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Battery Powered Light Bulb: CR4 Challenge (09/23/08)

Posted September 21, 2008 5:01 PM

This week's CR4 Challenge Question:

A light bulb is connected to the power line by a coiled wire. You insert an iron cylinder inside the coil of the coiled wire. How the light of the light bulb will be affected? Now you repeat this experiment, but this time you power the light bulb with a very powerful battery. What happens to the light of the light bulb?

Disclaimer: This is a theoretical challenge for discussion only – DO NOT ATTEMPT THIS EXPERIMENT! It is potentially dangerous.

And The Answer is....

In the first case, the light bulb will dim. Because the light bulb is connected to an AC source (the line voltage) the iron cylinder will produce a variable magnetic field which will produce a a voltage drop in the coil. Because, the power source is constant (110 V rms), by having a voltage drop in the coil, the voltage across the light bulb will diminish.

In the second case, the power source is DC voltage. The dc current flowing through the coil will not produce a variable magnetic field. Without a variable magnetic field in the coil no voltage drop will develop across it. Therefore, in this case the brightness of the light bulb remains unchanged.

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#1

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/22/2008 11:04 PM

With an AC source, the light bulb will dim, and the iron cylinder will warm up slightly. You've created an induction heater. With a DC source, the coil and iron cylinder will make no difference.

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#2

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/22/2008 11:23 PM

The coil of wire will add inductance so the circuit will look like an inductor in series with a resistor (total resistance being the resistance of the light bulb element plus resistance of the coil). If we assume the power line is an AC source then the inductor willl phase shift, store and release energy in its magnetic field which will change the reactance of the circuit but will not affect the 'real' power used, ie, the light bulb will not be any dimmer because of the coil. When the iron cylinder is inserted it will increase the inductance of the circuit which by itself will not affect the 'real' power dissipated by the light bulb, however, eddy currents will be generated in the iron cylinder due to the alternating current of the coil which will use up some of the energy so the light bulb will be dimmer with the iron cylinder inserted. If using a very powerful battery the difference is that it is a DC source and so there is no inductance and inserting the iron cylinder will not create any ongoing eddy currents so the bulb will not be affected. Note that there will be some effect while the iron bar is inserted as some energy is used in aligning the dipole's in the iron with the magnetic field lines of the coil, but this is only temporary.

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#8
In reply to #2

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 8:20 AM

I think a key here is what is meant by a very powerful battery. Voltage? Amps? My initial thought was that the filaments of the bulb would be burned out and the bulb would fail with the powerful battery in the circuit.

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#3

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/22/2008 11:35 PM

I agree with ken_fry on the AC scenario.

You are increasing the inductance by introducing the iron cylinder into the coil and hence increasing the impedance (Resistance of Light Bulb + Reactance of Inductor, 2.∏.f.L) of the overall circuit.

In the case of the DC, I believe the light will dim momentarily as you insert the iron core and then return to normal brightness. Some energy will be needed to magnetise the iron cylinder.

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#4
In reply to #3

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 12:40 AM

In the case of the DC, I believe the light will dim momentarily as you insert the iron core and then return to normal brightness.

I agree, although actually seeing the change might be tricky, depending upon the number of turns and size of the core.

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#5

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 4:24 AM

Some old theatre dimmers used variable inductors to dim stage lights by raising and lowering the core of an inductor to change its inductance. However inductors don't use iron cylinders as cores. An iron cylinder however will behave as a shorted turn so I would guess the coil would have little effect on ac as its inductance will be low.

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#6
In reply to #5

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 5:01 AM

The iron cylinder will definitely heat up.
But if the cylinder diameter is not too large, the inductance at 60Hz will still increase.
(The frequency at which inserting the inductor changes from an impedance increase to an impedance reduction depends on the diameter of cylinder.)

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#7

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 7:57 AM

Alternatively, someone will disregard the disclaimer and try to prove Edison was right about Westinghouse and Tesla. Probably will succeed, too (again, depending on the number of turns in the coil and the diameter of the rod).

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#9

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 9:24 AM

I thought one should also state that the light bulb would be brighter in the case of DC than in the case of AC given the dc equivalent voltage of the AC is the same as DC battery being used

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#10

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 1:35 PM

Using the idea that the Line potential (RMS) and the battery potential are the same, I contend that the bulb will burn just as brightly to the eye in both cases. Using very sophisticated measuring equipment you may be able to measure a slight light difference or a O'scope see and calculate a mild power difference As you insert the rod. Inserting the rod will have little effect assuming it can not touch in any way the coiled wire. There will be some "energy" loss to the iron coil in both scenarios. However that energy loss will be covered by the fact we are using a coil witch will resist a change in current and no connction to the rod for any change in voltage.

