Calculate New Power when adding a Resistor

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SUMMARY

The discussion focuses on calculating the total power dissipation when adding a second resistor, R2 = 20Ω, in series with an initial resistor, R1 = 23Ω, which dissipates power at a rate of 44W. The user initially calculated the current using the formula P=I²R, resulting in I=1.38A, and then found the equivalent resistance to be 43Ω. However, the calculated power dissipation of 82W was incorrect. The correct approach involves first determining the battery voltage, which remains constant, to accurately calculate the new power dissipation.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with power calculations (P=VI and P=I²R)
  • Knowledge of series resistor configurations
  • Basic electrical circuit analysis skills
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  • Calculate battery voltage using the initial power and resistance values
  • Learn about series and parallel resistor combinations
  • Explore the implications of power dissipation in electrical circuits
  • Study the impact of resistor values on total circuit performance
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Students studying electrical engineering, hobbyists working on circuit design, and anyone interested in understanding power dissipation in series circuits.

Apaullo13
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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?
 
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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!
 

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