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.
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!)
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.
Which is why you should never carry ‘spare’ batteries in your pocket along with loose coins.
It was the batteries not the e-cg as per the news report.
Don’t short out your battery.
You mentioned the internal resistance of the battery, which is the key to what happened to the OP:
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/R[sub]total[/sub]). 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.
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.
You need some form of resistance – a resistor, a light bulb, a motor, etc. – to lower the AMPERAGE in the circuit.
What’s the intended function of your circuit?
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.
Battery Exploder?
School clock project?
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.
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.
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:
I[sub]meter[/sub]R[sub]meter[/sub] = I[sub]plate[/sub]R[sub]plate[/sub]
I[sub]meter[/sub] is tiny, while R[sub]meter[/sub] is large. I[sub]plate[/sub] is large while R[sub]plate[/sub] is tiny.
Actually, I thought the voltage drop was across the copper plate, but I guess it’s actually in the battery (not sure why though).
Copper plate has low resistance. by comparison the battery has high resistasnce.
It’s in both.
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?