Calculate New Power when adding a Resistor

  • Thread starter Apaullo13
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  • #1
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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?
 

Answers and Replies

  • #2
vk6kro
Science Advisor
4,081
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Try calculating the battery voltage in the first calculation.

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