magnetism and currents
IamSmooth, Mon May 29 2006, 03:08AMWHen a current flows through a wire a magnetic field encircles the wire. Imagine two wires each with one amp flowing through them: one is connected to a voltage of X and the other to a voltage of 100*X. Can anyone explain how the magnetic field differs between the two wires?
The reason for the question is this:
Imagine two transformers with identical primary/secondary turn ratios but with different wire resistances. If the primary of thransformer1 has 10v @ 1amp and the primary of transformer2 has 100v @ 1amp, how do the secondaries "sense" the differnt voltages such that the output is 10x different for transformer2? I thought that the current travelling through the primary wire creates the magnetic field that cuts through the secondary to induce the secondary current.
Please don't quote that Vs = N*Vp. I know the basic theory. I am getting stuck on the physics behind the theoretic formulas.
Thanks.
Re: magnetism and currents
Simon, Mon May 29 2006, 06:17AM
Certainly: there is no difference.
The magnetic field depends purely on the current.
The turns ratio formulae can be derived by imagining the transformer coils share one magnetic field (i.e. perfect coupling - this is physics). The voltage induced in a coil is proportional to the amount of changing magnetic flux multiplied by the number of turns. Since the two share the magnetic field, that means the voltage induced in each coil is poportional to the number of turns, or the voltage ratio is the turns ratio.
The current is inversely proportional the number of turns because the current generates the field. The field is proportional to n x I so I is proportional to the field divided by n.
I hope this answers the question.
Simon, Mon May 29 2006, 06:17AM
IamSmooth wrote ...
WHen a current flows through a wire a magnetic field encircles the wire. Imagine two wires each with one amp flowing through them: one is connected to a voltage of X and the other to a voltage of 100*X. Can anyone explain how the magnetic field differs between the two wires?
WHen a current flows through a wire a magnetic field encircles the wire. Imagine two wires each with one amp flowing through them: one is connected to a voltage of X and the other to a voltage of 100*X. Can anyone explain how the magnetic field differs between the two wires?
Certainly: there is no difference.
The magnetic field depends purely on the current.The turns ratio formulae can be derived by imagining the transformer coils share one magnetic field (i.e. perfect coupling - this is physics). The voltage induced in a coil is proportional to the amount of changing magnetic flux multiplied by the number of turns. Since the two share the magnetic field, that means the voltage induced in each coil is poportional to the number of turns, or the voltage ratio is the turns ratio.
The current is inversely proportional the number of turns because the current generates the field. The field is proportional to n x I so I is proportional to the field divided by n.
I hope this answers the question.
Re: magnetism and currents
Bored Chemist, Mon May 29 2006, 08:29AM
Life gets a bit more complicated when you start to look at the resistance of the transformer windings.
The equation for V in and V out comes from assuming that no power is lost in the transformer. In that case the voltages have to be inversely proportional to the currents and the currents are related because they both related to the same magnetic field. More turns implies less current for a given field so the voltage is proportional to the number of turns.
OK that's the easy bit. What happens when power is lost as heat?
Well, the easy way is to realise that the primary can be modeled as a perfect (non resistive) transformer with a resistor in series with the primary. Then the voltage across the primary is due to the current and the inductance (Strictly, the reactance) of the primary. That current is limited by both the inductance and by the resistance so it's lower. With less current there's less field so there's less output.
I think the problem here is that you are considering the current in the primary of a transformer to be just a function of the resistance and the voltage. In fact, for most transformers the resistance is so low as to be forgotten about. The current depends on the load connected to the secondary.
If you had a perfect (superconducting) transformer and connected it to an AC supply the current wouldn't be infinite. With no load, the current would be given by the voltage divided by the reactance (omega L) of just the primary coil. The current and voltage would be 90 degrees out of phase and there would be no net power. Once you connect a load to the secondary coil then more current is drawn by the primary. The primary coil has to generate a back- emf that equals the applied voltage. To do that it needs some value for the magnetic field in the core. When you connect a load to the secondary a current flows through it and it produces a field which oposes the field from the primary. With a smaller field the back EMF at the primary no longer balances the applied voltage so the primary current rises until the balance is restored.
At least, that's my picture of it. You might want to ask a physicist.
Bored Chemist, Mon May 29 2006, 08:29AM
Life gets a bit more complicated when you start to look at the resistance of the transformer windings.
The equation for V in and V out comes from assuming that no power is lost in the transformer. In that case the voltages have to be inversely proportional to the currents and the currents are related because they both related to the same magnetic field. More turns implies less current for a given field so the voltage is proportional to the number of turns.
OK that's the easy bit. What happens when power is lost as heat?
Well, the easy way is to realise that the primary can be modeled as a perfect (non resistive) transformer with a resistor in series with the primary. Then the voltage across the primary is due to the current and the inductance (Strictly, the reactance) of the primary. That current is limited by both the inductance and by the resistance so it's lower. With less current there's less field so there's less output.
I think the problem here is that you are considering the current in the primary of a transformer to be just a function of the resistance and the voltage. In fact, for most transformers the resistance is so low as to be forgotten about. The current depends on the load connected to the secondary.
If you had a perfect (superconducting) transformer and connected it to an AC supply the current wouldn't be infinite. With no load, the current would be given by the voltage divided by the reactance (omega L) of just the primary coil. The current and voltage would be 90 degrees out of phase and there would be no net power. Once you connect a load to the secondary coil then more current is drawn by the primary. The primary coil has to generate a back- emf that equals the applied voltage. To do that it needs some value for the magnetic field in the core. When you connect a load to the secondary a current flows through it and it produces a field which oposes the field from the primary. With a smaller field the back EMF at the primary no longer balances the applied voltage so the primary current rises until the balance is restored.
At least, that's my picture of it. You might want to ask a physicist.
Re: magnetism and currents
IamSmooth, Mon May 29 2006, 11:10AM
e = N × dΦ/dt volts
This is from Faraday's law: changing magnetic flux times turns-ratio results in voltage.
I appreciate both of your answers
IamSmooth, Mon May 29 2006, 11:10AM
e = N × dΦ/dt volts
This is from Faraday's law: changing magnetic flux times turns-ratio results in voltage.
I appreciate both of your answers
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