# B Basics of transformers

1. Jul 12, 2017

### Vibhor

This is a snapshot of the diagram in the high school text book.

and here is the accompanying text

I would like to understand the "minus" signs in
- N1dΦ/dt and - N2dΦ/dt

This is my understanding -

When there is a crest of the sinusoidal voltage i.e top terminal of the input voltage is positive , then due to induced EMF in the primary coil of magnitude N1dΦ/dt , top end of the primary coil is positive and bottom end is negative .Now if we write the Kirchoff's law around the primary circuit then we get the desired equation as given in the text .

Is this reasoning correct ?

But in this case magnetic flux would be anticlockwise around the core , so now an EMF of magnitude N2dΦ/dt would be induced such that top end of secondary coil is negative
and bottom end is positive .

Is this reasoning correct ?

But then why is EMF in secondary coil written
as -N2dΦ/dt and not simply N2dΦ/dt ??

@ehild , @cnh1995 , @Merlin3189

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2. Jul 12, 2017

### cnh1995

The polarities of the top and bottom ends are decided by the "direction (or sense) of windings".
Look up "dot convention" in magnetically coupled circuits.
https://en.m.wikipedia.org/wiki/Polarity_(mutual_inductance)

The induced voltages can be either in phase or 180° out of phase, depending on the sense of windings.

This video might help.

3. Jul 12, 2017

### Vibhor

But why is there a minus sign in the two EMF terms ?

4. Jul 12, 2017

### Merlin3189

Seems ok to me.
Depends on the way the secondary is wound. It took me a while to be sure which way it is, as it looks to be wound differently from the primary. But I've slid it round to the same side and you can see it more clearly (I think).

Quite honestly I haven't worked out whether the primary or secondary should have the minus sign for the induced emf. You can do it by thinking of the wires and direction of field, but you have to be careful in translating from one diagram to another (at least I do!)
What I think is obvious from this diagram as redrawn, is that the secondary is a continuation of the winding of the primary - or is like the primary slid along the core. Therefore the induced emf's must both be in the same sense as shown.
But the E's must be in the senses I've drawn: the primary because it must oppose the driving source; the secondary because it must be in phase with the current for a resistive load.
So it seems they are correct that E1 = + N1dφ/dt and E2 = - N2dφ/dt

I tend not to bother with the signs in the calculation, just the magnitudes and use my understanding to know what the phase relations are. Quite often in transformer questions, there is no information about the sense of the windings (unless, as cnh says, they mark dots to show it) so you can't put it into the equation. Where the loads are complex (reactive) you do need to take phase into account in your equations, but you generally rely on the dots to indicate the sense of the induced emfs.

5. Jul 12, 2017

### cnh1995

Me too.
In fact, if both the windings do not have a common electrical ground, the minus signs don't make any sense. We can't say whether primary and secondary voltages are in phase or out of phase, because there is no common reference.
The minus sign in Lenz's law only highlights the fact that the induced emf "opposes" the cause of its induction.

In your image, the instantaneous current directions shown are correct.
The top of the primary is positive.
Current I1 is "entering" the primary from the top, hence using the right hand rule, the flux is anticlockwise.
The secondary flux due to I2 should be clockwise, hence, again using the right hand rule, the current I2 should be "leaving" the secondary from the top.
To make it so, the top of the secondary should be positive.

Hence, when top of the primary is positive, top of the secondary is also positive.
If you connected the bottoms of the two windings together, there will be a common ground to both the circuits.
So now with a common reference node, you can say that the voltages E1 and E2 are not 180 degrees apart, but are in phase.

6. Jul 12, 2017

### Vibhor

I really appreciate both of yours response . But while trying to understand transformer , I am bit confused about the direction of self induced EMF in an inductor .

In the above coil the current increases towards right . But the winding is such that when looked from the left terminal , the current moves counter clockwise .The flux increases towards left . In case the winding is such that when looked from left , current moves clockwise .The flux increases towards right .

