Electricity 101 Part 2: Ohm's Law and Fuses


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Electricity 101 Part 2: Ohm's Law and Fuses

Postby WingAdmin » Thu Oct 31, 2013 12:50 pm



Part 2: Ohm's Law and Fuses

I'm sure lots of people have heard of Ohm's Law before. It defines the relationship between amperage (current), voltage, and resistance.

As mentioned in part 1, voltage is the amount of force, or pressure, pushing the current through the wires. Amperage is the actual amount of current flowing through the wires. Anything that resists that flow is called...you got it, a resistor:

Relationship between voltage, amperage and resistance
Relationship between voltage, amperage and resistance


If we hooked up a wire between the positive and negative terminals of a battery, it would allow the current to flow more or less unrestricted - limited only by the size of the wire, and the internal resistance of the battery. This is a called a short circuit, and it is a Bad Thing. In our motorcycles, the battery can supply a massive amount of current - hundreds of amps - into a short circuit. The wire required to safely conduct hundreds of amps for any amount of time would be extremely thick. There are no such wires anywhere on our motorcycle. So the limiting factor becomes the size of the wire. It in effect becomes our resistor. As I mentioned in part 1, when something resists the flow of current, it heats up - because the electrons flowing through it bump into the atoms that make up the wire, causing friction. The more current flowing through, the faster and hotter the wire will get. How fast? I recently had a 12-gauge wire (which is about 2mm in diameter, about the size of a thicker spaghetti noodle) accidentally short across the battery terminals of a 12 volt motorcycle battery, while I was holding it in my fingers. Within a second, the wire melted, severely burning my fingers. I had this same thing happen when I was a teenager, only using a car battery - I still have the scar across my fingerprints to show for it. There is a lot of power in a motorcycle battery, a dangerous amount of power.

What about using a very thick wire to short out the terminals? Maybe a big thick steel bar? It would still heat up some, but the limiting factor would most likely be the internal resistance of the battery itself. The battery would heat up inside so quickly that the electrolyte would boil, causing the battery to explode, and spraying everything in the vicinity with extremely corrosive sulfuric acid. Did I mention that motorcycle batteries are dangerous?

So obviously we don't want this to happen. We don't want our batteries exploding, or the wiring in our motorcycles melting. When wiring in our motorcycle heats up to the point where it melts the plastic insulation surrounding it, the wires contact other wires, causing more short circuits, and increasing the amount of current flowing. More and more current flows, melting even more insulation, in a cascading disaster that frequently results in this:

Image

Why does this happen? Let's visit Ohm's law, which is nothing more than a simple equation:

Calculating Ohm's law
Calculating Ohm's law


E is electromotive force, measured in VOLTS
I is current, measured in AMPS
R is resistance, measured in OHMS

This triangle is a simple way of defining three equations:

E = I X R
I = E / R
R = E / I

Given any two of the values, you can calculate the third. Fortunately for us, we always have one of those values in our bikes: E is always 12 volts (or more technically, 13.8 volts when the bike is running).

Ever wonder how much current your fancy new halogen driving lights draw? Measure the light bulb with an ohmmeter. Let's say the light bulb measures 3 ohms. We want to find out current, or I:

I = E / R

in our case:

I = 12 / 3

I = 4 amps

So we know the light will draw 4 amps.

When we cause a short circuit, the resistance drops very, very low - let's say 0.1 ohms. What kind of current will we see then?

I = 12 / 0.1

I = 120 amps

120 amps is an awful lot of current.

There's one other equation you might want to know about:

Calculating Watts
Calculating Watts


E is electromotive force, measured in VOLTS
I is current, measured in AMPS
P is power, measured in WATTS

Just like the example with Ohm's law above, we can derive three equations from this:

P = I X E
I = P / E
E = P / I

In this case, P is the amount of power being consumed, in watts. In our light bulb example, we have a light bulb that is drawing 4 amps. We want to know how many watts it uses:

P = E x I

P = 12 X 4

P = 48 watts

So our light bulb is consuming (or dissipating) 48 watts. The vast majority of this is being given off as heat (only a small fraction is converted to light - filament-based light bulbs are very inefficient).

