# Calculate New Power when adding a Resistor

• Apaullo13
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.

#### 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?

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?

Try calculating the battery voltage in the first calculation.

This will stay constant for the second calculation.

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!