Short Circuiting the Secondary of a Transformer/Self Impedence

[Cardinalmont]

Thank you for reading my post.
I was watching the video this video on transformers and saw an amazing demo:


From 4:00-5:20 the guy does an experiment where he has a simple transformer circuit set up:

On the primary side of the transformer he has an AC power supply, and a lightbulb.
On the secondary side of the transformer there is only a lightbulb.

When he removes the lightbulb from the secondary side of the transformer, the lightbulb on the primary side goes out.

This is really hard for me to understand. I would think that since a transformer is essentially just wires, making the non-attached secondary into an open circuit would have only a brief effect on the primary as a result of the momentary change in flux when the secondary coil is no longer creating a magnetic field. It seems to me that after the brief effects of the change in flux, the transformer would just act as wires for the first circuit to flow through, thus continuing to light the bulb.

Why does the bulb on the primary side go out? The guy in the video mentioned self impedance but didn't go into it.

 

[anorlunda]

You're mixing things up. When he removes the secondary side bulb, he creates an open circuit. That is the opposite of a short circuit. No current flows through an open circuit.

The primary side bulb is in series with the transformer. It lights when current flows. When the current stops (or almost stops), the light goes out.

In real life, there is a small current through the primary side, even when the secondary is open circuit. The best way to visualize it is to look at the equivalent circuit for a transformer. There is a current labelled Io in the circuit that is nonzero even when Is is zero.

 

[Cardinalmont]

Thank you for your reply, but I still don't understand.
I understand that a lightbulb needs current to run through it in order for it to light.
I don't understand how removing the lightbulb in the secondary has a long term effect on the current in the primary.

 

[CWatters]

Why do you use the words "long term"?

When the secondary is open circuit it effectively increases the inductance of the primary reducing the primary current and the light goes out.

When the secondary is connected it effectively reduces the inductance of the primary, increasing the primary current and turning on the bulb.

 

[anorlunda]

OK, try this figure. It shows an ideal transformer. I added two light bulbs. The red bulb is in series, and the blue one in parallel.

At the start current flows to the load Zl and both bulbs are lit. Now we remove the load. Is=0. The red light goes out but the blue one stays lit. Do you see it now?

 

[Cardinalmont]

Why do you use the words "long term"?

When the secondary is open circuit it effectively increases the inductance of the primary reducing the primary current and the light goes out.

When the secondary is connected it effectively reduces the inductance of the primary, increasing the primary current and turning on the bulb.

I say "long term" because in the video the light doesn't just go out for a moment. It seems to permanently go out.

What about the primary makes it need the help of the secondary? Is the circuit just designed that way?

 

[Cardinalmont]

OK, try this figure. It shows an ideal transformer. I added two light bulbs. The red bulb is in series, and the blue one in parallel.

At the start current flows to the load Zl and both bulbs are lit. Now we remove the load. Is=0. The red light goes out but the blue one stays lit. Do you see it now?

View attachment 221542

I still don't see it.

 

[anorlunda]

What about the primary makes it need the help of the secondary?

What is your understanding how a transformer works? Can you write the equations for primary and secondary voltage and current?

 

[CWatters]

I say "long term" because in the video the light doesn't just go out for a moment. It seems to permanently go out.

Correct. Why do you think it should only be a temporary effect?

What about the primary makes it need the help of the secondary? Is the circuit just designed that way?

No it's down to the way transformers work.

With no load on the secondary the transftormer only draws a small current at the primary. This current is enough to set up a working flux in the core but not enough current to light the lamp in the primary circuit.

When a load lamp is connected to the secondary it draws current that creates an opposing flux that cancels out some of the flux in the primary. This effectively reduces the inductance of the primary causing it to draw more current lighting the lamp.

 

[Cardinalmont]

I'm not understanding the points either of you two are making and I think its because I don't fundamentally understand what makes the primary different than just a simple AC circuit. I'll try to restate what I was originally trying to ask.

Imagine this, I attach an AC power supply to a bulb with long wires. The bulb will light.
Then, without unplugging the circuit, I wrap those long wires around a soft iron ring. I imagine that the light bulb would still be lit.
In my head, this is essentially a primary with no secondary, yet I believe the bulb would not go out.
I don't see how my described secondary-less primary is fundamentally different than a primary attached to a open-circuited secondary.

If this is fundamentally different. How is it?
If it isn't fundamentally different and it would still go out, why would it?

