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

[OmegaFury]

Homework Statement

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.

Homework Equations

P_av=\frac{1}{2}I²_peakR

The Attempt at a Solution

I_rms=I_peak/\sqrt{2}
So, I_peak=16.7A x \sqrt{2}
Solving for R in the P_av equation: 2P_av/I²_peak=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 I_total=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.

 

[gneill]

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)?

 

[OmegaFury]

Okay, so I use P_av=ε_rmsI_rms
P_av/ε_rms=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?

 

[gneill]

Okay, so I use P_av=ε_rmsI_rms
P_av/ε_rms=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?

It looks good

 

[OmegaFury]

Thanks for the help