Using an acronym in learned in the service ELI the ICE man we know that Voltage leads current in a purely inductive circuit (bulb is a inductor as well) and current leads voltage in a purely capacitive circuit. Seeing that this is purely inductive the voltage at the bulb will be fairly constant. In either situation inserting the iron rod cause the field around our coil to start to collapse. The effect of the collapsing coil will keep the current at the bulb to be fairly steady as well. The bulb being an inductor as well it will also resist the change in current and the field will start to collapse trying to keep the current constant. With no connection to the iron core it will have no further drain on either circuit once it has reached it's heated and/or magnetized point (most likely microseconds).

Power (P)=Current (I) squared X Voltage(E)

So sit down with ELI the ICE man and have a piece of pie to solve this challenge?

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#11
In reply to #10

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 2:06 PM

My bad! P=IE or P=I squared X R can think of the third way just now.

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#12

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 6:09 PM

case 1: permanent dim

case 2: dim to bright subcase a: small time to come to full brightness, subcase 2: longer time to come to full brightness.

Explanation: V= -L di/dt. At switch closure for DC case, L is smaller without iron core and di/dt will induce counter-voltage and no voltage will appear across the bulb. With time, the voltage will reduce in response to lowered current (when V initially jumps across inductor, current goes toward zero). This process continues until steady state occurs. You might also ask what happens when you try to turn the switch off. Big surprise!

With core in place L is larger.

To see this mathematically, use Kirchoff voltage law for simple circuit. The solution is a first order differential equation and the voltage across the bulb is A(1-e(Kt)).

Do you really want more?

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#13

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 6:10 PM

There's another side effect that's only being hinted at by some posts. Everyone is focussed on the bulb and the current. (Per the original question)

The iron cylinder will become an electrmagnet as a function of number of turns and current. (IFF the coil is only one lead of the power source.)

I suspect that there is another sting in the question. All results so far are assuming that the coiled wire is only one current path. If the wire is like normal flex cord or household wiring, where supply and return path are in the same lead, then would the AC and DC situations be any different from the original condition?

My answer for the AC condition with conventional "wire" for electrical connection of lamps: No change to light output. (Magnetic field from current down the active exactly cancels magnetic field from return current in the neutral.)

My answer for DC condition with conventional "wire" for electrical connection of lamps: No change to light output. (Until the battery runs down.)

Let's see if that answer puts the cat amongst the pigeons.

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#15
In reply to #13

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 8:47 PM

Hi JaE,

" My answer for the AC condition with conventional "wire" for electrical connection of lamps: No change to light output. (Magnetic field from current down the active exactly cancels magnetic field from return current in the neutral.) "

Really interesting...

If this is the prerequisite of the poster then this means he/she wants to shoot for candidates of presidency and then i wonder why he/she posts here CR4 instead of more global sites applying IQ tests??

kind regards

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#17
In reply to #13

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 4:34 AM

That's a cunning interpretation. However, the challenge says "coil to power line". I for one wouldn't use that wording for connections to both power lines (nor to power and neutral lines)

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#14

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/23/2008 6:16 PM

Nothing happen with DC supply.

The Light bulb would be greatly dimmed and the iron cylinder also heats up greatly with AC.

I think the poster want to call attention to roasted fingers with disclaimer.

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#16

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 2:06 AM

I had to experiment (At reduced Voltage I might add!!!) to satisfy my curiosity .

I hooked up a 24 Volt AC (Measured 27.2V) isolating transformer output to a coil in series with 2 X 100ohm 5 Watt resistors.

It is evident there is a drop in the current (See pics below), albeit small for the coil I used, with the bolt going through the coil.

I then connected a DC power supply and adjusted the output until the current matched 129.9mA.

The DC voltage measured pretty much exactly 27.2 V as expected.

Pushing the bolt quickly into the coil caused the display to flicker momentarily down by 0.1 mA.

Pulling the bolt quickly out flickered the display up by 0.1mA

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#19
In reply to #16

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 8:03 AM

Great post! I love it when theory holds up in practice.

We all know that when meters of any kind are shown in posts, it is to prove the theory of energy extraction from the vacuum (aka "over unity"). This post demonstrates clearly that over unity devices work as advertised. On the right, the bolt blocks extra energy getting into the coil. Obviously, some of the extra energy can get in past the bolt, but for the purposes of discussion we can call 128.7 milliamps 100%. (A larger bolt would block more energy, making the demonstration even more convincing.)

