Calculate New Power when adding a Resistor

[Apaullo13]

As this is my first post please feel free to correct me if I am posting this is the wrong section or on improper etiquette in general. I'm having a bit of trouble with what should be a pretty simple problem.

Homework Statement

A resistor R₁ = 23Ω is connected to a battery that has negligible internal resistance and electrical energy is dissipated by at a rate of 44W.

If a second resistor R₂ = 20Ω is connected in series with R₁ , what is the total rate at which electrical energy is dissipated by the two resistors?

Homework Equations

P=VI=I²R, V=IR

The Attempt at a Solution

I first solved for the current I by using the equation P=I²R using my R₁. I found the current to be I=1.38A.

From there I found the equivalent resistance by adding my resistances as they are in series to get an equivalent resistance of 43Ω.

I then used the equation P=I²R plugging in the 1.38A and 43Ω.
This gave me a new power of 82W but the site says this is not correct. Any suggestions?

 

[vk6kro]

Try calculating the battery voltage in the first calculation.

This will stay constant for the second calculation.

 

[Apaullo13]

Thanks! I got it! I had it in my head that the current would be what remained constant since it was in series but wasn't quite sure about that. Thanks again!