Why does a battery die when connected on prallel?

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Connecting a 9V battery in parallel with a 2.kΩ resistor and a multimeter likely caused a short circuit, leading to excessive current draw and overheating of the battery. The initial current reading of 30mA was likely incorrect, as the multimeter may have been improperly set to measure current in parallel instead of in series, resulting in a much higher actual current. This misconnection can cause the multimeter to short-circuit the battery, leading to rapid battery depletion and heat generation. Additionally, the internal resistance of the battery suggests it could handle higher currents than initially measured. Proper measurement techniques are crucial to prevent damage to both the battery and the multimeter.
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During one of my labs we were using circuits. I connected a 9V battery in parallel with a 2.kΏ resistor and a multimeter. I meaured the current across the circuit and noticed it went really high at the beginning (about 30mA) and then it slowly went down and the battery became really hot and unusable. Why did that happened? I've been trying to find why the battery would get hot and stop working. Any help?
 
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Because you created a short across the battery terminals.

It would be like connecting a light bulb to the battery. Depending on the wattage of teh resistor, it may have gotten hot as well.
 
polaris90 said:
During one of my labs we were using circuits. I connected a 9V battery in parallel with a 2.kΏ resistor and a multimeter. I meaured the current across the circuit and noticed it went really high at the beginning (about 30mA) and then it slowly went down and the battery became really hot and unusable. Why did that happened? I've been trying to find why the battery would get hot and stop working. Any help?

The battery is obviously shorted.

What two devices were in parallel exactly? The resistor and the multimeter (as an ammeter)? What setting was the multimeter on?
 
polaris90 said:
During one of my labs we were using circuits. I connected a 9V battery in parallel with a 2.kΏ resistor and a multimeter. I meaured the current across the circuit and noticed it went really high at the beginning (about 30mA) and then it slowly went down and the battery became really hot and unusable. Why did that happened? I've been trying to find why the battery would get hot and stop working. Any help?

It sounds like you had a short circuit.
9 volts / 2000 ohms = 4.5 ma
This would occur if you connected your multimeter in parallel with your resistor while in the amps mode. It should be placed in series with the resistor when measuring current.
 
OmCheeto said:
This would occur if you connected your multimeter in parallel with your resistor while in the amps mode.
Yeah. That's where I was going too.
 
he did say ""I meaured the current across the circuit .."" and got 30 ma, presumably not through 2K..

Polaris, go to "Electrical Engineering and read this thread :

"" why must ammeter be connected in series?""
 
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polaris90 said:
During one of my labs we were using circuits. I connected a 9V battery in parallel with a 2.kΏ resistor and a multimeter. I meaured the current across the circuit and noticed it went really high at the beginning (about 30mA)
The current was much higher than 30mA. You were not reading the right scale. I surmise that you were not using a digital meter? 300mA possibly.
 
NascentOxygen said:
The current was much higher than 30mA. You were not reading the right scale. I surmise that you were not using a digital meter? 300mA possibly.

no, as Jim hinted at, I bet he was trying to read the current in parallel across the cct instead of in series and the poor meter was shortcircuiting the battery

Dave
 
davenn said:
no, as Jim hinted at, I bet he was trying to read the current in parallel across the cct instead of in series and the poor meter was shortcircuiting the battery

Dave

I think the current would have to be much higher than 30 ma with a short circuit. That would mean the internal resistance is 330 ohm, and only 0.27 Watt of power would be produced in the battery.
 
  • #10
davenn said:
no
yes

I bet he was trying to read the current in parallel across the cct instead of in series and the poor meter was shortcircuiting the battery
That premise was fundamental to my reply. That the meter was shorting the meter is not in question, a 9v battery is not going to get hot powering either a 330Ω load, or delivering 30mA.

Not only was the meter incorrectly connected to the resistor, but OP has not interpreted the meter reading correctly, regardless of whether it was digital or analog. That 30mA was not 30mA, maybe it was 300mA.

I suggested analog, simply because I find it difficult to see how a digital reading could be misread too low by a factor of 10. Maybe there was a blob of solder on the readout and this was mistaken for a decimal point?
 
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  • #11
here's short circuit tests on the common 9 volt 1604 "transistor radio battery" . Looks like they are capable of ~ 4 amps, briefly.

friedrichengineering.com/web_documents/9volt%20Battery.pdf
 
  • #12
Quite a few (cheap) digital multimeters have a separate socket for an unfused 10A range. You have to plug and switch range, so a misreading of a factor 10 (or even 100) is definitely possible.
 
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