But in both the cases , the direction of induced EMF would be such that it opposes the cause of change in flux This cause is the increase in current .So induced EMF would tend to oppose the increase in current .So left terminal would be positive and Right terminal negative .

So , is it correct to say that the polarity of induced EMF is independent of the windings of the coil ??

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7. Jul 12, 2017

### cnh1995

You are right when you say "polarity of the induced emf is such that it 'opposes the cause' of its induction."

Suppose you have two coils with opposite winding senses and they are not magnetically coupled.

If you increase the flux in both the coils, it will induce emfs in both the coils. But the polarities of the emf will be opposite. Because the induced emf will try to drive a current that opposes the increase in flux. If the winding senses are opposite, the induced current directions are also opposite, and hence, the polarities of the induced emf are opposite.

This means, the polarity of the induced emf depends on the sense of the winding. It is the 'flux from the induced current' that is independent of the winding direction.

8. Jul 12, 2017

### Merlin3189

As usual a simple question has required a lot more thought than I thought it needed! And my aside comment about working out the sign of the back emf, turns out to be nonsense!
I think your conclusion is quite right. We don't get into which way the flux is and whether dφ/dt is + or - , because it is immaterial. You drive a current through an inductor and the emf must oppose it. That's all we need to know.
Looking at your diagram above, I can't tell whether it is wound clockwise or ccw. But it doesn't matter. If it were wound in the opposite way, the flux would have the opposite sense, but the turns would go in the opposite direction, so the emf would stay exactly the same.

For the transformer, in many calculations we don't even mention the flux. All we need to know most of the time, is that the same flux links both coils, then the emfs are proportional to the number of turns in each coil and the currents inversely proportional.

By the time you get into problems where you need to know in detail about the flux in the core, you'll be the one answering the questions on transformers!

9. Jul 12, 2017

### Vibhor

Well , @cnh1995 doesn't seem to agree cnh1995 , what is your take on the above post ?

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10. Jul 12, 2017

### Vibhor

@Merlin3189 , in your post#4 , in the top figure , could you please explain the direction of E2 ? Why is it upwards ?

I think , since the flux is increasing upwards in the secondary coil , the induced EMF should be such that the bottom terminal is positive and top is negative .

11. Jul 12, 2017

### cnh1995

Well, I was talking about a different scenario where you increase the flux using some external mechanism. I'll explain it later (I can't draw figures right now).
This is correct. Because in your inductor example, if the current is increasing towards right, its left terminal is obviously hooked up to the +ve terminal of a voltage source. So the induced emf has to be independent of the winding direction, otherwise it would violate KVL, wouldn't it?

12. Jul 12, 2017

### Vibhor

OK.

How is flux clockwise ? And we are looking from top or bottom ?

Sorry , I am still not clear why the top of secondary coil should be positive .Please see my post 10 .

13. Jul 12, 2017

### cnh1995

The secondary current should be such that it produces a flux in clockwise sense i.e. downwards in the secondary limb.
Applying right hand rule, the current direction comes out to be the one that is shown in the diagram i.e. I2 leaves the secondary from its top.
This means the top should be +ve.

Imagine the secondary open circuited, and the flux is increasing upwards. Induced electric field will be indeed from bottom to top, but that would drive the electrons from the top towards the bottom. This would nullify the electric field inside the coil and will make the top +ve and bottom -ve.

14. Jul 12, 2017

### cnh1995

Use the right hand rule for primary and you'll see the primary flux is counterclockwise. So to oppose this flux, the secondary flux should be clockwise.

We are looking at the diagram from outside the plane of the paper, neither from the top or the bottom.

15. Jul 12, 2017

### Vibhor

By primary flux , you mean flux due to induced EMF in the primary coil . Right ?

Why should the secondary flux oppose the primary flux and not the flux due to magnetic field in the core ?

16. Jul 12, 2017

### cnh1995

This is what I was talking about.
The coils are not connected to any source. The magnetic flux in the core is increasing towards right via some external arrangement.

