Basic electricity question

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  #1  
Old 12-27-04, 03:57 PM
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OakIsland
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Basic electricity question

Basic electricity question

Something still a bit mysterious to me is how something *capable* of great energy (my car battery) won't blow up a little light I put in series with the 12V battery. It's like the little light determines how much energy it needs, and only that much flows.

Anyway, there's a related thought I want your help with. I want to know "what will happen if..."

If I connect just the little light and tiny wires (18 or 20 gauge) to the battery, I anticipate that the little light will shine normally--nothing will melt down or blow up, right?

But what would happen to the little light and tiny wires if I disconnect the positive and put the little light between the positive terminal and the disconnected heavy wire and turn the headlights on? Will the little light be the "restrictor" and prevent the headlights from turning on? or will the headlights demand more juice and overwhelm the little light?

Thanks,
OakIsland
 
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  #2  
Old 12-27-04, 05:09 PM
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hex2k1
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In a dc series circuit we know that current is your constant and the sum of all voltage drops should equal the source. so if you have two lights in series one being the headlights and the other your little light the current in the circuit is going to be the same at any point in the circuit. so if your 18 or 20 awg wire can handle the current the circuit will work but you will have a really bright little light if the wattages are different which im pretty sure they are. and if thats the case the little light will eventually burn out if it doesnt happen immediately. thats why you try to connect everything to your battery in parallel not series which makes it possible to use different size wires according to the circuit you are making. and just for info purposes i know alot of people that get confused with this if you take a wire from the - post and one from the + post and hook up a light in between you are technically making a series circuit but that series circuit is in parallel with the source thats why youre able to use small wire with no problems. hope this helps out
 
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Old 12-27-04, 07:36 PM
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Ohm's law at work.


Lets assume both bulbs you are talking about are rated for 12 volts.
Lets assume the little bulb is a 12 watt, and the big one is 120 watt.

The little bulb has a resistance of 12 ohms, and on 12 volts draws 1 amp.
The big bulb has a resistance of 1.2 ohms, and on 12 volts draws 10 amps.


SO, if you put the 2 bulbs in series and connect them to a 12 volt circuit, you have 12 volts and 13.2 ohms, so the current is about 0.91 amps.

The little bulb would be fairly bright, seeing 90% of normal amps. The big bulb would be very dim, seeing less than 10% of normal amps.
 
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Old 12-27-04, 09:08 PM
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OakIsland
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hex2k1 and 594tough, thanks for your replies! Great info in each.

hex2k1, interesting point about why the little bulb survives directly on the battery: it's in parallel.

OakIsland
 
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Old 12-27-04, 10:11 PM
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The little bulb survives the "great energy" of the battery not because it is in parallel, but because that "great energy" is only at 12 volts. And the "little bulb" stands a much greater chance of surviving than a big bulb, because the little bulb typically has much higher internal resistance than the big bulb.
 
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Old 12-27-04, 10:28 PM
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OakIsland
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Hmm! Thanks for your comment.

Well, here's my real project so that you can all advise on the best solution. I'm investigating why my car battery was drained although just a couple months old. It may turn out to be the alternator, but I'm still curious to improve upon this little light bulb scenario further.

Considering that there might be a short in the vehicle that drains the battery, I was following advice from the car manual to insert a bulb between the positive post and the lead (after disconnecting the car radio) and see if the bulb lights. If it does, there's some sort of short.

That started me on this journey, thinking about bulbs. Another fix-it article mentioned having a little bulb just to check points along the path, say, to the brake lights, finding out where the wire is no longer hot (when appropriately switched on). Okay, cool, I want to get a bulb and wires, I told myself--I want to try some of these little tests for my own experience.

I bought a tiny hobby light (14V, 200 mA) and 18 gauge wire. Then I remembered that the first test I noted above was a series test...and thus this thread of discussion.

At the moment, I'm thinking the little bulb is JUST FINE for parallel tests (e.g. touching a point I expect to be hot and grounding the other wire to see if the bulb lights) and it's just the series test that's in question.

So, now that you know the 14V/200mA part, do you still think it has enough whatever to handle being in series with headlamps on?

Thanks,
OakIsland
 
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Old 12-27-04, 10:38 PM
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Yes, it can certainly be in series with anything you might throw at it. It's designed for 14 volts, so 12 volts is no problem. You could put it directly across the battery terminals (i.e., in series with a dead short). And putting it in series with anything else (such as the very low resistance headlights) reduces the current through it even more, so it's even less of a problem than if you put it directly across the battery.

Note that the headlights will probably be quite dim when in series with this tiny bulb.
 
  #8  
Old 12-27-04, 10:47 PM
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OakIsland
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Originally Posted by John Nelson
putting it in series with anything else (such as the very low resistance headlights) reduces the current through it even more, so it's even less of a problem than if you put it directly across the battery.
Ha! Of course! If it survives with no extra load, of COURSE it could survive being in line with something SHARING the load. (Why didn't I think of that...but it's so great that you are all helping me think through this!)

You make a point about 14V vs 12V. Why does 14V rating help it do just fine vs if it were, say, a 6V rated light? And, kinda related, how does the 200mA rating determine what the light can/can't handle?

Thanks!
OakIsland
 
  #9  
Old 12-27-04, 11:02 PM
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The 200mA merely tells you how much current it draws at 14 volts. From that, you can figure out the internal resistance if you want. R=V/I, so the resistance is 14/0.2, which is 70 ohms.

When you put that bulb across 12 volts, it will only draw I=V/R or 12/70, which is about 170mA. Since power is proportional to the square of the current (I*I*R), the bulb will be 27% dimmer than it was designed for. In series with the headlights, it would draw less current and be even dimmer (depending on how much extra resistance was added by the headlights).

If you were to put a 6 volt bulb across 12 volts, it would draw twice the current it was designed for, and would probably have a very short life (maybe just a fraction of a second).

If you were to run this 14-volt 200mA bulb from a transformer, the 200mA figure tells you what power transformer you would need. You'd need one capable of producing at least 200mA. A typical 10-watt 16-volt doorbell transformer, for example, is capable of producing about 625mA at 16 volts.
 
  #10  
Old 12-27-04, 11:15 PM
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OakIsland
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Very cool. Every detail was helpful for me to read and absorb. My confidence has likewise increased!

Thanks much,
OakIsland
 
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