 

[anorlunda]

It sounds like you don't get the idea of a transformer. You would need a derivation of the equations of mutually coupled circuits to satisfy your curiosity. What is your background? calculus? differential equations? electric circuits? fields & Maxwell's Equations?

There are some threads here on PF that explore the internal workings of transformers in exhaustive depth. But that it not a good strategy for learning. Neither is asking questions on the Internet a good strategy for learning. A textbook or an online course is usually a better choice. We might get some recommendations for you on PF, but we need to know what level is appropriate.

 

[CWatters]

Imagine this, I attach an AC power supply to a bulb with long wires. The bulb will light.
Then, without unplugging the circuit, I wrap those long wires around a soft iron ring. I imagine that the light bulb would still be lit.

So you think that inserting a large inductor has no effect on the current in an AC circuit?

 

[cnh1995]

I don't see how my described secondary-less primary is fundamentally different than a primary attached to a open-circuited secondary.

A secondary-less primary is same as a primary with open secondary. But what you described in #10 does not sound like a transformer,since the power is not being transferred via mutual induction.

A transformer with closed secondary is totally different from that with an open secondary.

Why does the bulb on the primary side go out? The guy in the video mentioned self impedance but didn't go into it.

https://www.physicsforums.com/posts/5637789/
(Please read the entire thread. I corrected one of my own posts in the end).

 

[Cardinalmont]

https://www.physicsforums.com/posts/5637789/
(Please read the entire thread. I corrected one of my own posts in the end).

I read through this and it gave me a lot of the information I was missing. Thank you so much! This new understanding lead me to be able to do the research I needed in order to fill in some more of the missing pieces. I think I get it now:

The problem I was having was that in my head, the voltage of the system was equal to the product of the resistor's (light bulb's) resistance and the current running through it. I figured the coil would have negligible resistance. $$V=IR$$ but really, the inductor that the copper coil of negligible resistance is wrapped around, the inductor connecting the primary and secondary resists the ever changing ac current supplied thus using most of the voltage supplied and not leaving enough for the lightbulb. $$V=IR +L (ΔI/Δt)$$ When the secondary is a full loop, the induced current helps to reduce $L (ΔI/Δt)$ in the primary thus raising the primary resistor's value of $IR$, thus lighting the bulb.

 

[Cardinalmont]

At the start current flows to the load Zl and both bulbs are lit. Now we remove the load. Is=0. The red light goes out but the blue one stays lit. Do you see it now?

View attachment 221542

So I think I get this now, the inductor will use most of the circuits voltage thus causing the red light to go out, but since the inductor is in parallel with the blue light, they will both be given the same amount of voltage, thus making the blue bulb bright. The high inductance of the inductor hurts the red bulb but helps the blue bulb.

 

[Cardinalmont]

OK, try this figure. It shows an ideal transformer. I added two light bulbs. The red bulb is in series, and the blue one in parallel.

At the start current flows to the load Zl and both bulbs are lit. Now we remove the load. Is=0. The red light goes out but the blue one stays lit. Do you see it now?

View attachment 221542

Interestingly enough, I just built this circuit and tried it out and I didn't receive the results you described.

When I built the circuit that is shown in the video, which is the same circuit as yours without the parallel bulb in the primary, it worked as expected! When the secondary was a full loop, both the light in the primary and secondary lit. When I removed the light from the secondary, the light in the primary went out!

When I attached a bulb in parallel to the inductor as shown in your diagram it very brightly lit up both bulbs in the primary and put out the bulb in the secondary. Even when I disconnected the bulb in the secondary, the two bulbs in the primary remained brightly lit. I believe this is because the resistance of the parallel bulb was so small in comparison to that of the inductor that virtually no current was flowing through the inductor. I attached some ammeters and confirmed it!

 

[CWatters]

OK, try this figure. It shows an ideal transformer. I added two light bulbs. The red bulb is in series, and the blue one in parallel.

At the start current flows to the load Zl and both bulbs are lit. Now we remove the load. Is=0. The red light goes out but the blue one stays lit. Do you see it now?

View attachment 221542

That's wrong. With no secondary current both bulbs see half the supply voltage so will both be on at roughly half brightness.

 

[anorlunda]

That's wrong. With no secondary current both bulbs see half the supply voltage so will both be on at roughly half brightness.

On second thought, you're correct, thanks for pointing out my error.

My example was flawed because one of those bulbs would short the transformer primary. What I was trying to illustrate is that the voltage across the primary would be the same as supply voltage with no current. I should have shown a high impedance voltmeter across the primary rather than a second bulb.