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#1
03-01-2016, 07:43 AM
 greggamma Guest Join Date: Mar 2016 Posts: 1
What happened? Electrical circuit education needed

I hooked a 9 Volt battery using 2 alligator clips to a piece of copper plate. Here is what happened 1) My volt meter read nothing. 2) There was a pop sound 3)The battery heated up quickly...less than 15 seconds.
Now A) What can I do to prevent that? B) WHAT happened? The battery tests at 8.6 volts with my multimeter c)If a battery has 9 Volts of potential why does it overheat when basically taking in what it puts out? Its the same voltage and current. Since the metal is resisting it I would have thought there was even LESS energy going back to the battery. D) Do I just need a resistor in the wire to lower the voltage going to the plate?
Thanks for any info.
#2
03-01-2016, 07:47 AM
 Seren359 Guest Join Date: May 2015 Posts: 86
If you connected direct to a copper plate it has almost zero resistance so the battery is discharging at it's maximum CURRENT.
This is why it heats up (and can catch on fire!)
#3
03-01-2016, 07:58 AM
 Hari Seldon Member Join Date: Mar 2002 Location: Trantor Posts: 12,215
Yes and there is almost 0 voltage because the resistance of the copper plate is much lower than the internal resistance of the battery. Not recommended.
#4
03-01-2016, 10:07 AM
 Machine Elf Guest Join Date: Feb 2007 Location: Challenger Deep Posts: 11,476
Quote:
 Originally Posted by Hari Seldon Yes and there is almost 0 voltage because the resistance of the copper plate is much lower than the internal resistance of the battery. Not recommended.
You mentioned the internal resistance of the battery, which is the key to what happened to the OP:

Quote:
 Originally Posted by greggamma I hooked a 9 Volt battery using 2 alligator clips to a piece of copper plate. Here is what happened 1) My volt meter read nothing. 2) There was a pop sound 3)The battery heated up quickly...less than 15 seconds. Now A) What can I do to prevent that? B) WHAT happened?
If you're drawing a circuit diagram, you can model the battery as a voltage source PLUS a resistor, arranged in series (the resistor represents the battery's internal resistance to current flow). Add another resistor in series to model the copper plate. So now your circuit diagram looks like just a voltage source with two resistors in series, (a voltage divider). As Hari Seldon notes, the copper plate's resistance will be much lower than the internal resistance of the battery. This means most of the battery's voltage will be dissipated across its own internal resistance; this is why the meter reads very close to zero volts (you're measuring +9 volts from the battery's chemical activity, together with -9 volts due to current flowing across its internal resistance), and why most of the power being produced by the battery is actually delivered to the battery. As heat, in that internal resistance. This configuration represents pretty much the lowest possible circuit impedance, so the battery is delivering as much current as it possibly can (I = V/Rtotal). So the battery is producing as much power as it can, and dumping all of that power into itself as heat.

You can't change the battery's internal resistance, so if you want to avoid roasting/exploding it, you need to not move so much current through it. To do that, increase the total circuit impedance. You can add resistors in series to make that happen. Note that power will be dissipated in those resistors, so they either need have high resistance (so as to make the current through them small), or they need to be rated for power dissipation (instead of little rice-sized resistors, you'll have a rectangular block of ceramic stuff maybe the size of your pinky). If you get low-value resistors, you'll still be moving a lot of current through your battery, and it will still get warm.

Last edited by Machine Elf; 03-01-2016 at 10:08 AM.
#5
03-01-2016, 08:49 AM
 bob++ Guest Join Date: Jan 2013 Location: Worcestershire UK Posts: 6,253
Which is why you should never carry 'spare' batteries in your pocket along with loose coins.
Quote:
 Robbie, 24, had been carrying some spare batteries for his e-cigarette in his trousers when he said one exploded and set his entire right leg on fire. He managed to reach into his pocket to push out one battery which had allegedly burnt through the material of his trousers. The other battery then melted on to his thigh - so Robbie said he pulled it off and threw it away before putting out the fire that had engulfed his entire leg.
It was the batteries not the e-cg as per the news report.
http://www.mirror.co.uk/news/uk-news...s-mans-7440075

Last edited by bob++; 03-01-2016 at 08:50 AM.
#6
03-01-2016, 09:47 AM
 74westy Guest Join Date: Jan 2004 Location: Regina, SK, Canada Posts: 1,841
Quote:
 Originally Posted by greggamma A) What can I do to prevent that?
#7
03-01-2016, 10:16 AM
 Gary T Member Join Date: Mar 2002 Location: KCMO Posts: 11,301
Quote:
 Originally Posted by greggamma Since the metal is resisting it...
Don't know where you got that idea but it's wrong. Electrically the metal plate is pretty much the same as a jumper wire but with even less chance of offering any resistance.

