# Applying Ampere's Law to a Transformer

• Master1022
In summary, the discussion revolves around applying Ampere's Law to a transformer. The concept of magnetizing current is introduced, which is the current required to produce flux and counter-emf. In an ideal transformer, magnetizing current is zero and the primary and secondary amp-turns oppose each other, resulting in a net current of zero. The transformer model on Wikipedia is referenced as a good example to understand this concept, with the XM term representing the non-ideal components of a real transformer.
Master1022

## Homework Statement

I was wondering if and how we can apply Ampere's Law to a transformer?

I am just thinking about the bog standard transformer with a core, primary coil, and secondary coil.
1) How do I deal with the current from both coils?
2) Do I need to think about the opposing magnetic field created by the secondary coil? - the eventual goal is to go and find the flux density and total flux...

## Homework Equations

Ampere's Law:
$$\oint \vec H \cdot d \vec l = I_{enclosed}$$

For an efficient transformer:
$$I_{p} \times V_{p} = I_{s} \times V_{p}$$

## The Attempt at a Solution

The main question I have is about the current enclosed. I understand that if there was 1 coil, then we would have the enclosed current as N x I. With two coils, however, do I need to worry about the fact that the currents will be in opposite directions. I know that current is treated as a scalar quantity, but it does obey some vector properties...

If I attempt to go down the route of treating current as a vector, I end up with the enclosed current being equal to 0 because:
$$\frac{N_p}{N_s} = \frac{I_s}{I_p}$$
and thus:
$$N_{p} \times I_{p} = N_{s} \times I_{s}$$

but with I_s in the opposite direction, so the 'enclosed current cancels out'??

What have I misunderstood to arrive at this problem? Any help is greatly appreciated.

Master1022 said:
If I attempt to go down the route of treating current as a vector, I end up with the enclosed current being equal to 0 because:
NpNs=IsIpNpNs=IsIp​
\frac{N_p}{N_s} = \frac{I_s}{I_p}
and thus:
Np×Ip=Ns×IsNp×Ip=Ns×Is​
N_{p} \times I_{p} = N_{s} \times I_{s}

but with I_s in the opposite direction, so the 'enclosed current cancels out'??

What have I misunderstood to arrive at this problem? Any help is greatly appreciated.

A common stumbling block.
You are right on with your thinking.
A quick off the cuff explanation of the basic idea follows

It's deceptive when first starting out.
Consider a one winding coil, just an inductor.
Indeed current is required to produce flux and counter-emf.
You might call that "Magnetizing Current " .
With inductors we are usually interested in creating some specific impedance at some specific current so we make the magnetic path with enough reluctance that it provides the desired counter-emf at that current.

In a power transformer we are after efficiency
and any current spent just magnetizing the core is a parasitic loss
so we optimize the core to require not very much magnetizing current at all.
which means we shoot for high inductance
Infinite inductance would be great if we could achieve it because it would mean magnetizing current is zero.

Magnetizing current of an ideal transformer would be zero
and that's the assumption behind those Ip/Is ratios being inverse to the turns ratios.

Ip really includes magnetizing current
which we design to be small enough compared to load current that the turns ratios give results close enough for practical work.

That's a brief qualitative answer
i'm not dismissing you

just right now i don't have in hand links to the old threads where we went into this at length.

Summary" Ampere's law holds
volts per turn is a measure of flux and is the same in both primary and secondary windings (neglecting leakage)
secondary amp-turns oppose primary amp turns
leaving you with zero which is the mmf required to magnetize an infinitely permeable zero reluctance core.

In your equations just add to primary current the amps necessary to establish flux in your core and you will have the concept.

Wikipedia's transformer model is actually pretty good

https://en.wikipedia.org/wiki/Transformer

it has at its heart an ideal transformer on the right that needs no magnetizing current
and that's a good concept to have in the back of your mind

i circled in red the deviation from ideal
XM is the less-than-infinite inducrance of a real transformer
and RC represents the heat losses from hysteresis eddy currents et al
Usually they're small enough to ignore.i hope above gets you past this step in your thinking.
If not i could revisit this evening.

old jim

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Master1022 said:

## Homework Statement

I was wondering if and how we can apply Ampere's Law to a transformer?

I am just thinking about the bog standard transformer with a core, primary coil, and secondary coil.
1) How do I deal with the current from both coils?
2) Do I need to think about the opposing magnetic field created by the secondary coil? - the eventual goal is to go and find the flux density and total flux...

