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My question pretty much sums up everything that I want to ask.

How is the electrical power in the secondary the nearly same as that in the primary coil of a transformer?

Electrical power lost must be as low as possible in a transformer. How is the current in the secondary coil due to induced voltage of the same power as the primary coil? What insures that the electrical power in the secondary coil would be of nearly the same magnitude as that in the primary coil?

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    \$\begingroup\$ The primary and secondary communicate by way of magnetic flux in the core. They use that to negotiate a solution that satisfies Ohm's law at the primary and secondary output terminals. \$\endgroup\$
    – user57037
    Jan 9, 2022 at 6:06
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    \$\begingroup\$ Energy conservation law is a good start. The nature simply refuses to work otherwise. \$\endgroup\$
    – fraxinus
    Jan 9, 2022 at 9:24
  • \$\begingroup\$ the induced current in the secondary coil - voltage is induced and current flows because of the induced voltage and any load connected. Current IS NOT induced. \$\endgroup\$
    – Andy aka
    Jan 9, 2022 at 9:29
  • \$\begingroup\$ Transformers are designed not to store energy but rather transform power with low leakage inductance and higher impedance than load for efficiency \$\endgroup\$ Jan 9, 2022 at 12:03

5 Answers 5

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Have patience and read it, don't skim through.

The same question I had a few weeks ago, I got the answer after several hours of brain-storming and searching answers in internet, so I want to share it.

  1. Beginner level explanation First:

Lets say the transformer is 100% efficient (no heat losses, no hysteresis losses, no eddy current loss, nothing).

The transformer ensures the voltage on the secondary is N2/N1 times the voltage on the primary due to the phenomenon of mutual induction (explained later), i.e. a changing current in the primary results in a voltage in secondary (and a secondary current if the circuit is complete).

Then, say if the load side (secondary) is not connected to any device (say here a resistor), the current in the secondary is bound to be zero as I = V/R (period.) Therefore since the secondary current is zero, and thus the energy dissipated in the secondary = VI = Vs*0=0.

Now according to conservation of energy the wnergy dissipated in secondary must be equal to power in primary * efficiency of transformer and since efficiency is 1 here (100%) therefore the power input in primary is 0 and thus the current in primary is zero.

"So the current in primary doesn't determine the current in secondary, It is the current in secondary (which is in our hand) which determines the current in primary" enter image description here

You can see this video (by Khan Academy) too if the above concept did not click by now: https://youtu.be/VrbxUQxu0l0

Think about this a few times, it must click in your mind that the above quoted statement is true, make sure you hear the click sound before you read more ;).

But conservation of energy is used so "we" can "calculate" the primary current but what is the "mechanism" that ensures the transformer to automatically manage it and maintains the current draw from primary to feed the required current in secondary?

  1. More detailed explanation (the mechanism):

The mechanism causing the Vs to be N2/N1 Vp and which lets the primary "know" how much current to pass through it so the secondary current (in our hands) is maintained, is mutual induction, it is a very very long topic if you want to go in depth, I will explain it in short, and give links to some videos which explain it in this question's perspective.

Mutual induction is a phenomenon in which a change in current in one conductor causes a change in current (or development of a voltage) in other conductor by the mechanism of changing magnetic flux linkage causing electric field, governed by Maxwell's 3rd equation or simply Faraday's Law.

To make sure you are in the Same Page:

A Change in Magnetic Flux results in a development of a emf(voltage) across a conductor and if the circuit is complete, there is a current induced.

Now see this video by ElectroBOOM: https://youtu.be/ySx84Ca7BFQ (from 6:02 if you have less time)

Note: See the above video and then read the below conclusion, else it will be very confusing.

As already mentioned in the video, the phenomenon of mutual induction in transformers:

a. There is a changing current in primary which causes a changing magnetic field around the primary and also wrapping the secondary.

b. This changing magnetic field around the primary and secondary induces a voltage and this voltage causes a reverse current in the primary (due to this changing magnetic field) causing the primary current to reduce but it also induces a voltage in the secondary (as it is wrapping it too) , and if a load is connected, there is a changing current induced in the secondary due to this voltage induced in secondary .

c. The changing current in secondary causes a changing magnetic field opposite to the first changing magnetic field and thus stopping the reverse current in primary and also the induced current in secondary and as they decrease the fields come back and this mutual induction balance settles at a point of dynamic equilibrium when the primary current is exactly such that it draws exactly the needed current from primary that can maintain the secondary current we are drawing, obeying conservation of energy.

