# Wattage question

#### PiRsq

A 100w lamp in a 110 v circuit produce how much current?

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

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Originally posted by PiRsq
A 100w lamp in a 110 v circuit produce how much current?

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

Howdy, PiRsq!

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 and voltage. The wattage is rated for a given voltage level, so if your 100-watt lamp was intended for 110v and you screw it into a 220v circuit, well, it will probably burn out rather quickly. My guess is that you aren’t expected to worry about that at this time, but simply take it on good faith that the bulbs are rated for the voltages being applied.

Does that help?

#### Chi Meson

Homework Helper
One thing to always remember about lightbulbs in every problem in elemtary physics: There is only one factor that remains constant for that bulb (assuming the temperature is not changing).

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)

#### PiRsq

Ok so I get the value of 0.91 for current when a 100 watt bulb is plugged into a 110v circuit. And I got 0.45 for a 100 watt bulb plugged into a 220 v circuit. But in this book im doing the problems from, it says when 100 watt bulb is plugged into 220v circuit, I should get 1.82 amperes which is double of what I got. I did P=EI so I=P/E. Am I mistaken somewhere?

Well, then the catch would be that the problem said you’d be using the same bulb in the 220 volt circuit. In that case they are determining the internal resistance of the bulb using the 110v example then dividing 220v by that value to solve for current in the second problem. So after solving for current in the first example, divide 110v once more by .91a and you’ll have that constant which Chi Meson mentioned, and dividing the higher voltage by that constant should give you the 1.82.

#### PiRsq

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

That power draw of 100 watts came from multiplying the value of current passing through the lamp by the value of the voltage across its terminals. What that means is the value of 100 watts can only be true for one of the line voltages, not both of them. Part of your job might be to figure out which voltage the 100W applied to, but careful reading of the wording of the problem suggests 100W is the power consumption when using the 110v line, not the 220v. So, this means you can’t just divide 100w by 220v to solve for the current in the second problem (although you could simply double the current from the first problem and have your answer). There is also a constant value the lamp possesses which is independent of the line voltage. That constant is the fixed resistance internal to the lamp. Solving for that gives you another way to determine the current in the 220v circuit.

#### PiRsq

When I think about it, it makes sense to double current when doubling voltage. But mathematically, I can't seem to solve it. So I have to deal with some other equations before I get to the P=EI equation right?

When I think about it, it makes sense to double current when doubling voltage. But mathematically, I can't seem to solve it.
Try this for lamp 1;
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 not applicable to the 220V circuit. The reason it makes sense for the current to double is because we know there is some value possessed by the lamp that remains fixed. That value is the internal resistance of the lamp (R) and the formula is: I=E/R. It is easy to see that if E doubles then R will divide into it twice as many times, giving you twice as large a value of I.
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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#### PiRsq

Thank you so much BoulderHead and Chi Meson, btw, is electricity any interesting because I want to study into creating Zero Point Energy and stuff and I was wondering if electricity would help. What other subjects should I look into?

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