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## Homework Statement

A resistor R

_{1}= 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 R

_{2}= 20Ω is connected in series with R

_{1}, what is the total rate at which electrical energy is dissipated by the two resistors?

## Homework Equations

P=VI=I

^{2}R, V=IR

## The Attempt at a Solution

I first solved for the current I by using the equation P=I

^{2}R using my R

_{1}. 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=I

^{2}R 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?