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I need help understanding the formula Power loss= I^2 R

  1. Nov 19, 2008 #1
    The power loss is given by I ^2 R where I is the current and R is the resitance.

    I know why these variables are in this equation.
    V = I R
    P = I V

    P = I² R

    What I don't understand is the practical explanations for this. Why are there two current variables? Are there two currents in the wire? If so then how come?

    Thanks for all the efforts in advance
  2. jcsd
  3. Nov 19, 2008 #2


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    Good question!

    There aren't two currents, what the equation means is that if you double the current you end up with 4 times the power loss. It's like the area of carpet you need for a room - if you make the room twice as long and twice as wide you need 4x as much carpet.

    The physical explanation is that the voltage difference along a wire depends on the current - more current flowing with a resistance means more voltage (pressure of electricity if you like) is built up. This extra voltage means more power.
    So if you double the current your would double the power, but you also double the voltage which doubles the power agian = 4x as much power.
  4. Nov 19, 2008 #3


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    The practical explanation for I^2R, is that the power (measured in Watts) is equal to the single variable of current "I", (which becomes squared in this equation, as you have confirmed mathematically) times the resistance (R) along that path of a circuit. So as long as you keep the resistance small, you can minimize the power consumed or lost as heat.

    So, if your resistance becomes large (for example by longer paths between components in a circuit), you consume more power. Inside an IC (integrated circuit), the lead lengths between components are very small, compared with an analogous circuit using discrete components. So a circuit using ICs consume less power (and powered using smaller batteries).
    Last edited: Nov 20, 2008
  5. Nov 19, 2008 #4


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    Yes this is why you want thicker wires in longer extension cords.

    Lower R means less wasted power.
  6. Nov 19, 2008 #5
    But if more current flowing then it means more voltage is built up then why can't we double the ratio between the current and the resistance so it will be I^4 R instead?
  7. Nov 19, 2008 #6
    Here I offer you a full explanation, let's begin with its first principles.

    Current is defined as the rate of flow of charge,

    I = Q/t

    Rearranging this and obtaining:
    Q = It

    Subsequently, the potential difference across any 2 points in a circuit is the work done on the charge, per unit coloumb.

    Work = VQ

    Power is the rate of doing work,

    Power = VQ/t

    Since Q = It,

    P = VI

    Resistance is defined as the ratio between the potential difference across any 2 points of the circuit and the current flowing through them. Hence,

    R = V/I

    Rearranging this gives us:

    V = IR


    P = VI = (IR)I = I2R
  8. Nov 20, 2008 #7


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    The OP already follows the math (see original post), the question was to offer practical discussion to gain insight into the mathematical representation.
  9. Nov 20, 2008 #8
    I am not so sure about this.
    I am afraid he is confused about the meaning of the square in the I^2.
    Many high school students have this problem, unfortunately.
    Sorry if I am wrong, but this is my feeling from the way the questions are formulated.
  10. May 8, 2011 #9


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    I am a high school student and this is how I understand I2R power loss:

    So you know Plost=I*(delta)V. I say (delta)V because it is helpful to remember that the energy lost is potential energy in the form of voltage. If you took the voltage of a wire transmitting power from point A to point B, its voltage at A would be higher than at B because some potential energy is dissipated in the form of heat because the wire has some resistance. You also know that V=IR; this is Ohm's law. However, when you apply Ohm's law you are also assuming that the resistance R is constant and that voltage is solely a function of current. V(I)=IR. y=mx where m, the slope, is constant R. Thus when you graph V(I)=IR, if Ohm's law applies you get a nice straight line with a constant slope of R. If the slope of such a graph changes (is curvy) then Olm's law does not apply. Now lets go back to our equation Plost=IV. If Olm's law is valid and R is constant then we know Voltage is a function of current so Plost(I)=V(I)*I. As you can see we have just expressed power lost completely as a function of current. Thus when Olm's law applies power loss is dependent on current. When we actually finish the substitution we get Plost=I2R.
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