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If we open-circuit the secondary winding of a transformer and put a constant current through the primary, flux $$B=\mu H$$ will be generated by the coil in accordance with Ampere's law. The flux should travel around the core even though the secondary is open, as shown below:

enter image description here However, given that the secondary current is zero (open-circuit), the ∮H⋅dℓ=Is equation cannot be valid. This seems to be a paradox - what am I missing here?

Edit: An alternative, perhaps clearer formulation of the question: Under DC at steady-state, a real transformer with an open secondary acts as an inductor - so obviously flux is generated in the primart winding; I presume this flux flows around the core, therefore there is flux through the secondary winding - so why is there no current?

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    \$\begingroup\$ A constant current? Like DC current? \$\endgroup\$ – tangrs May 31 '17 at 3:24
  • \$\begingroup\$ Yeah - perhaps I should have said 'current source' instead. \$\endgroup\$ – Sam Barrett May 31 '17 at 6:40
  • \$\begingroup\$ If you put the current source then this circuit is forbidden. You get a current transformer that has to have secondary connected to load or short circuit otherwise it blows. Have look at current transformer use. \$\endgroup\$ – Marko Buršič May 31 '17 at 7:38
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However, given that the secondary current is zero (open-circuit), the ∮H⋅dℓ=Is equation cannot be valid. This seems to be a paradox - what am I missing here?

Amperes law states that the integrated magnetic field is related to an electric current.

\$∮H⋅dℓ=\mu_0I_enc\$

In the definition the line integral must be closed, when you open the circuit and no current flows, you violate this integral. It is no longer a closed line integral and you can't use it to preform integration and the math breaks down. The current loop needs to be a closed loop. A lot of people just simplify the line integral, but if you wanted to preform a true integration, you would follow the wire out around the load and back in your line integral, since this is negligible compared with the current generated through the secondary people often leave it out.

I presume this flux flows around the core, therefore there is flux through the secondary winding - so why is there no current?

The loop is broken, current has have conductance to flow (vacuum and air make poor conductors). However, a voltage is still generated across the secondary, the electrons are being pushed up "against the wall" of the broken conductor, ready to flow if you decided to put a load on the secondary.

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  • \$\begingroup\$ Thank you laptop2d, it was the "...if you wanted to preform a true integration, you would follow the wire out..." part that was causing the confusion for me. Great answer! \$\endgroup\$ – Sam Barrett Jun 8 '17 at 16:14
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Ampere's law is applied to closed circuits only, the statement itself says:

For any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop. (Hyperphysics)

Now when you are saying open circuit there is no question that we are applying the Ampere's law on it.

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    \$\begingroup\$ I believe the 'closed loop' phrase above refers to the loop around which the magnetic field exists, not the 'closed loop' of the circuit. For example, the 'solenoid' picture and the 'right-hand screw rule' pictures don't depict a closed circuit (though obviously it would have to be closed at some stage to allow current to flow). \$\endgroup\$ – Sam Barrett May 31 '17 at 6:30
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The circuit is not clear. You say you put a constant current but draw an ideal voltage source.

You have two options:

1) You connect an ideal voltage source to the primary. Since the secondary is disconnected, at steady state the primary current is zero (considering an ideal transformer with no core losses).

2) You connect an ideal current source to the primary: That connection is forbidden. By definition an ideal current source cannot be connected to an open circuit, the same way that an ideal voltage source cannot be connected to a short circuit.

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  • \$\begingroup\$ Perhaps I'm not stating the problem very clearly so apologies for that. Under DC at steady-state, a real transformer with an open secondary acts as an inductor - so obviously flux is generated; I presume this flux flows around the core, therefore there is flux through the secondary winding - so why is there no current? \$\endgroup\$ – Sam Barrett May 31 '17 at 6:33
  • \$\begingroup\$ That's not how transformers work. You can't induce current (minus the initial blip) on the secondary if there's no alternating current on the primary. \$\endgroup\$ – tangrs Jun 1 '17 at 2:27
  • \$\begingroup\$ @SamBarrett there is flux, and that flux is constant. Current can only be generated on the secondary under variations of flux, not under a constant flux. \$\endgroup\$ – Claudio Avi Chami Jun 5 '17 at 6:41
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Given a transformer is a linear system, operating with open-circuit on secondary is allowable condition. And driving the primary with (AC?) current source is allowable condition.

What you have is ---- just an inductor.

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There is no current in the secondary winding because it is an open circuit.

When you apply a voltage to the primary of a transformer and have no load on the secondary (open circuit) you will see current flowing in the primary winding for 2 reasons.

  1. The primary winding has an impedance and therefore a portion of the current you will see will be due to this small impedance. This portion of the current is typically so small due to the magnitude of the impedance seen by the magnetic circuit that it is negligible.

  2. The magnetic circuit requires an exciting current in order for flux to flow in the core steel. The flux is dependent on the supplied voltage, frequency and number of turns in the primary winding or whichever winding you are applying the voltage to. As you increase the flux density of the core steel, more current is required.

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Faraday's law tells you what open circuit voltage is induced in the secondary and, if you take current from the secondary, then that current is not "induced" but is present as a result of the open circuit voltage appearing across the load impedance i.e. ohm's law.

Any transformer produces magnetism in the core ONLY as a result of primary voltage, frequency and primary inductance. Secondary currents and there respective additional primary currents do not cause any change in the magnetism in the core.

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There is no paradox to me. You shall integrate H along the whole red line. Then you will get current connected with this H loop.

That H loop must be closed for the formula to work. Electrical current loop may be open, though.

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