Question for Speedy-Petey

[Speedy Petey]

Although I used a different program for that calculation, I have LectriCalc™ for my Palm T/X PDA and Treo phone.

Really cool program for quick and accurate calculations, and you get a version for your Window's computer as well.

Yes, I know how to do all these things by hand, but WHY bother!??

[596]

No Push.

60w@ 240v = 30w@120v

I over though it and confused myself over something extremely simple.

Sorry to break in here, but this discussion has gone off the track! The above quoted statement is very very wrong.

The mistaken assumption in this discussion is that the POWER number 60 watts was a constant, or that power varied directly as volts. BOTH of these assumptions are wrong.

Power varies as the SQUARE of voltage. The ONE constant in this picture is the RESISTANCE of the bulb. SO, we first calculate that using the known starting spec of 60 watts at 240 volts. This gives us a bulb resistance of 960 ohms. Now, if we drop the voltage to 120 volts, the amps = 120 divided by 960, or 0.125 amps. The power can be calculated from I x E, which gives us 15 watts.

[Speedy Petey]

That was my initial impression, that the resistance was the same and I come up with the same 15 watts using the same formula.
I will be the first to admit I am not best at theory, but this is pretty basic stuff and I thought I had a handle on it. I hate looking the fool but in this thread it's too late.

[pushkins]

Soooooooo, here I was confused, then unconfused and now confused again.

So a 60 watt bulb @240 Volts would give 60 watts of brightness but if the power was dropped down to 120 volts on the same bulb it would glow at 15 watts ?

[normel]

Power varies as the SQUARE of voltage. The ONE constant in this picture is the RESISTANCE of the bulb. SO, we first calculate that using the known starting spec of 60 watts at 240 volts. This gives us a bulb resistance of 960 ohms. Now, if we drop the voltage to 120 volts, the amps = 120 divided by 960, or 0.125 amps. The power can be calculated from I x E, which gives us 15 watts.

Actually the power varies as the square of the current, not the voltage (P=I^2R), but 15W is the correct answer.

[596]

P=I x E

P=E² ÷ R

P=I² x R

In this example, the known, fixed, value was the R. And based on the given spec that it was a 60 watt, 240 volt bulb, we know its value is 960 ohms.
Then, we mutilied the voltage by ½, so power is multiplied by (½)² , which is ¼ . The resulting current at any point in this exercise was the result of the fixed resistance and the GIVEN voltage.