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Old Aug 2nd 2011, 11:49 PM   #1
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Generator question

I'm having trouble understanding one aspect of q6 https://sites.google.com/site/xtheunknown0/physics-1

If the light blows, why does the current in the primary decrease? Can you explain it without inductance and back EMF?

xtheunknown0
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Old Aug 5th 2011, 01:13 PM   #2
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Originally Posted by xtheunknown0 View Post
I'm having trouble understanding one aspect of q6 https://sites.google.com/site/xtheunknown0/physics-1

If the light blows, why does the current in the primary decrease? Can you explain it without inductance and back EMF?

xtheunknown0
You can think of the bulb in the secondary as something that is allowing the conversion of potential energy to kinetic energy. The potential energy comes from the primary and is delivered to the secondary. With the bulb not blown some of that PE is converted into KE moving the little electrons around through the bulb doing work.

Work done is energy taken. So the primary trying to keep the PE in the secondary the same has to source a little more current into that side to deliver the additional energy that is being used by the bulb.

When the bulb blows that side acts as an open and no current can flow. So no energy will be converted into KE. Therefore the primary side does not need to deliver as much energy to the secondary side to keep that side at some PE value. Therefore a little less current needs to flow within the primary side.

Isolation transformers like this deliver electrical energy from one place to another without using any wires. They do so completely through the energy that is contained within the magnetic fields that flow within the core.

I have a very good book written by some brilliant EE who uses a full form of Maxwell's equations and describes magnetism like this as a magnetic current. He can show mathematically that the magnetic current when the bulb is not blown is inducing an electrical current that is doing work within the bulb. Therefore power is being used by the secondary which comes from the primary. When the secondary no longer draws any power from the primary, the primary need not "work" as hard.

Does that help?
Craig
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Old Aug 5th 2011, 10:56 PM   #3
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Hello again,

I was thinking about it some more and thought of basically another "intuitive" way of thinking about this problem.

When the bulb is connected and working, you can kind of think of this as another load in parallel with the primary load. Instead of electrons delivering KE to the load themselves, instead they are delivering energy magnetically to the secondary load. When anything with less than an infinite impedance is in parallel with a load (the primary coil in this case), the overall impedance seen by the source is lower than with the load alone.

When the bulb opens, burns out, it acts as an open and so that path is no longer a finite load in parallel with the load impedance (again that of the primary coil).

So the overall impedance to the source increases when that bulb "burns out", and so when the impedance increases with the same input potential voltage, the current must decrease according to Ohm's law. The impedance increases because there is one less additional way energy can be used by the circuit.

Essentially this is the case. Instead of being electrically connected (the secondary) to the primary load, it is instead magnetically connected. Either way energy is being delivered to the secondary from the source through the primary.

When that "path" opens it is no longer a non-infinite impedance in parallel with the primary load impedance. It is now an infinite load impedance in parallel with the primary load impedance, so the impedance seen by the source is only that of the primary. When it was working it had a finite impedance that when in parallel with the primary impedance is lower than the primary impedance alone, which must therefore draw more current from the source.

As long as there is a way in which PE can be converted to KE into the electrons to move and do work within some element, that PE converted to KE must come from the source.

Within the primary, the source moves electrons doing some work in the process converting some PE into KE which in turn through magnetic fields is transferred into PE in the secondary. When PE is converted into KE to the electrons, you might as well say that power will be drawn. If more energy is required by the secondary to do work within the bulb in the form of KE within the electrons doing that work, then the more is asked from the source.

This is because the KE of the electrons within the primary are "creating" a PE in the secondary, some of which is being converted into KE within the secondary doing work within the bulb. When that blows, no more energy is needed or can even be transferred from PE to KE within the secondary so less is asked from the source. The source only need keep the secondary at a certain PE level, but there is no more converting of the secondary PE to KE that needs to be "replenished" by the source into the secondary.

I do not know if that helps or not. I am trying to avoid using inductance like you asked. Although speaking about the inductance seen by the source when a current can flow within the secondary and primary is different that the inductance seen by the source when only the primary has a current flowing through it. Inductance being part of the imaginary portion of impedance is greater when the bulb is blown then when it is working. This greater inductance when the bulb is blown creates a larger impedance to the flow of current within the primary. The opposite is true when the bulb is working. The mutual inductance is such that it is lower than the self inductance of the primary alone and so the impedance magnitude is lower, drawing more current from the source when the bulb is working.

Again I will leave inductance out of this and hope what I have said so far may have helped.

Many Smiles,
Craig
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