# Calculate New Power when adding a Resistor

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?

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?