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Due to the question being discussed in length on the comment section, I would like to address my question further with a new post linking from the discussion here: Transformers confusion with output voltage and number of iron cores used

So with the help from this community, I understand that more iron cores means a greater magnetic permeability and less magnetic leakage. But does this mean that the magnetic field increases?

And assuming this is true, if we have a transformer with many iron cores in the center as compared to air, then the one with more iron cores will increase the current in the primary coils. So what I am failing to see here is, why is a greater magnetic field established (due to more iron cores used) increase primary current of the transformer?

I know there there is a magnetic field established when there is a current carrying wire, but I do not understand the relationship between magnetic field and current established. that is why does a stronger magnetic field established increase current in the primary coils is my key source of confusion at the moment with transformers.

Eventually I want to try to understand why more iron cores used increase the out voltage value of a transformer(secondary voltage). This is my understanding at the moment with many

$$\text{more iron cores}\rightarrow \text{stronger current in primary coils}\rightarrow \text{less current in secodary coils}\rightarrow\text{stronger secondary voltage }$$

I am not sure if this pathway to my understanding is correct, but I am stuck on the first arrow from more iron cores means stronger current in primary coils.

I would appreciate the help and clarification.

EDIT: This is somewhat what the experiment the professor was doing in front of the class today. enter image description here

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  • \$\begingroup\$ The iron (or ferrite or whatever) shortens the magnetic path length, dramatically. The cost of this, relative to air or vacuum, is the possibility of core saturation and energy loss mechanisms. The benefit is that it really does tend to concentrate the magnetic lines better so they don't "bloom out" as much. Your other descriptions about "stronger current in primary coils" is difficult to follow for me. But I think it (and some assumptions underlying it) is misguided. Look up anything from the Unitrode Seminars. (Sign-in required, sadly.) \$\endgroup\$ – jonk May 24 at 5:49
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Without a diagram, I can only assume that 'adding cores' to a transformer means increasing the effective area of iron within it. This interpretation is also consistent with the behaviour you describe.

Let's assume it's some form of teaching transformer like this (image from betterequipped.co.uk)

enter image description here

where you alter the iron area core you introduce into the magnetic circuit.

You can start with no iron, then 'add cores' by putting a few thin laminations into the magnetic loop, then go on to increase the area of iron to fill the entire 'iron area' with iron.

You need to make a distinction between magnetic field (which is a per-area measure) and magnetic flux (which is a total measure within the centre of each core).

As you add more iron area, you increase the magnetic flux passing through the cores, and it's the flux, not the field, that controls how much voltage is generated per turn. More flux, more volts.

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  • \$\begingroup\$ All the above answers however detailed seem too complicated for my very illiterate physics knowledge I have. I just do not understand where is the original magnetic field coming from. I asked something like this on the physics stack exchange but I have no answer: physics.stackexchange.com/questions/480578/… I would appreciate further clarification. \$\endgroup\$ – Aurora Borealis May 24 at 8:31
  • \$\begingroup\$ @AuroraBorealis yes, grokking transformers is tricky. I'll try to expand this answer in the next few days. \$\endgroup\$ – Neil_UK May 24 at 9:26
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does this mean that the magnetic field increases?

Yes because Inductance increases with permeability

the one with more iron cores will increase the current in the primary coils.

No because primary current depends on I=V/Zin and Zin increases with inductance
But it means you can use shorter lower resistance wire to create a no-load inductance < 10% or rated load at rated voltage. This excitation voltage generated current is necessary to create mutual coupling in the shared core to transfer power from primary to secondary windings, yet not saturate the core (Bmax=1.5 T typ. in large transformers or 2 Tesla in special materials or 0.7T in ferrite cores) at absolute max voltage under any condition.

A stronger magnetic field can improve coupling which otherwise adds series impedance to the power transfer and thus poorer load regulation. (=dV/dZ)
Essentially it makes a stronger magnetic bridge.

? relationship between the magnetic field and current established.

NO B field is controlled by maximum input voltage only and not current. ( but also winding inductance with core gain factor for permeability)

( think magnetic bridge again)

When higher voltage begins to saturate, where the threshold may be defined by a 10% drop in inductance, it also starts to generate 3rd and 5th harmonics on the input current and possibly others. Once the bridge is established, power can be transferred and load current does not affect B field. When primary inductance drops, current rises and it gets hotter (Ps=I²R) from saturation winding losses and harmonic losses.

more iron cores used increase the out voltage value

No THink magnetic bridge again. More core material allows a strong bridge for more powerful transfers. ( You also need heavier gauge wire and maybe flat copper straps in layers for MVA sizes)

  • Voltage ratio is purely winding turns-ratios on a shared core, with 1 core per phase which come in different shapes.
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  • \$\begingroup\$ thank you for the effort to explain, but my knowledge on impedence.. inductance reluctance is almost 0... \$\endgroup\$ – Aurora Borealis May 24 at 8:31
  • \$\begingroup\$ Wikipedia ought to raise that a few % \$\endgroup\$ – Sunnyskyguy EE75 May 24 at 8:32
  • \$\begingroup\$ I tried to understand from the knowledge i have by making use of Faraday's law of EM induction through this question here physics.stackexchange.com/questions/480578/…, I am not sure how much this is valid. \$\endgroup\$ – Aurora Borealis May 24 at 8:34
  • \$\begingroup\$ I have added a picture in my original post. \$\endgroup\$ – Aurora Borealis May 24 at 8:39
  • \$\begingroup\$ Your link only repeats the same false assumptions. Read my answer again \$\endgroup\$ – Sunnyskyguy EE75 May 24 at 12:28
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So with the help from this community, I understand that more iron cores means a greater magnetic permeability and less magnetic leakage. But does this mean that the magnetic field increases?

No, permeability is a material constant and is unrelated to the size or amount of the material just like the density of (say) water is 1000 \$kg/m^3\$ irrespective of the size or shape or amount of water.

Yes to less magnetic leakage but only up to a certain point. With a thin core the magnetic reluctance (the ability to "conduct" a magnetic field) is higher/better than air and this means that a coil wrapped around the core will pass a fairly high percentage of the flux through the core and not through the air. As the core gets thicker, the reluctance falls and so does the leakage but it's a relationship of diminishing returns.

But does this mean that the magnetic field increases?

You are talking about transformers in your question and so the answer must be a certain "no". As you thicken the core, you make the reluctance of the core smaller (it has an ability to pass more flux for a given current in the winding), the inductance of the winding increases and so the natural tendency is for the current to reduce and, given that current is the "creator" of magnetic fields this means that flux reduces.

Eventually I want to try to understand why more iron cores used increase the out voltage value of a transformer(secondary voltage).

Transformer output voltage is fundamentally dependent on coupling the flux produced by the primary winding into the secondary winding. With no coupling you don't have a transformer so, if you make coupling better then you get more secondary voltage. Better coupling means less leakage and this means a lower reluctance core which in turns means a thicker core and/or one with higher permeability.

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