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OK, so if you directly connect the two terminals of a battery, there is very little resistance in the circuit and you will drain the battery very quickly. However if you have a very simple circuit, for example a light bulb connected to the battery, the battery will drain slowly based on the watts used by the light bulb. My question is, if this light bulb provides only a little resistance, wouldn't a lot of current flow through and therefore drain the battery quickly?

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  • \$\begingroup\$ Complexity has little to do with it. A digital watch with thousands of transistors can run for years on a tiny battery, but a flashlight will drain a much bigger battery in a few hours. \$\endgroup\$ Feb 8, 2015 at 18:43

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You are correct. If the light bulb only has a small resistance the battery will be drained fast.

That is true for any circuit, be it simple or be it complex. It's a matter of total resistance.

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  • \$\begingroup\$ So by this reasoning, wouldn't a 100W light bulb with its greater resistance drain a battery more slowly that a 5W light bulb which would resist less? \$\endgroup\$ Feb 8, 2015 at 19:44
  • \$\begingroup\$ It depends on the voltage rating of the light bulb. A 220V 100W light bulb has a resistance of (220V)^2/100W = 484 \$\Omega\$. A 5W 220V bulb has (220V)^2/5W = 9680\$\Omega\$. So the less wattage bulb has a higher resistance. And will drain the battery more slowly. \$\endgroup\$
    – Arsenal
    Feb 8, 2015 at 19:48
  • \$\begingroup\$ OK, I think this is starting to make sense. One remaining point of confusion, do you get more or less resistance if you add additional 5W bulbs in series. Guessing less (to make the math add up) but this doesn't seem intuitive. \$\endgroup\$ Feb 8, 2015 at 20:38
  • \$\begingroup\$ A light bulb is basically just a resistor, adding them in series also adds the resistance. If you apply the same voltage across them, less current will flow and the bulbs won't be as bright as before. \$\endgroup\$
    – Arsenal
    Feb 8, 2015 at 20:39
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You may be a little confused by the resistance of a light bulb. Typically, a "cold" (that is, not glowing) bulb filament will have a resistance about 1/10 its resistance when hot. So an operating (glowing) bulb will discharge a battery much less quickly than you might think if you just go by what a meter says the resistance is.

Once it's glowing, another effect comes in to play. For a given desired brightness, you can get better efficiency by overdriving the bulb. That is, if you take a bulb and drive it at higher voltage, it gets even hotter than it would normally, and its resistance gets even greater. Since it is hotter, a greater fraction of the energy being dissipated by the filament is given off as visible light as opposed to infrared. See "blackbody radiation" for an explanation. Of course, this means the filament will burn out more quickly, but you can't have everything.

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  • \$\begingroup\$ Makes sense. Now obviously more light bulbs would equal more power consumed from the battery. I am wondering why though, as more light bulbs means more resistance, means less voltage / current. In other words, how do we go from direct wire draining the battery very quickly, to a small bulb draining it very slowly, to several bulbs draining it quickly once more. It's like the relationship between resistance and load flips at some point. \$\endgroup\$ Feb 8, 2015 at 19:54
  • \$\begingroup\$ Sorry, but you are confusing series and parallel connections. If you put several light bulbs in series, they will glow much less brightly, and drain the battery more slowly. If you put them in parallel, all will glow at normal brightness, and the battery will discharge faster. It's like draining water from a tank - put several hoses end to end and the water will drain more slowly. Cut holes for more spigots and attach more hoses, and the tank will drain quickly. \$\endgroup\$ Feb 8, 2015 at 20:56
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If you connect a high-power (bright) light between the battery terminals it will drain the battery faster than a lower-power (dim) light, because the bright light will draw more current than the dim one.

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