What if our light bulb failed somehow, and short circuited, so instead of 3 ohm load, it presented a 0.1 ohm load? All of a sudden we have a draw of 120 amps trying to force its way through wires intended to carry maybe only 10 amps. Of course, it won't actually draw 120 amps, because the size of the wire will limit just how much current can be drawn. But when this happens, the wire will heat up very quickly, to the point where it and its insulation will melt.

So how do we prevent this from happening? That's what fuses are for. A fuse is a small piece of wire or metal, encased inside a protective enclosure that allows you to see the wire or metal, but protects it from touching anything. This piece of wire or metal is sized so that when a certain amount of current flows through it for a certain amount of time, it heats up and melts, or "blows." Once it melts, the circuit is opened - just as if you turned a switch off - and the power stops flowing.

You will typically encounter two types of fuses: "Slow blow" and "Fast blow" (or fast acting). Just as their names imply, slow blow fuses will tolerate excess current for a short amount of time before they melt, or blow. Fast blow fuses will blow almost instantly when the current exceeds the rated amount.

A 20 amp slow-blow fuse may tolerate 23 amps of current for 15 seconds before blowing. Of course, they work on heat, so if it has had 21 amps across it for 15 seconds, and then the current jumps to 23 amps, it may take only another 3-4 seconds before it blows. Of course, a surge of 30 amps will blow the fuse instantly, regardless of whether or not it is a slow-blow type.

Also, because they work on heat, a 20 amp fuse will tolerate more excess current if the ambient temperature is near freezing, than if the ambient temperature is scorching hot.

Fuses serve two purposes: Their main purpose is to protect wiring, but they also protect what is connected at the end of the wire. Let's look at a fuse connected to a radio. The radio is a relatively low-powered device, and might draw at most 3 amps of current. The wiring would be sized appropriately to power the radio, and a 5 amp fuse would be installed, to give a little headroom and avoid nuisance fuse blowing.

Let's say the radio developed a short circuit inside of it. Because it has a short circuit, it would attempt to draw a tremendous amount of current. If this current were actually allowed to flow into the radio, the internal parts of the radio would be melted or destroyed. Instead, the fuse blows, protecting the radio.

Similarly, let's say the wire leading to the radio developed a short circuit. Once again, a tremendous amount of current would flow instantaneously, blowing the fuse, and protecting the wire from melting.

This is why it is very important that the fuse be physically located as close as possible to the source of power. In your house, the fuse or breaker panel is located at the service entrance for the power. In your bike, the fuses should be located right next to the battery. The idea is to have the absolute minimum amount of wire exposed that is UNFUSED - that is to say, wire that is not protected by a fuse.

It's also why it is VERY important to never put an oversized fuse in place of one that keeps blowing. If a fuse is blowing, there is a reason - something is drawing more power than the wiring is designed to handle. Let's say you have a 10 amp fuse on a circuit that keeps blowing. The wiring on that circuit will also be rated for about 10 amps. In our example, the fuse keeps blowing because the circuit is faulty, and is drawing 15 amps. To "fix" the problem, you put a 20 amp fuse in place of the 10 amp fuse. Guess what - the "fuse" now becomes the wiring itself. After a few minutes of 15 amps of current being drawn, the wiring will melt - and now you have a huge problem - somewhere, wrapped inside a tiny bundle of wires, deep inside your bike, you have a melted wire. How are you going to find it and fix it? And that's assuming it didn't start melting other wires, and end up causing a fire! Please don't put oversized fuses in your fuse box!

The last thing we'll talk about in this part is unwanted resistance. Let's use our 3 ohm light bulb as an example. What if we put two 3 ohm light bulbs in sequence, so one comes after the other (this is called in SERIES)?

Battery and two light bulbs
Battery and two light bulbs


In this case, we add the resistance of both light bulbs together. Instead of 3 ohms, we have 6 ohms. How much current will be flowing now?

I = E / R

I = 12 / 6

I = 2 amps

Instead of 4 amps, we now have only half the current flowing, 2 amps. Let's look at the power dissipated:

P = E x I

P = 12 X 2

P = 24 watts

We have only 24 watts, but this is spread across two bulbs, so that's 12 watts per bulb. The bulbs will be quite dim. This is not something we would normally do on our motorcycles. But here's something we DO find that causes an extra load in series: connectors.