Isn't it beautiful?

Hope it helped you, but you need to wrap your brain around it by watching the video and the answer several times, and unleashing your imaginative power to understand this phenomenon and mechanism properly, if you want to of course.

Reality:

But we know nothing is 100% efficient according to the 2nd law of thermodynamics and indeed for transformers it is true too, so the primary current is a bit higher than value we calculate assuming the transformer to be 100% efficient such that the (extra bit current * voltage at primary) accounts for all the inefficiencies in the transformer.

Please correct me in comments, if I said something inaccurate or incorrect.

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    \$\begingroup\$ b. This changing magnetic Field induces a reverse current in the primary causing the primary current to reduce but it also induces a changing current in the secondary Changing currents induce voltages, you're answering a noob, please be consistent and use correct terms. Otherwise, valiant attempt. Missing fact that for constant input voltage, flux and so magnetising current stays constant, magnetising current reduces as core inductance increases, to zero and infinity for an ideal transformer. \$\endgroup\$
    – Neil_UK
    Jan 9, 2022 at 6:28
  • \$\begingroup\$ @Neil_UK Ok I have edited it, Apologies, I went with the flow, and assumed the reader to figure somethings by themselves. and I have explained the concept with Mutual Induction alone (which might be complete by itself, for this question atleast), and dont want to introduce the concept of Inductance as it might be confusing and the answer will be even longer (already it is very long) , If you still want me to do it, Please let me know. Thanks for the Suggestion. \$\endgroup\$ Jan 9, 2022 at 6:44
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    \$\begingroup\$ No you haven't: b. This changing magnetic Field induces a reverse current - please fix this. Voltages are induced by changing currents. But (b) is also very confusing here and ditto (c). Not so beautifully worded here I'm sorry to have to point out. This is why it's usually problematic to talk about mutual induction. I talk about induction (not mutual) and then consider the fluxes rather than the currents because it's easier to explain but, it's your answer and you have to find it beautiful if others don't. \$\endgroup\$
    – Andy aka
    Jan 9, 2022 at 9:43
  • \$\begingroup\$ @Andyaka "but it's your answer and you have to find it beautiful if others don't." sorry to hear that you didnot like my answer, The points a b c are conclusion points from the video I mentioned, and asked the OP to see before proceeding further, and "Voltages are induced by changing currents" , I have explained , that voltages are induced by changing B flux and if the circuit is complete, a current flows due to that voltage. If you want me to change my answer further, please let me know. \$\endgroup\$ Jan 9, 2022 at 12:37
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    \$\begingroup\$ I am not asking you to do anything. I'm just pointing stuff out that you can take on-board or not. It's not easy to explain transformers and, it's not really good to link to YouTube (or other) videos in doing so. Answers ought to be self-contained as much as possible. \$\endgroup\$
    – Andy aka
    Jan 9, 2022 at 12:43
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I'm drawing on Fitzgerald/Kingsley/Umans's Electric Machinery, 6th ed. and paraphrasing portions from there. I additionally recommend picking up this book for yourself as it discuss a number of electromagnetic and electromechanical systems rather well.

The primary and secondary windings are on the same magnetic circuit, and hence are coupled. For the simplification of this explanation, assume that the core permeability is much better than that of free space to the point that magnetic leakage is negligible; this means that the same flux is seen in the primary and secondary. Furthermore, suppose that the primary has M turns and the secondary has N turns, AC voltage \$V\$ is applied to the primary, and a load is attached to the secondary.

When an AC voltage is applied to the primary, that voltage leads to a small magnetization current which creates a flux in the core, running through both the primary and secondary; this flux must induce an opposing voltage in the primary. This establishes the voltages seen on each winding, of course proportional to the number of turns -- the changing magnetic field induces an electromotive force given by \$\iint\vec{B}\cdot d\vec{A}\$ per turn.

At the same time, the magnetic circuit consisting of primary MMF, and secondary MMF, must be considered. When the secondary voltage is applied to the load, the load begins to conduct a current, which is also seen in the secondary winding. Since the flux must continue to induce a compensating voltage in the primary, the MMF resulting from secondary current (magnitude N*I_secondary) must be cancelled by an equal and opposing MMF from the primary winding. Since the magnitude of the secondary MMF is N*I_secondary, and that of the primary MMF is M*I_primary, we can see that I_primary = N/M * I_secondary.

