Which resistor produces the most heat?

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The forum discussion centers on determining which resistor generates the most heat in a circuit analysis problem. The conclusion reached is that the 5 ohm resistor dissipates the most power, calculated using the formula Power = I^2R, resulting in a maximum output of 48.05W. Participants debated the necessity of circuit analysis techniques, with some asserting that the answer is evident without extensive calculations. The discussion highlights the importance of understanding the principles of superposition, voltage division, and current division in electrical circuits.

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  • Understanding of electrical circuit analysis techniques, including KCL and KVL.
  • Familiarity with the principle of superposition in circuit analysis.
  • Knowledge of power calculations using the formula Power = I^2R.
  • Basic concepts of series and parallel resistor configurations.
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  • Study the application of the principle of superposition in more complex circuits.
  • Learn about voltage and current division techniques in circuit analysis.
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Students and professionals in electrical engineering, circuit designers, and anyone interested in understanding heat generation in resistive components within electrical circuits.

  • #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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