Removing the bolt (left picture) gives the coil better access to the vacuum energy, increasing flow to 129.9 milliamps. This is substantially over unity (129.9/128.7 = 155%), but the more reasonable interpretation is that just a .1 milliamp change (from 128.7 to 128.8) should be considered "over unity." Therefore a 1.2 milliamp change is actually 12 times over unity (1200 percent!).

Replicators of this experiment will want to place the coil in a partial vacuum to provide even better access to the vacuum energy.

This can only be taken as strong proof that Exxon, Ford, Microsoft, and Intel are keeping these secrets from the public. Microsoft, in particular, has played a central role in promoting "solid state" technology while discrediting the more valuable vacuum tube technology to prevent the public from realizing energy from the vacuum. Last time I checked, there was not a single vacuum-tube-based computer supported by Vista! Not one!

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#21
In reply to #19

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 8:28 AM

"Replicators of this experiment will want to place the coil in a partial vacuum to provide even better access to the vacuum energy."

Why stop with a partial vacuum, says I? Let 'em take it down to zero atmospheres in the whole lab, that way they'll have all the free energy possible available (and none of that pesky energy-absorbing oxygen to foil their attempts)! Those O2 molecules must be the bane of their existence. Soaks up free energy like a bastard, it does!

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#18

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 6:34 AM

Inductors generate a voltage proportional to the Rate of flux change, and so the current in the AC circuit will depend on the frequency of the AC and the "reluctance" in the circuit. The reluctance depends on how many magnetic field lines are cutting the coil which in turn depends on the degree of magnetic linkage between the ferro-magnetic core and the coil. So AC case - the light gets dimmer dependent upon how far the magnetic core is inserted into the coil.

DC case, the frequency is 0 and there is no reluctance. BUT as the magnetic core is inserted, it disturbs the magnetic field in the core whilst it is moving. this generates a voltage to oppose the action (Lenz's Law) and so the current through the bulb reduces. As long as the core is stationary, it has no effect on the current. As the core is removed The voltage is in the opposite sense to insertion and current through the bulb increases. So DC case - the light gets dimmer on core insertion; brighter on extraction to a degree determined by the speed of the core movement. This phenomenon is reported in MPM's excellent post #16

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#20

Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 8:14 AM

My E & M professor in college once said that if you connected a large enough inductor in series with your house A/C wiring, the effect would be to create a phase shift in the 60 hertz power that would cause the electric meter to read nearly zero. You'd be paying practically nothing for your electricity.

The problem with this plan, he cautioned, is that if the A/C power went out the energy stored in the inductor would be released and create a massive surge through your house wiring.

/No, I haven't tried it. Yet.

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#22
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Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 8:32 AM

Did he happen to mention just how large that induction coil would need to be? You'd HAVE to pay nothing for the electricity, the cost of all that copper wire would tap out your budget!

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#23
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Re: Battery Powered Light Bulb: CR4 Challenge (09/23/08)

09/24/2008 11:56 AM

Did your E and M professor not understand how domestic power meters work, or was his presentation merely unclear? (Given that he thought that power line failure would damage your house wiring, I'm afraid that I suspect the former) .

Starting with the meter: so long as the Voltage reference coil is connected to the same potential as the input to the circuit (whether this be before or after the inductor), the meter will in principle continue to read power correctly. The only effect would be that the awful power factor would increase the costs for the generation company due to increased line losses. What you would need to do is change the phase of the Voltage source relative to what is passing through the current coil. As the impedance of the current arm of the meter is quite small, you won't get far by connecting an inductor around the current sensor. You can fool the meter by connecting an inductor in series with the Voltage sensor input, but this would have to be quite large (as the sensor has quite a high impedance). As you have to rewire the meter to do this, you might just as well disconnect the Voltage arm or provide a shunt around the current arm. The only place I can imagine this makes sense is you made the Voltage sensor part of a resonant circuit, to provide the "correct" Voltage amplitude on the sensor coil but at the wrong phase, but this would require a capacitor as well as the inductor. Please do not try this at home - apart from being illegal it can be dangerous if either you are not very skilled or you don't fully know what you are doing (and if you know what you are doing, you don't need my advice anyway).
Note: the effects are the same with eletronic power meters - but the terminology would have to change

Now as to what happens if the AC power fails (I'm only considering te original proposed arrangement here - the problems for the phase-shifted Voltage reference are slightly different). If the inductor is in series with your circuit, it will simply try to continue passing whatever current it was passing when the mains failed - your load should be well capable of handling that. What risk there is is if the power line goes open circuit. This will allow a large Voltage to be generated across the inductor on the side remote from your load, so there could be a fire hazard.
There would be more of a risk to your equipment if you turn off a large proportion of the load at the same time. Again, the inductor would attempt to maintain constant current, so the short-term Voltage surge could damage the wiring and whatever equipments remain connected.

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