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

1. Apr 4, 2012

### OmegaFury

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
Pav=$\frac{1}{2}$I2peakR

3. The attempt at a solution
Irms=Ipeak/$\sqrt{2}$
So, Ipeak=16.7A x $\sqrt{2}$
Solving for R in the Pav equation: 2Pav/I2peak=R
(2 x 800W)/(16.7A x $\sqrt{2}$)2= 2.87 ohms.

I'm assuming that in an outlet, the heater and the dryer are in parallel, so V=V1=V2 and Itotal=I1+I2
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 I2, solve for R, and find the power of the heater.

Last edited: Apr 4, 2012
2. Apr 4, 2012

### Staff: Mentor

I think you can avoid the average and peak conversions and stick to rms values.

What's the rms current drawn by the hair dryer if it uses 800W at 120V (rms)?

3. Apr 4, 2012

### OmegaFury

Okay, so I use PavrmsIrms
Pavrms=800W/120V=6.67A. The rms current of the heater would be 16.7A-6.67A= 10.03A. Thus, the power rating of the heater would be 120V x 10.03A= 1203.6W. Is that correct?

4. Apr 4, 2012

### Staff: Mentor

It looks good

5. Apr 4, 2012

### OmegaFury

Thanks for the help