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Differences in formulas (Ohm's law, power)

  1. Sep 9, 2014 #1
    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.
     
  2. jcsd
  3. Sep 9, 2014 #2
    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?
     
  4. Sep 9, 2014 #3
    Thank you for the reply, especially for:
    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: Sep 9, 2014
  5. Sep 9, 2014 #4
    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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