resistance of a light bulb?

Question

Please remind me how to work out the resistance (or impedance if you wish) given say 40w and 240v?

Answer 1

Using P = VI

40W = 240 x I, so I = 40/240 = 1/6

Also, V= IR

So 240 = (1/6) x R. So R = 6 x 240 = 1440 ohms.

Answer 2

Light Bulb Resistance

Answer 3

R=v^2/p

Answer 4

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The heat is power dissipated. Power is I^2R (current squared times the resistance), so clearly you have to consider both the resistance and the current going through the device when you talk about power. In the first half of your statement about thinner/thicker wire, what you said is not true IF the wire is the load, but is true IF there is a load that the wires are supplying, and that load has a much higher resistance than the wire supplying it. If the wire is the load (has the majority of resistance in the circuit and therefore drops most to all of the voltage), then it is just like the light bulb, and a thicker wire will get hotter with lower resistance. Draw a circuit out and go through the math. Let's say that the voltage supply is 100 volts and is connected to a 100W (at 100V supply) light bulb with 100 feet of #10 wire. The resistance of 100 feet of #10 is about 0.10 ohms. If the light bulb is 100W with 100V applied, obviously its resistance is 100 ohms (100W/100V = 1 amp, and 100V/1 amp = 100 ohms). So total circuit resistance is the sum of the resistances, 100 ohms for the light bulb and 0.10 ohms for the wire. The current then is 100V/(100 + 0.10) = 0.999A The actual power dissipated by the light bulb is I^2R (current squared times the resistance) - (0.999A)^2 * 100 ohms = 99.8W The power dissipated by the wire is calculated the same way - (0.999A)^2 * 0.1 ohms = 0.0998W, a small fraction of the wattage in the light bulb. Now halve the size (which will double its resistance) of the wire by replacing it with #13 and do the same calculations - Total current I = 100V/(100ohms + 0.2ohms) = 0.998A (a slight decrease)... Power in light bulb W = (0.998A)^2 * 100ohms = 99.6W (a slight decrease) Power in the wire W = (0.998A)^2 * 0.2ohms = 0.1992W (about double the previous power, but still a small amount). So you can see that changing the wire resistance had little effect on the circuit because the resistance of the wire is very small compared to teh resistance of the light bulb. When you changed the wire resistance by a factor of 2, it only made a tiny difference. BUT changing the light the light bulb resistance by a factor of 2, and you change the current and power by nearly a factor of 2. So in summary and as I said above, the difference is based on who is dropping the most voltage in the circuit.

Answer 5

With heating appliances such as toasters, kettles, soldering irons and so on the resistance measured with a continuity tester should be the same as that calculated from the wattage rating printed on the appliance. For example a 25 Watt 250 Volt soldering iron will draw 1/10 Amp and have a resistance of 2500 Ohms.
This does not apply to tungsten filament light bulbs as the resistance is much lower when cold than when at the working temperature.
I assume you have tried this and wondered why the measured resistance was different from your calculated value.

Answer 6

Standard output of socket in North America is 15 Amps (unless you're dealing with appliances like fridges, stoves, washing machines, driers, and stoves ( I believe 25 to 50 amps).
E=V^2/W
so 57600/40 = 1,440 Ohms under direct current - probably somewhere in Europe!
Do a search on the Internet for the Ohm's Law circle table- an excellent tool for calculation reference involving equations like this.

Answer 7

Use the formula: Power = V^2/R
So R = V^2 / Power.

This will give you the OPERATING impedance of the light bulb based on its rated power. Its "off" impedance will be lower than this because the resistance of the filament is proportional to its temperature. i.e. the resistance increases when it gets hotter.

Cheers.

Answer 8

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Answer 9

Power = Volts X Current so 40w = 240v X 0.167a
Resistance = Volts/Current so r = 240/0.167
= 1440 ohms

Answer 10

RE:
resistance of a light bulb?
Please remind me how to work out the resistance (or impedance if you wish) given say 40w and 240v?