A connector connects two segments of wires to each other. They do so by pressing pieces of metal (usually copper) tightly together, in hopes that current will flow with the least amount of impedance. However, things get in the way - dirt, oil, grime, dust - all things that impede, or resist the flow of current. Presto, our connector has become a resistor!

Let's say we put a connector in series with our light bulb. The connector hasn't been cleaned in a while, and it is presenting a load of 0.5 ohms to the circuit. That, along with the 3 ohms of our light bulb, presents 3.5 ohms of resistance.

I = E / R

I = 12 / 3.5

I = 3.4 amps

Notice instead of 4 amps, we're only drawing 3.4 amps. The amount of power being dissipated will be less as well, although I won't get into the more complex calculations determining which item is dissipating what power. We know however, that both the light bulb and the connector are dissipating power, or energy. Because some of that energy is being dissipated by the connector, and the overall amount of current is less, the light bulb will be dimmer.

We can't just get rid of energy - Einstein tells us that it has to go somewhere. In the case of the light bulb, electrical energy is transformed into light energy and heat energy. But we're also dissipating power in the connector as well! Where does that energy go? It turns into heat as well. And over time, that heat causes the contacts to move, and lose their connection. This increases the resistance, which increases the heat, which increases the resistance which increases the heat...and the result:

Image

And THAT is the reason to keep connectors clean, and in the case of the four-cylinder Goldwings, cut the stator connectors out entirely and wire them in directly.

There are two more practical applications for this in our bikes: Headlights and, in the four cylinder Goldwings, regulators.

Think about the path power takes from the battery to your headlight: It goes through the fuse block (two connectors), through a long wiring harness into the ignition switch (four more connectors), to the high beam switch (two more connectors) to the headlight area (two more connectors), and finally to the headlight (two more connectors). That's an awful lot of connectors, and an awful lot of potential resistance.

Now let's say we took a wire, hooked one end of the wire to the battery, and the other end to the headlight, bypassing all the middle men. That's a lot of connectors we've gotten rid of, which means a lot less resistance, which means a lot more current - which means a much brighter headlight. Now that's obviously not a realistic way to run your headlight, but we can take a wire from the battery, through a fuse, to a relay, to your headlight. A relay is a switch that is controlled by another electrical circuit, instead of by your finger. In our case, the relay would be hooked up to the wire that originally powered the headlight. Now we have almost a direct connection from the battery to the headlight, and it still works like the factory installed version - and we have a much brighter headlight as a result.

In the four-cylinder wings, the regulator (the device that controls the amount of voltage being used to charge the battery) gets its "reference" voltage from the ignition switch. It uses this voltage to determine how much voltage to convert from the stator. If it sees the reference voltage drop down, it boosts its output to compensate. If the reference voltage is too high, it reduces its output. The ideal is for it to keep the voltage at a precise 13.8 volts at the battery posts.

However, there are lots of connectors in between the battery, ignition switch and regulator. On older bikes, these old connectors create resistance, so the amount of voltage the regulator "sees" is lower than the actual amount at the battery. In response, the regulator attempts to "compensate" for what it thinks is too low a voltage, and it instead ends up boosting the voltage up to 15, 16 volts and even more. This will boil the electrolyte out of the battery, and shorten its life.

Using the same method as we used for the headlight, we can run a direct (fused!) wire from the battery, switch it with a relay, and plug that into the regulator. Now the regulator sees a valid, correct voltage level, and correctly regulates the voltage to 13.8 volts.



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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby dadztoy » Thu Oct 31, 2013 3:23 pm

:D :D :D for part 2 - again - pretty much the way It was taught to me nearly 5 decades ago.... BTW - I also learned vacuum tube theory as well as solid state theory at the time...

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby Mag » Thu Oct 31, 2013 10:30 pm

I've read it twice, to no avail, waiting until the weekend so I can pop open a cold brew and really mull over this. I am one of those 'afraid' of electronics people but want to learn, so keep going!