Orange arrows show the induced current. You can see that the polarity of induced emf changes with the direction of winding.

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17. Jul 12, 2017

### cnh1995

The image you posted is of an ideal transformer. They have neglected the magnetizing current (the very basic transformer model at highschool level). Hence, in your image, there is no magnetic field in the core. Primary mmf and secondary mmf cancel each other exactly. I think you should study how a practical transformer works (with some assumed idealities).
https://www.physicsforums.com/posts/5696172/
See if this helps.

In a real transformer, flux in the core is established by the magnetizing current. Secondary induced voltage drives the secondary current and to counter the secondary mmf, an extra current flows through the primary. It is called as 'reflected load current'. So in reality, it is the additional primary current that cancels the secondary mmf and maintains the original flux established by the magnetizing current. Magnetizing current is often neglected in the basic model because it is very small compared to the reflected load current (3 to 5%), thanks to the high permeability, low reluctance steel core.

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18. Jul 12, 2017

### Merlin3189

you're right.
I put E2 in the same direction as their current flow through the resistor. I put E1 to oppose the current flow from the source. These are arbitrary decisions. If the emf or current turns out to be in the opposite sense to your arrow, it just comes out as negative.

Then in the bottom diagram I showed the induced emf in coil 2 in the same direction as the induced emf in coil 1, because they are wound the same way when looked at from the (arbitrary) flux direction through both coils.. (This bottom diagram was just an easy way to see the relationships between the induced emfs.)

If you look at my updated picture below, you can see that indeed the "polarity" of the secondary (based on an assumed polarity in the primary) is as you say.

So, as they say in the text, E1=N1.dφ/dt and E2=-N2.dφ/dt
meaning that i2 is negative, ie.in the opposite sense to the arrow and so producing an opposite flux to that produced by the primary (since both coils are wound in the same direction - cw viewed from the top in my lower diagram.)

This gets confusing - "Flux due to induced EMF"
Flux is flux. It is caused by current, not by EMF. There is a current flowing in the primary and a current flowing in the secondary. Caused by what, does not matter. These currents cause flux.
For a given transformer, the flux produced by each coil would (ideally) be proportional to current x turns and the net flux is the vector sum of that produced by each coil.

Since the transformer currents marked on the diagram are inversely proportional to the turns, these fluxes do cancel out, as cnh has pointed out.
$φ_1 ∝ N_1 i_1 \text { and } φ_2 ∝ N_2 i_2$ and
since $\frac {i_1} { i_2} = \frac {N_2 } { N_1 }$
then $i_1= i_2 \times \frac {N_2 } { N_1 }$
So $φ_1 ∝ N_1$ $\times { i_2 \times \frac {N_2 }{ N_1 } } = i_2 N_2 = φ_1$
So both fluxes are equal in magnitude and, because the currents wind in opposite directions, cancel out.

When the primary is not connected to a load and no secondary current flows, the ideal model suggests no primary current flows. But as cnh has explained, there is still a small magnetising current in the primary(which is commonly ignored). So $i_1 \ and \ i_2$ might better be labelled as $Δi_1 \ and \ Δi_2$ being the extra currents which flow when a secondary load is connected.

19. Jul 12, 2017

### cnh1995

@Vibhor, I think I've made a mistake in applying the right hand rule to the secondary.
Using right hand rule, to set up a clockwise flux, the current should leave the secondary from the bottom, meaning the bottom should be +ve and the top should be -ve.
And yes, then E1 and E2 are 180 degrees apart.

I took the secondary winding sense wrong (maybe due to my phone's small screen).

So if the bottom is positive, the current I2 should be leaving the secondary from the bottom.
Sorry if I caused any confusion!

20. Jul 12, 2017

### Merlin3189

Very difficult to keep track of the direction of winding from the diagram. That's why I redrew it, to help me. And I actually got it wrong at first attempt, because I translated it rather than rotating to get it round the core to the other side.