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

## 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

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

## 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.

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gneill
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)?

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?

gneill
Mentor
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?
It looks good

Thanks for the help