's true that for a trans former V_1 I_1 must be equal to V_2 I_2 because of conservation of energy, but the conclusion that I_2 must go up if V_2 goes down is only valid if V_1 I_1 doesn't change.

Now with the transformer plugged into a voltage source, such as a wall outlet, V_1 doesn't change, but what happens to I_1 is determined by what happens on the secondary side.

You must use ohms law there to determine the secondary current, and then use I_1 N_1 = I_2 N_2 to find the primary current.

I_1 = primary current
V_1 = primary voltage
N_1 = number of turns of primary.
I_2 = secondary current etc.

 

[CWatters]

What willem2 said.

It's better to think of the output current I₂ as being controlled by the load resistance R_Load

So Ohms law says...

I₂ = V₂/R_Load

Then the Input Current I₁ is determined by

I₁ = N₂/N₁* I₂

 

[sophiecentaur]

Can I suggest you take care to get your Cause and Effect in the right order?

Vout is governed by Vin and Iin is governed by Iout. Thats the basic order of cause and effect.

Stick with this, step by step - it will get there in the end...
The following can be looked upon as being in order Cause - effect - cause - effect -...
The Volts across the Primary are fixed (let's assume a perfect supply and a perfect transformer).
The Volts across the Secondary are related to the Primary Volts by the turns ratio (that's the first thing we learn). If the secondary circuit is an open circuit then no current flows and the self inductance of the (perfect) transformer primary let's no primary current flow. This is because there is a back emf across the primary, due to its high (enough) self inductance. (Just as if there were no secondary windings). The secondary volts, when applied across a finite Load will cause a Secondary Current to flow.

Because of the laws of induction, we find that the secondary current is 90 degrees out of phase with the volts. The current in the secondary then induces a voltage across the Primary which happens to be in phase with the primary voltage (laws of induction again) This cancels out some of the back emf across the Primary and allows some current to flow in the primary winding.

The actual amount of current that flows in the Primary is governed by the (inverse) turns ratio and the Secondary Current. Power In has to be the same as Power Out. It cannot be more in a lossless transformer and it cannot be less because there's no more power available.
Soooo Vin times Iin equals Vout times Iout - the well known Transformer Equation, which can be re-arranged to produce whichever quantity is unknown.

 

[omega_minus]

If the secondary circuit is an open circuit then no current flows and the self inductance of the (perfect) transformer primary let's no primary current flow. This is because there is a back emf across the primary, due to its high (enough) self inductance. (Just as if there were no secondary windings).

This is like saying inductors never draw current. A transformer (ideal or not) is a coupled inductor, which follows the law V=IZ. There will be a primary current regardless of the existence of a secondary winding or not.

 

[sophiecentaur]

This is like saying inductors never draw current. A transformer (ideal or not) is a coupled inductor, which follows the law V=IZ. There will be a primary current regardless of the existence of a secondary winding or not.

That goes without saying (for the purposes of learning the basics, one at a time) - which is why I used the word "enough". It's good enough, like a practical voltage our current source is good enough for the spec demanded of it. If we are talking about ideal devices - which is the way to start most of these things - then I reckon that 'no current' is near enough.
From the nature of the original question, I should have thought it was obvious what level of answer was appropriate. Would you also think that copper losses and core losses should be included at the start?

 

[omega_minus]

Actually after more thought, Sophiecentaur is right. Reflecting an infinite impedance into the primary will still give an open circuit.

 

[omega_minus]

That goes without saying (for the purposes of learning the basics, one at a time) - which is why I used the word "enough". It's good enough, like a practical voltage our current source is good enough for the spec demanded of it. If we are talking about ideal devices - which is the way to start most of these things - then I reckon that 'no current' is near enough.
From the nature of the original question, I should have thought it was obvious what level of answer was appropriate. Would you also think that copper losses and core losses should be included at the start?

That all occurred to me a bit after the fact, Thus my follow up post. Not trying to confuse anyone here...

 

[Griffin Kowash]

Thank you everyone! This makes much more sense than before. Sorry if I bored anyone with such a basic question...thanks for bearing with me!