(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A heater is plugged into the same 120-V AC outlet as an 800-W hair dryer. If the total rms current drawn is 16.7A, then calculate the power rating of the heater.

2. Relevant equations

P_{av}=[itex]\frac{1}{2}[/itex]I^{2}_{peak}R

3. The attempt at a solution

I_{rms}=I_{peak}/[itex]\sqrt{2}[/itex]

So, I_{peak}=16.7A x [itex]\sqrt{2}[/itex]

Solving for R in the P_{av}equation: 2P_{av}/I^{2}_{peak}=R

(2 x 800W)/(16.7A x [itex]\sqrt{2}[/itex])^{2}= 2.87 ohms.

I'm assuming that in an outlet, the heater and the dryer are in parallel, so V=V_{1}=V_{2}and I_{total}=I_{1}+I_{2}

Using ohms law V/R=I, 120V/2.86 ohms= 41.81 A. Since this is too high, I know I'm looking at this problem completely wrong. I wanted to use that to find I_{2}, solve for R, and find the power of the heater.

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# Homework Help: Power rating of a heater in the same outlet as a hair dryer

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