Which resistor produces the most heat?

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Discussion Overview

The discussion revolves around identifying which resistor generates the most heat in a given circuit. Participants explore various methods of analysis, including the principle of superposition, circuit analysis techniques, and the application of power formulas. The conversation includes both theoretical considerations and practical implications of resistor behavior in circuits.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the power output of a 5 ohm resistor to be 48.05W, suggesting it produces the most heat based on their analysis.
  • Another participant expresses skepticism about the high current through the 5 ohm resistor but agrees it generates the most heat.
  • Some participants discuss alternative methods of analysis, such as loop analysis, to find voltages and currents with both sources active.
  • There is mention of using power formulas like P=V^2/R or P=I^2R, suggesting that calculations may not be necessary in some cases.
  • One participant points out that the behavior of ideal voltage and current sources affects the analysis of resistors in the circuit.
  • Another participant raises a question about the minimum calculations needed to determine which resistor generates the most heat, considering the arrangement of resistors in series and parallel.
  • There is a discussion about the conditions under which resistors in parallel or series will generate more heat, with some participants suggesting that the analysis could be more complex than initially thought.

Areas of Agreement / Disagreement

Participants generally agree that the 5 ohm resistor produces the most heat based on the calculations presented, but there is no consensus on the methods used to arrive at that conclusion. Multiple competing views on analysis techniques and the implications of circuit configurations remain unresolved.

Contextual Notes

Some participants express confusion regarding the application of circuit analysis techniques, indicating that their understanding is still developing. There are also references to specific conditions under which certain resistors may generate more heat, but these conditions are not fully resolved or agreed upon.

  • #31
CWatters said:
If you mark up the drawing correctly and apply KVL you should get an equation like this..

+15 + 10 + 10 + ? = 0

where ? is the voltage across the current source. It's not going to be -15V.

PS...

KVL says the voltages around a loop sum to zero. I find I make far fewer errors if I always write an equation that explicitly adds up to zero (eg V1+V2+V3+V4 = 0) and avoid trying to guess the polarity and write something like V1 + V2 + V3 = -V4

:oops: Thx, I get it now. So e.g. going round the outside circuit from bottom left corner potentials are successively 0, -15, +20, +10. Over the current source current is in the direction of voltage increase not decrease.

But no one to say the resistance there is -35 Ω - if voltage across it changes, current is still 1 A. A non-ohmic element. (Nohmic? Gnomic?)

It just needs dropping a few mental habits leaned in circuits with constant voltage supplies.
 
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  • #32
epenguin said:
Thx, I get it now. So e.g. going round the outside circuit from bottom left corner potentials are successively 0, -15, +20, +10. Over the current source current is in the direction of voltage increase not decrease.

That's not what I get at all.

By bottom left corner I take it you mean the junction between the 20R and the 15R ?

You didn't say which way around you are going either and that changes the sign of all the voltages.

If you go clockwise you get +10 +10 -? +15 = 0 where ? is the voltage across the current source. Solve to give ? = -35V. As you are going clockwise that means the top end of the current source is +ve wrt the bottom.

If you go anti clockwise you get.. -15 + ? -10 -10 = 0. Solve to give ? = +35V. As you are going anti clockwise that means the top end of the current source is +ve wrt the bottom which is the same thing.

Power.jpg
 
  • #33
We are in substatial agreement now, I was going anti-clockwise just as far as the junction where the voltage is + 10 V same as the potential from battery, I.e. two routes to same point where you do a circuit in one direction, perhaps preferable but equivalent.
 

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