Differences in formulas (Ohm's law, power)

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Discussion Overview

The discussion revolves around the calculations related to Ohm's law and power in an electrical circuit, specifically examining the relationships between voltage, current, resistance, and power for a single-phase electrical appliance. Participants explore different methods to calculate operational current and the implications of the given parameters.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes three methods to calculate operational current "I" using voltage "U," resistance "R," and nominal power "N," but questions why the results differ.
  • Another participant argues that the parameters provided are inconsistent, stating that 230V across 26 ohms cannot dissipate 2600W, suggesting a recalculation of power using the formula P = V^2/R.
  • A third participant expresses confusion about the relationship between resistance and power dissipation, asking if knowing voltage and nominal power is sufficient to calculate current.
  • A later reply asserts that knowing voltage and power is indeed enough to calculate current and resistance, emphasizing that the relationships are governed by two fundamental equations: P = I*V and V = I*R.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the initial parameters and calculations. While some assert that the formulas should yield consistent results, others highlight the contradictions in the provided values, leading to unresolved questions about the relationships between the variables.

Contextual Notes

The discussion reveals limitations in the assumptions made about the electrical appliance's specifications, particularly regarding the relationship between voltage, resistance, and power. There is also an indication of potential misunderstandings about the physical principles involved.

Who May Find This Useful

This discussion may be useful for students and individuals interested in understanding the relationships between voltage, current, resistance, and power in electrical circuits, as well as those exploring the implications of conflicting parameters in theoretical scenarios.

nuuskur
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For the sake of simplicity, assume we have an electrical appliance, single phase.
Powered by U = 230V from the wall, its nominalpower N = 2600W and the resistance R = 26 ohm.

If I wanted to calculate its operational current "I" (not entirely sure about the English terminology - the amount of the current's intensity in the circuit), I could go about it in 3 ways:
1) I = U/R = 230/26 = X amps
2)N = UI => I = N/U = 2600/230 = Y amps
3)U = RI, N = UI => N = I2R => I =√[N/R] = √[2600/26] = Z amps

Seemingly all of these Should give us the correct answer, but why do they differ and which would be the correct one? It's not homework, just curious.
 
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But you have just made things up... right? I could say I am taller than you, you are taller than mike and mike is taller than me. That is inconsistent, its just made up and not realistic. That is the problem with making up problems, they are harder to be realistic than students often realize.

The answer is that 230V dropped across 26 Ohms will not dissipate 2600W. With V and R known the power is V^2/R = (230^2)/26 ~ 2035W. You have over specified the situation with contradictory information (Like I did with the height situation.) Does that make sense?
 
Thank you for the reply, especially for:
ModusPwnd said:
The answer is that 230V dropped across 26 Ohms will not dissipate 2600W. With V and R known the power is V^2/R = (230^2)/26 ~ 2035W.
I'm a math student, clearly wanting to understand physics the wrong way. The quote explains something, but now I have more questions - does the resistance Have to dissipate, with 230V across it, so-and-so watts? Is there any specific material you would suggest in regards to this topic?

If I had the appliance's nominalpower and the 230V from the wall as known values, would that be enough to calculate I? If we observed an actual electrical motor, for example, with all the values written in a brochure, could we calculate I and arrive at the exact same results with the three formulas I wrote earlier?

Thanks in advance
 
Last edited:
Yes, it's enough to know the voltage and the power to calculate both current and resistance.
And you will get the same values no matter in what order you do it or with which formulas.
There are basically two formulas:
P=I*V
and
V=I*R (Ohm's law).
and 4 variables.

If you know 2 of them the other two are completely determined by the two independent equations. It is a math problem, isn't it? :)
 

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