Timeline for Can you calculate internal resistance of battery by shorting it?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2018 at 11:43 | comment | added | Jasen Слава Україні | it depends on the battery and the meter if you can measure the short circuit current, if the battery is an AA alkaline and the meter doesn't have a fuse on the 10A range, you might be lucky. also short circuit current may not be representative of the resistance of the battery at lower currents. | |
| Jul 14, 2018 at 11:43 | vote | accept | AznBoyStride | ||
| Jul 14, 2018 at 11:42 | comment | added | AznBoyStride | I understand things a lot more now! One quick small question: is it still impossible to measure given an perfect meter and perfecf battery? ( which does not exists ) | |
| Jul 14, 2018 at 11:37 | comment | added | User323693 | @AznBoyStride Some batteries will slightly get heated up to the extent one can't touch.. Some batteries may just silently explode, if somebody shorts the battery directly. Assuming everything is safe, still it is impossible to measure the internal resistance with a short across it. Only thing you can measure is current which certainly varies heavily over time for a short | |
| Jul 14, 2018 at 11:35 | comment | added | AznBoyStride | Basically implying that measuring the current using a meter is not accurate unless it’s an ideal meter. And it’s more accurate to measure the voltage drops instead using an external resistor! I understand it now, thank you | |
| Jul 14, 2018 at 11:32 | comment | added | AznBoyStride | I know how to calculate the internal resistance with an external resistor like you suggest, I’m just trying to understand why I can’t calculate the internal resistance by just shorting the battery and using a multimeter to get the current going through the internal resistance | |
| Jul 14, 2018 at 11:31 | comment | added | Ignacio Vazquez-Abrams | If you're using a meter to measure current then the resistance is inside the meter, and is affecting your measurement. | |
| Jul 14, 2018 at 11:30 | comment | added | AznBoyStride | If that’s the case; then what kind of current am I measuring with a multimeter when I short the battery? | |
| Jul 14, 2018 at 11:29 | comment | added | AznBoyStride | Are you implying that you can’t measure the current through the internal resistance with a meter, if there is no other external known resistance ? | |
| Jul 14, 2018 at 11:28 | comment | added | Ignacio Vazquez-Abrams | Potential dividers are an application of Ohm's Law, and you should understand them as well. | |
| Jul 14, 2018 at 11:27 | comment | added | AznBoyStride | Or maybe I’m wrong, since I have no idea about potential dividers yet, I’m just trying to understand things solely on my knowledge of ohms-law. If ohms-law is not enough to explain my confusion, then I might have to read on potential dividers first D: | |
| Jul 14, 2018 at 11:25 | comment | added | Ignacio Vazquez-Abrams | Current is calculated by measuring the voltage drop across a known resistance. | |
| Jul 14, 2018 at 11:23 | comment | added | AznBoyStride | Can you please elaborate on why you can’t get the current without using additional resistors? If I short the circuit, doesn’t that mean that the current I’m measuring is running through the internal resistance? And therefore I would be able to calculate that internal resistance given the current I measured and the voltage of the battery. Unless of course the meter will READ the wrong current running on the internal resistance somehow? | |
| Jul 14, 2018 at 11:22 | answer | added | User323693 | timeline score: 2 | |
| Jul 14, 2018 at 10:57 | comment | added | Ignacio Vazquez-Abrams | Yes, proportional to the ratio created by the potential divider created by the two resistances. | |
| Jul 14, 2018 at 10:56 | comment | added | AznBoyStride | I’m sorry if I’m so wrong as this is not my field of studies, but doesn’t voltage drop over the internal resistance as well? | |
| Jul 14, 2018 at 10:53 | comment | added | Ignacio Vazquez-Abrams | You can't get the current without providing a resistance for voltage to drop over, at which point you may as well just get the voltage drop with a reasonable resistance and do that math instead. | |
| Jul 14, 2018 at 10:42 | history | asked | AznBoyStride | CC BY-SA 4.0 |