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Is it because high resistance will stop the filament from burning out? Also is there any relation of the fact that tungsten has high melting point?

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  • \$\begingroup\$ Power= V^2/R, Yes. \$\endgroup\$ – RoyC Feb 15 '17 at 12:11
  • \$\begingroup\$ To glow it has to be high temp, for it to be high temp it there needs to be power dissipated. And high resistance is a nebulous term, I always thought that light bulbs were low resistance. \$\endgroup\$ – Voltage Spike Feb 15 '17 at 17:19
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You don't want a high resistance. You want the correct resistance for the power rating. e.g. a 100W bulb in a country with 110V power has an operating current of just under 1A and a resistance of 121 ohms.

For a given wattage and supply voltage R = V2/W

Note that is the resistance at operating temperature, at room temperature the resistance will be lower. That is why bulbs often break when you switch them on rather than after they have been on for a while, the initial current when they are cold is a lot higher.

Why the high melting point? Because it's going to be hot. Incandescent bulbs generate light because they are hot following the laws for black body radiation. By necessity they are operating with a filament temperature of around 3000K (around 5000F if you use weird temperature measurement units).

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  • \$\begingroup\$ recalling an old article in "electronics world" magazine, the resistance changes 10:1 as the bulb heats up. This resistance ratio is measurable: ohmmeter for cold R; ampmeter for hot R. \$\endgroup\$ – analogsystemsrf Feb 15 '17 at 14:03

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