Calculate New Power when adding a Resistor

In summary, the conversation discusses a problem involving a resistor connected to a battery and the rate at which electrical energy is dissipated. The problem is solved by first finding the current, then the equivalent resistance when a second resistor is added in series. The conversation also mentions the suggestion to calculate the battery voltage and the realization that the current remains constant in series.
  • #1
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


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?
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  • #2
Try calculating the battery voltage in the first calculation.

This will stay constant for the second calculation.
  • #3
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!

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