Power, Resistance, Voltage, Current

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Discussion Overview

The discussion centers around the relationship between power, resistance, voltage, and current in electrical circuits, specifically using a tungsten filament lamp as an example. Participants explore the implications of different scenarios involving these variables, addressing potential confusion arising from the application of formulas in varying contexts.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant calculates power using the formula P=V^2/R for a tungsten filament lamp with a specified resistance and voltage, yielding a high power value that seems unrealistic.
  • Another participant clarifies that the two scenarios presented are fundamentally different, as they involve different known variables (voltage vs. power) and lead to different results.
  • A third participant suggests a revised example where the resistance and voltage yield consistent results, emphasizing the importance of using compatible values across calculations.
  • One participant notes that the resistance of a filament changes with temperature, which could affect calculations if not considered.
  • A later reply points out that in the second scenario, the voltage is not the same as in the first, further highlighting the differences between the two cases.

Areas of Agreement / Disagreement

Participants generally agree that the two problems are different and that using consistent values is crucial for accurate calculations. However, there remains some uncertainty regarding the interpretation of the resistance value and its implications in the context of the problems presented.

Contextual Notes

Limitations include the assumption that the resistance value given in the textbook applies under specific conditions, and the potential for confusion arising from different operational states of the filament (cold vs. hot).

skepticwulf
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Suppose I have a tungsten filament lamp with a resistance 0,32 ohm
If it operates on 220V line, its power is P=V^2/R=151250W !
However, If the problem is given like , "you have a 100W with resistance 0,32 ohm what's the current? that'd come up as
I=Sqroot(P/R)=17,6A
First result doesn't make sense, or doesn't look "realistic" but the second one does. But they both stem from the same formula: P=IV
Why does it seem to me that these two formulas does not reconcile with each other?
 
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They arent the same problems. In A you give both the voltage and resistance, finding the power. But in B you give the power and then find the current. The only variable the same between the two is the resistance. The voltage, current, and power are wildly different. Use the same values for all your variables and you'll find that the formulas work great.
 
+1 to what Drakkith said. They aren't the same problem. The data in the first implies something more like a searchlight than a regular light bulb.

Suppose I rewrite your OP so they are the same and more realistic...

I have a tungsten filament lamp with a resistance 484 ohms
If it operates on 220V line, its power is P=V^2/R=100W.

If the problem is given like, "you have a 100W with resistance 484 ohm what's the current? that'd come up as
I=Sqroot(P/R)= 0.454 A.

As a check... P = I*V = 0.454 * 220 = 100W

All consistent now.

Aside: If you are tempted to measure the resistance of a filament be aware that it changes with temperature. The resistance when the filament is cold will be somewhat different to when it's glowing white hot. The calculations above assume the hot value for the resistance.
 
Thank you but in my textbook's example the resistance of a tungsten lamb was given as 0.32 ohm.
Maybe that confused me and maybe I should have taken it like this: "For the resistance value of 0.32 ohm on a 220V line you will get a maximum power of 151250W.
But if you connect a 100W lamp to the line with 0.32 ohm resistance you will get 17.6A of current". That's the notion of being two problems "different" Am I correct?
 
Last edited:
In the second case the voltage is not 220 V but around 5.5 V. The two situations are different, even though the lamp may be the same.
 
Thank you :)
 

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