## Homework Equations

Ampere's Law:
$$\oint \vec H \cdot d \vec l = I_{enclosed}$$

For an efficient transformer:
$$I_{p} \times V_{p} = I_{s} \times V_{p}$$

## The Attempt at a Solution

The main question I have is about the current enclosed. I understand that if there was 1 coil, then we would have the enclosed current as N x I. With two coils, however, do I need to worry about the fact that the currents will be in opposite directions. I know that current is treated as a scalar quantity, but it does obey some vector properties...

If I attempt to go down the route of treating current as a vector, I end up with the enclosed current being equal to 0 because:
$$\frac{N_p}{N_s} = \frac{I_s}{I_p}$$
and thus:
$$N_{p} \times I_{p} = N_{s} \times I_{s}$$

but with I_s in the opposite direction, so the 'enclosed current cancels out'??

What have I misunderstood to arrive at this problem? Any help is greatly appreciated.
The relation V1*I1 = V2*I2 holds only for an ideal transformer. In an ideal unloaded transformer, inductance and therefore input impedance are infinite, input current is zero, flux is zero. It's a situation of zeros times infinities to get secondary voltage V2 = V1*N2/N1. Ampere's law is irrelevant since current is everywhere zero.

In a real but still unloaded transformer, input inductance is finite, input current I1 ("magnetizing current") is finite, flux is finite, Ampere's law holds. Magnetomotive force mmf is generated = N1*I1 = H*d where d is the mag. path length. If you now load the secondary (say with a resistor) there is current in the secondary and by Lenz's law this current generates a current resulting in a secondary flux opposing the primary flux. I1 goes up, I2 goes up but total mmf does not change: N1*I1 + N2*I2 is unchanged. So H and flux are also unchanged.

Bottom line, primary and secondary currents don't cancel exactly because the magnetizing current is there irrespective of loading the secondary.

jim hardy said:
leaving you with zero which is the mmf required to magnetize an infinitely permeable zero reluctance core.

rude man said:
In an ideal unloaded transformer, inductance and therefore input impedance are infinite, input current is zero, flux is zero.

i slightly disagree , though i suspect it was just a typo.. or perhaps i misunderstood...
Were flux zero so would be its derivative and hence counter-emf as well.

I think instead flux = ##\frac {0 mmf} {0 Reluctance} ## = undefined
so it will be whatever flux counters applied emf

rude man said:
...flux is zero. time integral of voltage.
It's a situation of zeros times infinities divided by zero , zero mmf divided by zero reluctance to get flux that results in secondary voltage V2 = V1*N2/N1.

but it could be @rude man meant infinite permeability which would be a multiplication .

old jim

jim hardy said:
i slightly disagree , though i suspect it was just a typo.. or perhaps i misunderstood...
Were flux zero so would be its derivative and hence counter-emf as well.

I think instead flux = ##\frac {0 mmf} {0 Reluctance} ## = undefined
so it will be whatever flux counters applied emf

but it could be @rude man meant infinite permeability which would be a multiplication .

old jim
Agreed. emf = - dφ/dt, emf is the applied voltage, so φ is correspondingly finite. Thanks for the observation.
So scratch "zero flux".

jim hardy

## 1. What is Ampere's Law?

Ampere's Law is a fundamental principle in electromagnetism that relates the magnetic field around a closed loop to the electric current flowing through that loop.

## 2. How is Ampere's Law applied to a transformer?

In a transformer, Ampere's Law is used to determine the magnetic field produced by the primary coil, which then induces a voltage in the secondary coil. This allows for the transfer of electrical energy from one coil to the other.

## 3. What is the relationship between Ampere's Law and the number of turns in a transformer?

Ampere's Law states that the magnetic field is directly proportional to the number of turns in a coil. Therefore, the more turns in the primary coil of a transformer, the stronger the magnetic field and the greater the induced voltage in the secondary coil.

## 4. How does Ampere's Law affect the efficiency of a transformer?

Ampere's Law plays a crucial role in determining the efficiency of a transformer. If the number of turns in the primary and secondary coils are not proportional, it can lead to energy losses and reduce the overall efficiency of the transformer.

## 5. Can Ampere's Law be used to calculate the output voltage of a transformer?

Yes, Ampere's Law can be used, along with other principles such as Faraday's Law, to calculate the output voltage of a transformer. By knowing the number of turns in the primary and secondary coils, and the input voltage, the output voltage can be determined using Ampere's Law.

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