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby redial » Thu Oct 31, 2013 11:05 pm

Would you mind covering "How to join two wires together", please. If you use a soldering iron, a gas solder gun, the way to apply shrink insulation, and what are the benefits of doing a good job.

Thanks, although I am not electrically illiterate, I do appreciate the revision, and thank you for taking the time to present Electricity 101 Parts I and ....
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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby medicrider » Fri Nov 01, 2013 12:03 pm

Thanks for a great article and for your website! I'm new to the goldwing community and have learned sooo much. Keep up the great work!

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby cap10paul » Fri Nov 01, 2013 12:21 pm

Thanks for taking the time to put the basics in "leyman" terms. I unfortunately am aspiring to one day reach "leyman" status and will read this over and over hoping to retain some of it. I have a 94' SE awaiting my ability to find a short when everything is turned off. As I attach the battery terminals a small spark generates even though everything is turned off. Pulling one fuse out at a time hasn't isolated it in fact it sparks with all fuses pulled out. Frustrating as she is such a fine old bike.

Paul

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby hotwire » Fri Nov 01, 2013 7:15 pm

Good job on all the text explaining all, especially the resistance part. From my electrical education, I remember one calculation from the physics: it takes a specific amount of electrons flowing thru a wire, (any size wire, doesn't matter), and that number is exponential: 6.28 billion billion !! With that many electrons flowing , much friction happens, and of course, friction creates heat, and smaller wires can tolerate less than bigger wires. Most electrical using components are specifically designed for a particular amount of current draw, so instead of designing circuits to be protected on the premise of that appliance's current draw, the overload protection has to be designed to protect the wire, and that is why it is paramount to do as has already been said, USE FUSES!!! that are rated for the wire size, and not the for the rating of the accessory. protect the wires and if the fuse blows, your lights or whatever that circuit feeds, is suspected then to be shorted. Number 14 (AWG) American Wire Gauge, is rated for 15 amps, Number 12 is rated 20 amps, and number 10 is rated for 30 amps. When in doubt to wire size, a smaller fuse is always good safe policy.

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby hotwire » Fri Nov 01, 2013 7:20 pm

Sorry. I left out a needed part of explanation about current flow: It takes 6.28 billion billion electrons flowing thru any given size wire to CREATE ONE AMPERE of current draw. Apologize for leaving out that part, 'nuff said.

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby MNAspencade » Sun Nov 03, 2013 10:02 pm

Many Thanks for the nice explanation on electricity - I think it will be very helpful to many of us Wing owners. I too would like to see the basic information keep flowing. The idea of how / what to use to track down electrical issues would be a Huge resource to help us in repairing and or relating information on the forums.

Keep up the great work !!

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby mervk » Sun Nov 03, 2013 10:15 pm

Magic post, and will be useful many times in the year to come. Thanks for taking the time to write it. Could it be extended to include a description of how the audio system works? Eg: Component functions, interaction with other components, location, what will or won't work if a particular component fails etc. Would certainly help me fix up the 2 way installation in my 1500. Cheers.
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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby wing rider 2012 » Wed Nov 27, 2013 5:36 pm

virgilmobile wrote:If I'm allowed..I'll add a little too....If not..by all means...delete this post...

The common idea is the "red" wire is the "hot" wire....In DC electronics reality,, it is not...
Electricity(moving electrons) is provided from the negative terminal of the battery...The electrons actually flow through the frame to all the individual grounds,into the "load"..(ex.light bulb) and then return to the positive lug of the battery....

The main reason to ensure all the "grounds" are clean and tight....There the source of the "power" needed to operate the load....

One other thought...electrons travel best on the surface of a conductor,not up the middle...So in reality a #12 solid copper wire will carry less "power" than a #12 fine stranded copper wire...
Strange but true...the #12 fine strand wire has much more surface area for the electrons to travel on....
Negative(ground)= excessive electrons
Positive(hot)= lack of electrons
Electrons travel from negative to positive to complete the circuit...


Virgle, not to question your electronics knowledge but what you are describing about stranded wire verses solid wire is called "skin effect" and it is almost nonexistent in a DC circuit, this is a condition that exist in AC and high frequency. If a person was to use solid strand wire in a DC circuit the cross section view of electron flow would encompass the entire wire. The higher the frequency the greater the skin effect, this is why most fixed position high frequency conductors are hollow. Stranded wire is mainly used due to it being flexible and withstands vibration better than solid wire.