Quote:
 Do I just need a resistor in the wire to lower the voltage going to the plate? Thanks for any info.
You need some form of resistance -- a resistor, a light bulb, a motor, etc. -- to lower the AMPERAGE in the circuit.
#8
03-01-2016, 12:28 PM
 Mops Member Join Date: Dec 2002 Location: Germany Posts: 2,829
What's the intended function of your circuit?
#9
03-01-2016, 03:41 PM
 bobot Member Join Date: Jan 2009 Location: Chicago-ish Posts: 6,796
Quote:
 Originally Posted by Mops What's the intended function of your circuit?
Battery Exploder?
#10
03-03-2016, 02:13 PM
 Morgenstern Guest Join Date: Jun 2007 Location: Southern California Posts: 11,866
Quote:
 Originally Posted by Mops What's the intended function of your circuit?
School clock project?
#11
03-01-2016, 03:35 PM
 umop ap!sdn Registered User Join Date: Sep 2003 Location: Hellizona Posts: 2,227
Your voltmeter reads zero because electricity is following the path of least resistance, i.e. through the copper plate with its negligible resistance, and not through the meter. Even if you had used a resistor, you could still cause the resistor and/or the battery to overheat if the resistance is too low because the amperage would be too high.
#12
03-03-2016, 03:47 PM
 Chronos Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 80,086
Quote:
 Quoth umop ap!sdn: Your voltmeter reads zero because electricity is following the path of least resistance, i.e. through the copper plate with its negligible resistance, and not through the meter.
Other way around. A voltmeter is supposed to only allow a negligible amount of current through it, and thus should have a resistance much greater than the circuit does (which, in this case, it would). The reason it read no voltage was because almost all of the voltage drop was within the battery, not outside of it in the copper.

And I have to reiterate Mops' question: What is it you're trying to do? Whatever it is, you're doing it wrong.
#13
03-03-2016, 07:31 PM
 Dr. Strangelove Guest Join Date: Dec 2010 Posts: 7,364
Quote:
 Originally Posted by Chronos Other way around. A voltmeter is supposed to only allow a negligible amount of current through it, and thus should have a resistance much greater than the circuit does (which, in this case, it would).
That's exactly what umop ap!sdn said.

To be clear, most multimeters have two basic modes:
- Voltage measurement. The two leads have high resistance and the meter is connected in parallel with the device to be measured. All but a tiny amount of current flows through the device.
- Current measurement. The leads have negligible resistance between them and the meter is connected in series. All of the current flows through the meter, and there is a negligible voltage drop from the meter's small internal resistance.

In this case, we have:
ImeterRmeter = IplateRplate

Imeter is tiny, while Rmeter is large. Iplate is large while Rplate is tiny.
#14
03-07-2016, 10:31 PM
 umop ap!sdn Registered User Join Date: Sep 2003 Location: Hellizona Posts: 2,227
Quote:
Originally Posted by Dr. Strangelove
Quote:
 Originally Posted by Chronos Other way around. A voltmeter is supposed to only allow a negligible amount of current through it, and thus should have a resistance much greater than the circuit does (which, in this case, it would).
That's exactly what umop ap!sdn said.
Actually, I thought the voltage drop was across the copper plate, but I guess it's actually in the battery (not sure why though).
#15
03-07-2016, 10:54 PM
 Snnipe 70E Guest Join Date: Oct 2008 Location: San Jose, CA Posts: 3,919
Copper plate has low resistance. by comparison the battery has high resistasnce.
#16
03-07-2016, 11:31 PM
 beowulff Member Join Date: May 2001 Location: Scottsdale, more-or-less Posts: 16,098
Quote:
 Originally Posted by umop ap!sdn Actually, I thought the voltage drop was across the copper plate, but I guess it's actually in the battery (not sure why though).
It's in both.
#17
03-07-2016, 11:59 PM
 octopus Guest Join Date: Apr 2015 Posts: 7,676
Don't do that with a larger battery like a car battery. Internal resistance is lower and things go bad quick. What's the point of your circuit by the way?
#18
03-03-2016, 05:42 PM
 bob++ Guest Join Date: Jan 2013 Location: Worcestershire UK Posts: 6,253
since when were laptops grounded? Every one that I have ever seen only has a hot and return (so to speak) two wires. The law in the UK requires that it is fitted with a three pin plug, but the ground pin is not connected to anything. If I saw a spark there, I would assume that there was some crossover inside the plug itself.

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