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I'm about to replace the batteries of my UPS. For safety, I'd like to drain the old batteries completely before disposing, but I don't have a large 12v load at my disposal.

Can I wire it to a 220V incadescent bulb (a.k.a. a 'lamp') to drain it?

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  • \$\begingroup\$ You might be better off using a car headlamp as it would drain the battery faster. \$\endgroup\$ – Thomas O Oct 4 '11 at 11:29
  • \$\begingroup\$ Well I don't have a car headlamp lying around. \$\endgroup\$ – Dave Van den Eynde Oct 4 '11 at 11:54
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Yes. The current will be higher than at 230V (230, not 220) though, because the lower voltage won't heat the filament as much as the 230V would. And a less hot filament is a lower resistance (the filament is a PTC resistor).

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  • \$\begingroup\$ As a follow-up, do you think it's a good practice to use a lamp for this purpose? \$\endgroup\$ – Dave Van den Eynde Oct 4 '11 at 9:03
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    \$\begingroup\$ @Dave - I think it is. Shorting would be an alternative, but this could cause a high pressure inside the cell. A bulb is a good idea if you don't have a low-resistance power resistor at hand. Or you could use a higher resistor value and leave the battery connected for a looooong time. \$\endgroup\$ – stevenvh Oct 4 '11 at 9:23
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    \$\begingroup\$ Actually the bulb current will be lower at 12V than at 230V. The relationship is non-linear but still monotonic. Think of it this way, if the current were higher it would heat the bulb to a higher temperature than 230V would. The steady state has to be lower. \$\endgroup\$ – Olin Lathrop Oct 4 '11 at 13:23
  • \$\begingroup\$ @OLin - A 60W bulb draws 0.25A at 230V. Same current at 12V would mean the bulb consumes only 3W. I haven't tested this, but my gut feeling says the filament doesn't heat enough to represent a 50\$\Omega\$ resistance. Heck, I can't even test it; all my 60W incandescent bulbs have been replaced with CCFLs! \$\endgroup\$ – stevenvh Oct 4 '11 at 13:47
  • \$\begingroup\$ But think of driving the bulb with a 250mA current source. If what you are saying is true, then the voltage will get stuck at 12V. This means a bulb exhibits hysteresis if true. Is that really the case? \$\endgroup\$ – Olin Lathrop Oct 4 '11 at 14:56

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