Compute Joule Heat in Heating Coils

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SUMMARY

The discussion focuses on calculating Joule heat generated in two heating coils with resistances of 12.0 ohms and 6.0 ohms, connected in both parallel and series configurations to a 115V source. For the parallel connection, the effective resistance is calculated as 4.0 ohms, resulting in Joule heats of 9953W and 4976W for the respective coils. In the series connection, the effective resistance is 18 ohms, yielding Joule heats of 490W and 245W. The use of the formula P = V²/R for calculating power is emphasized as a more straightforward approach.

PREREQUISITES
  • Understanding of electrical resistance and Ohm's Law
  • Familiarity with Joule's Law for calculating power
  • Knowledge of series and parallel resistor configurations
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the application of Joule's Law in different circuit configurations
  • Learn about effective resistance calculations in parallel and series circuits
  • Explore advanced power calculation techniques using P = V²/R
  • Investigate the thermal effects of electrical components in practical applications
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and anyone interested in understanding the principles of electrical heating and power calculations in circuits.

BillJ3986
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Homework Statement


Two heating coils have resistances 12.0 ohms and 6.0 ohms, respectively.
a)What is the Joule heat generated in each if they are connected in parallel to a source of emf of 115V?
b) What if they are connected in series.

Homework Equations


Am I doing this question properly?

The Attempt at a Solution


Since the resistors are in parallel. In order to find R effective I use the formula 1/12.0 +1/6.0=1/4 so my R effective is 4.0 ohms. I then needed to find the current, I. So I used I=Emf/R, I=115/4.0=28.8A, I think that is a little high.I use P=I^2(R), P=28.8^2(12)=9953W to find the joule heating in resistor 1? and then I did the same for the second resistor in parallel. P=28.8^2(6.0)=4976W.
For part B I had to find the current so I did I=Emf/r+r and they are in series so I just added them up to 18 ohms for my R effective. I found I, using I=115V/18ohms= 6.39A. P=6.39^2(12)=490W, second resistor p=6.39^2(6)=245W
 
Last edited:
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Welcome to PF!

Hi BillJ3986! Welcome to PF! :wink:

Yes, that's all correct :smile:

but … why did you bother with finding the current in part a) ? :confused:

you know the voltage across each resistor, so just use the formula P = V2/R :wink:
 
Thank you,

Yeah I'm looking over my notes now and I see that formula. I'm Glad I found this place, hopefully I can get my physics grade up a little higher. Thanks again for your input
 
Last edited:

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