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Also, the same 100w lamp in a 220 v circuit produces how much current?

How did you get your answers?

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- Thread starter PiRsq
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- #1

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Also, the same 100w lamp in a 220 v circuit produces how much current?

How did you get your answers?

- #2

BoulderHead

Howdy, PiRsq!Originally posted by PiRsq

Also, the same 100w lamp in a 220 v circuit produces how much current?

How did you get your answers?

First, I’d recommend dropping your use of the word “produce” because it doesn’t accurately depict what is going on. The production process likely takes place at the local utility company, rather than inside the bulb.

Ok, if you are aware that (current)(voltage)=power, then all you need to do is substitute in the two values you were given and solve for the unknown. Are you familiar with using mathematical manipulations to isolate an unknown variable?

Hint; dividing both sides of that equation by voltage would leave you with:

current = power/voltage

Incidentally, those bulbs are likely to have TWO markings on them; wattage

Does that help?

- #3

Chi Meson

Science Advisor

Homework Helper

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It's not voltage (that's controlled by the outsdie power source)

It's not current (that's proportional to the voltage)

It's not power (That's proportional to current and voltage)

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- #5

BoulderHead

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Im sorry but im still a little confused. Why is it not according to P=EI, 100=220i, where I=100/220?

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BoulderHead

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- #9

BoulderHead

Try this for lamp 1;When I think about it, it makes sense to double current when doubling voltage. But mathematically, I can't seem to solve it.

100W/110V=0.91A (you have already done this part, of course)

Now if you were looking at the equation above and thinking “..Yeah, but if I replace the 110V with 220V I will halve the value of I instead of doubling it.”, then you were thinking wrong because the 100W is

So for lamp number one you determine the current using I=P/E, then you determine the resistance using R=E/I, and wahlah, you now have something to take into the second problem and solve for current using I=E/R.

If you had been taking a test, running out of time, and only needed to write down the value of I for the second problem in order to get the points, then it might have been possible to forgo solving for R. You could have just understood that the current of problem 1 would be doubled and quickly answered the problem, thus being able to move on to the next one…

…but, methinks it better to do the long way until you get the hang of it.

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