Wingadmin, very good article.
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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby jwh20 » Mon Dec 01, 2014 1:47 pm

Very interesting article but I do need to point on one serious error. In the text it says:

In this case, P is the amount of power being consumed, in watts. In our light bulb example, we have a light bulb that is drawing 4 amps. We want to know how many watts it uses:

P = E x I

P = 12 X 4

P = 48 watts

So our light bulb is consuming (or dissipating) 48 watts. The vast majority of this is being given off as heat (only a small fraction is converted to light - filament-based light bulbs are very inefficient).

Read more: posting.php?mode=reply&f=19&t=20239&sid=99301ce30d04a4c0c57fff9028e8f5dd#ixzz3Kfn66jhH


While this technique is valid in some circumstances, it's NOT valid for incandescent (i.e. white-hot filament) light bulbs. These devices exhibit what is called a "positive temperature coefficient" behavior which means, simply, that the current draw will go DOWN as the temperature of the filament goes up. So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit larger than what the bulb will draw once it gets hot.

It's best to use the manufacturer's specifications for power consumption but if you must measure it, you want to measure the current with an ammeter when the bulb is ON.

Also note that other lighting technologies such as LED or HID are even trickier to measure since they are complex electronic devices and not simple electrical ones like a incandescent bulb.

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby WingAdmin » Mon Dec 01, 2014 1:56 pm

jwh20 wrote:Very interesting article but I do need to point on one serious error. In the text it says:

In this case, P is the amount of power being consumed, in watts. In our light bulb example, we have a light bulb that is drawing 4 amps. We want to know how many watts it uses:

P = E x I

P = 12 X 4

P = 48 watts

So our light bulb is consuming (or dissipating) 48 watts. The vast majority of this is being given off as heat (only a small fraction is converted to light - filament-based light bulbs are very inefficient).

Read more: posting.php?mode=reply&f=19&t=20239&sid=99301ce30d04a4c0c57fff9028e8f5dd#ixzz3Kfn66jhH


While this technique is valid in some circumstances, it's NOT valid for incandescent (i.e. white-hot filament) light bulbs. These devices exhibit what is called a "positive temperature coefficient" behavior which means, simply, that the current draw will go DOWN as the temperature of the filament goes up. So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit larger than what the bulb will draw once it gets hot.

It's best to use the manufacturer's specifications for power consumption but if you must measure it, you want to measure the current with an ammeter when the bulb is ON.

Also note that other lighting technologies such as LED or HID are even trickier to measure since they are complex electronic devices and not simple electrical ones like a incandescent bulb.


That's correct - but being that I was using the bulb as an analogy for a generic load in order to explain basic Ohm's law, I figured I wouldn't get into that level of complexity. I thought using "bulb" rather than "resistor" would make it more applicable to motorcycle owners. :)

Also, I think you misspoke: "So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit larger than what the bulb will draw once it gets hot." The resistance will be lower when it is cold, and higher when it is hot. Current draw will be higher when cold, and lower when hot.

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby bigbird1 » Mon Dec 01, 2014 2:01 pm

"Also, I think you misspoke: "So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit larger than what the bulb will draw once it gets hot." The resistance will be lower when it is cold, and higher when it is hot. Current draw will be higher when cold, and lower when hot."

Absolute zero= no resistance= superconductor

Now I have a shot at the $100 giveaway for posting. :D

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Re: Electricity 101 Part 2: Ohm's Law and Fuses

Postby jwh20 » Mon Dec 01, 2014 2:15 pm

bigbird1 wrote:"Also, I think you misspoke: "So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit larger than what the bulb will draw once it gets hot." The resistance will be lower when it is cold, and higher when it is hot. Current draw will be higher when cold, and lower when hot."

Absolute zero= no resistance= superconductor

Now I have a shot at the $100 giveaway for posting. :D


Yes, I stand corrected on that issue. I meant to say:

"So if you measure the resistance with a meter, you'll get the COLD resistance which will be quite a bit LOWER than what the bulb will draw once it gets hot.


My bad!!




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