You can see that power is conserved - the primary power is V * (N/M * I_secondary), and the secondary power is (V * N/M) * I_secondary. Note that magnetization current isn't seen here, since it's out of phase from the supply voltage and hence contributes to purely reactive power.

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  • \$\begingroup\$ additional flux is created within the core as a result of the secondary MMF - that is wholly untrue \$\endgroup\$
    – Andy aka
    Jan 10, 2022 at 9:39
  • \$\begingroup\$ @Andyaka Since your comment was rather vague, I've gone ahead and rewritten that section by paraphrasing directly from Fitzgerald/Kingsley/Umans's Electric Machinery, 6th ed. If you still have a complaint with the answer, please be more clear. \$\endgroup\$
    – nanofarad
    Jan 10, 2022 at 15:26
  • \$\begingroup\$ That's improved but I don't like your use of induce or induction now. When it comes to magnetic components that word is reserved for the induction of voltage. So, it grates against me to see people saying that current (or MMF) is induced. Current flows due to induced voltage and a load. the secondary current induces an MMF of magnitude NI_secondary.... - should be The secondary MMF of NI_secondary is cancelled by a primary (and opposing) MMF of equal magnitude. Now it's going a little away from what you said but it's the best I can come up with in a comment. \$\endgroup\$
    – Andy aka
    Jan 10, 2022 at 15:38
  • \$\begingroup\$ @Andyaka I'll revise again when time permits; also looking for a good synonym that doesn't imply induction of voltage \$\endgroup\$
    – nanofarad
    Jan 10, 2022 at 15:42
  • \$\begingroup\$ Well a battery doesn't induce current in a resistor so, once voltage is induced, any current that flows is due to the established voltage (by induction or otherwise) and a load. I will say that some folk think I'm being hyper sensitive about ring-fencing the term induction to voltages (but, they are of course wrong LOL). \$\endgroup\$
    – Andy aka
    Jan 10, 2022 at 15:46
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It's not the transformer that ensures that what you pay for in real energy will mostly be delivered to the secondary coil but the power company, by taking back the surplus energy stored in the magnetic flux every cycle and not making you pay for it (apart from line losses in the house but not beyond). If you have too many idle transformers running, eventually the power company will complain.

The transformer is designed for running on a stable voltage and frequency regime: if there are significant changes, it may overheat because of not being able to recycle the energy stored in magnetic flux in the manner it is designed for.

Audio transformers are built for a significantly larger frequency and voltage regime but have higher percentual losses and nowhere near the energy efficiency as mains transformers, let alone the energy transferring capability.

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Electrical power must not be lost in a transformer.

Incorrect. Some power must be lost in the transformer: in the primary winding, in the core, and in the secondary winding. Note that these are three separate inefficiencies. You can have very efficient (low resistance) windings with a relatively low performance core (cheap steel instead of grain-oriented steel).

Faraday's Law of Induction and Ampere's Law define the basic properties at work in a transformer, but the actual design of a transformer is a management of a complex array of design tradeoffs. For example, beyond the power losses in the three basic elements, power also is lost in the transfer of energy from the primary winding to the core and from the core to the secondary winding.

What ensures that the power in the secondary coil is as great as possible? It is not one physical thing, but it is one thing: the skill of the designer.

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What insures that the electrical power in the secondary coil would be of nearly the same magnitude as that in the primary coil?

Since ideal inductors dont consume power they only store and release by conservation of energy the output power must be equal to input power.

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    \$\begingroup\$ 0 for 2, almost a strike out. The output power never is equal to the input power. \$\endgroup\$
    – AnalogKid
    Jan 9, 2022 at 4:18
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    \$\begingroup\$ The OP hasnt mentioned anything about losses he assumes the transformer is ideal. \$\endgroup\$
    – Miss Mulan
    Jan 9, 2022 at 4:23
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    \$\begingroup\$ Incorrect. In both the title and the body of his question he says "nearly the same", and explicitly mentions "power lost". \$\endgroup\$
    – AnalogKid
    Jan 9, 2022 at 6:03

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