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I know that 40Watt,110V incandescent bulb has about 302 ohm resistance of tungsten filament (which is metal and of course have low resistivity) comparing it to 40Watt, 220V bulb which has about 1200 ohm. My question is --- how that much big resistance is produced in tungsten filament? I know about R=resistivity X length/Area but is that enough to get such higher resistance

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  • \$\begingroup\$ Your 144Ω are wrong. For 40W@110V, it has to be 302Ω. P = U²/R → R = U²/P \$\endgroup\$ – Janka Jan 17 '17 at 1:00
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    \$\begingroup\$ The answer is tungsten is a PTC resistor. Nearly all metals are. It has much higher resistance at 3000K than at 300K. \$\endgroup\$ – Janka Jan 17 '17 at 1:03
  • \$\begingroup\$ @Janka thanks for correction actually i was studying for wikipidea and it was for 100W@110V \$\endgroup\$ – Vaibhav Jain Jan 17 '17 at 1:03
  • \$\begingroup\$ @Janka but is it possible to generate that much temperature in bulb \$\endgroup\$ – Vaibhav Jain Jan 17 '17 at 1:05
  • \$\begingroup\$ It's glowing yellowish-white, that means the filament is at 3000K. See en.wikipedia.org/wiki/Planck%27s_law \$\endgroup\$ – Janka Jan 17 '17 at 1:07
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The different resistances of those incandescent bulbs are made with different length and different diameters of tungsten wires. A very long and thin wire is wound to a filament, sometimes even a double filament. But the values calculated from power and voltage are the resistances in hot condition. The resistance at cold condition is much lower. I have here a bulb with 120 W at 230 V and a resistance at room temperature of 29 Ohm. The resistance in hot condition at 2950 K is 440 Ohm. If we assume a diameter of 0.1 mm for that tungsten filament wire, we can calculate a length of 4.3 m for a resistance of 29 Ohm. I used the specific resistance of tungsten 0.0528 Ω ⋅ mm²/m for that calculation.

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