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Secondly, when the current starts to rise in the secondary wire, the magnetic field it generates is multiplied by the loop count. Obviously if we had lots of loops, a very small current will quickly generate enough magnetic field to cancel the inducing field out and there is no more force to raise the current. In other words: high voltage transformer will impose strong voltage on the secondary side, but this strength is quickly depleted.

This is a statement from this answer referring to why the current decreases in the secondary coil of a Transformer. The problem that I saw in this statement is that he said the induced current cancels out the magnetic field from the primary coil. Doesn't that mean there will be no more magnetic fields? And thus no more induction?

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  • \$\begingroup\$ Why is the text so large? \$\endgroup\$ – Chu Dec 11 '18 at 15:59
  • \$\begingroup\$ Lenz's law - an induced current opposes the field that caused it. \$\endgroup\$ – Chu Dec 11 '18 at 16:02
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    \$\begingroup\$ I suspect a poor translation from some other language. The phrase "strong voltage" is not normally used in this context. \$\endgroup\$ – Elliot Alderson Dec 11 '18 at 16:27
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    \$\begingroup\$ @Andyaka FYI, I found the source: physics.stackexchange.com/a/443775 \$\endgroup\$ – Kevin Reid Dec 11 '18 at 22:41
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    \$\begingroup\$ That answer is basically wrong and I’ve left a comment on it. \$\endgroup\$ – Andy aka Dec 11 '18 at 22:51
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In a normal transformer there are two windings called primary and secondary. The primary refers to the driven winding i.e. the winding where AC power is fed. The secondary has a voltage induced and the turns ratio determines how big or small that secondary voltage is relative to the primary voltage. So, if the turns ratio is 1:1, and you fed the primary with 100 volts RMS, you would expect to see 100 volts RMS on the secondary.

In addition, the secondary voltage is in-phase with the primary voltage unless you swap the secondary wires to make it the inverse (180 degrees out of phase). Faraday's law ensures this.

If you then applied a 100 ohm load to the secondary, you would get 1 amp RMS flowing from the secondary and 1 amp RMS flowing into the primary. In other words power in equals power out (ideal transformer).

What actually makes a transformer "tick" is another current (called magnetization current). This is in addition to the 1 amp load current mentioned above. If you want to simply measure that magnetization current, you could remove the secondary load and use a current-shunt on the primary and what you would see is the magnetization current.

That magnetization current (for a sinewave applied voltage) lags the primary applied voltage by 90 degrees and is basically the current that would naturally flow into the primary should there be no secondary load (or the secondary winding were removed). It's the current flowing in an inductor and that inductance is the primary winding inductance.

So, there are two currents going into the primary when there is a secondary load: -

  • Magnetization current lagging by 90 degrees (always there) and
  • Primary referred secondary load current and, for a resistive load on the secondary, this current is in phase with the primary and secondary voltages.

The upshot of all of this (when you analyse a 1:1 transformer), is that the current entering the primary winding (due to the secondary load) is the same current as that leaving the secondary and flowing into the load. Given that the two windings are wound together, the net magnetic fields that they produce is big fat zero.

But there is still the good old magnetization current in the primary and that is unaffected by both secondary load current and primary referred secondary load current. And, importantly, it is that magnetization current that generates the secondary voltage so, any talk of load currents cancelling the magnetic field is largely from those who know little or believe they know more than they actually do..

Just consider this very simple example of applying a 1 volt step to the primary when the secondary has a load of 1 ohm: -

enter image description here

Picture from here.

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  • \$\begingroup\$ My problem is that also why Current decreases in a step-up transformer, I know its just as simple as plugging in the P=V. I formula, thus if Voltage rises, Current decreases in a step up transformer. What I don't understand is that why the voltage increases and current decreases, why can't current increase while voltage decreases, the fact that it would give the same exact answer. * Two person I read online said the windings 'traps/congest' the current, so the more the turns, the more 'congested' it will be, is this true? \$\endgroup\$ – acmilan Dec 12 '18 at 14:30
  • \$\begingroup\$ In any transformer whether it is step-up or step-down, the secondary current depends entirely on the secondary voltage and the attached load. That is ohm's law. Congestion is a word used by fools that are trying to describe transformer action. \$\endgroup\$ – Andy aka Dec 12 '18 at 15:09
  • \$\begingroup\$ For the answer you mentioned at first, you were using a 1:1 transformer, is it ? Could you use a 2:1 or 1:2 transformer, so that I could analyze it better? \$\endgroup\$ – acmilan Dec 13 '18 at 2:39
  • \$\begingroup\$ @acmilan This is a simple question and answer site and not a learning development programme. If you have a question then either ask it as a comment or, if it might be too lengthy, ask a new question. \$\endgroup\$ – Andy aka Dec 13 '18 at 9:28
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Any fluctuations in a magnetic field will tend to induce currents in conductors that counteract that fluctuation. This is how the power transfer works...it's kind of like Sushi Station, where taking dishes off the conveyor lightens the conveyor, until it gets back to the kitchen where they add more dishes. Even more interesting is that a magnet moving past a conductor will create a magnetic field that opposes its motion, thereby supporting conservation of energy. Try sliding a neodymium magnet down a 45 degree ramp of plywood, then see if slows down when the ramp is aluminum.

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    \$\begingroup\$ Voltages are induced as per Faraday's law. Current may arise due to the induced voltage and some load impedance. Currents ARE NOT induced. \$\endgroup\$ – Andy aka Dec 11 '18 at 16:28
  • \$\begingroup\$ @Andyaka, you're right that the current is a secondary value...my point is that if the current is caused by the magnetic flux, it will flow in such a way as to oppose that flux. \$\endgroup\$ – Cristobol Polychronopolis Dec 11 '18 at 19:16
  • \$\begingroup\$ In a transformer, the changing flux produces the secondary voltage and that voltage is at right angles to that flux hence, any secondary load current is also at right angles (for a standard resistive load). This means that the secondary current does not flow in a way that opposes the flux. The ampere turns in the secondary perfectly cancel the primary referred secondary ampere turns and the net result is no alteration to the magnetisation flux and the secondary continues to produce a voltage and your answer is wrong. Sorry to say. \$\endgroup\$ – Andy aka Dec 11 '18 at 19:30

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