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What exactly is Joule's law of heating?

  1. Oct 9, 2016 #1
    Hi everyone,

    I'm struggling to understand what I think is a very basic concept: Joule's law of heating. Allow me to explain my confusion:

    We know that voltage can be expressed as:

    V = I · R

    And power can be expressed as:

    P = I · V

    Making power equivalent to:

    P = I2 · R

    According to my textbook, Joule's law of heating can be written as:

    E = I2 · R · t

    Where E is the amount of electrical energy that gets converted to thermal energy. What confuses me is that – since this equation is equal to P · t – it is equal to simply measuring the total amount of electrical energy. To me, this says that it is always the total amount of electrical energy that gets converted to thermal energy, and that doesn't really make sense to me. I would think that part of the electrical energy dissipates as heat; not all of it!

    If you are reading this and think you might be able to help, I would be very grateful!
  2. jcsd
  3. Oct 9, 2016 #2

    Stephen Tashi

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    Are you thinking of an example such as a battery powering both a heating element and an electric motor in series ?

    The "V" in the equation refers to the voltage drop across the heating element, not the total voltage drop across both the heating element and the motor. So not all the electrical energy is dissipated as heat, but all the electrical energy that causes the voltage drop across the heating element is converted to heat - or radiant energy etc.
  4. Oct 9, 2016 #3
    Hey Stephen – thanks for getting back to me!

    I am pretty new to this stuff, so the different concepts (especially voltage) are still a little blurry to me. The example I am thinking of is simply passing a current through a resistor that – to some extent – converts the electrical energy to heat. Are you saying that the voltage implicit in I2 · R · t is actually the voltage drop between point A and point B (point A being before the resistor, and point B being after the resistor). I sense that this is an extremely basic question, but once this is clear to me, I think I am much better suited to proceed with my studies.

    Thanks so much for your help so far!
  5. Oct 9, 2016 #4

    Stephen Tashi

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    Yes, that is correct.
  6. Oct 9, 2016 #5
    I think your hangup is that you are wondering how we represent useful work in a circuit diagram.

    Let's look at a simplified circuit for a series wound DC motor


    The source on the right represents the back EMF of the motor. The current of this circuit is easy enough to calculate (12-11.5)/2= 250mA. Notice that the current flows out of the cathode V1 but into the cathode V2. Power is still (P=VI). So, V1 is emitting 3W, and V2 is absorbing 2.875W which is being converted to mechanical work.

    This is just one way that we can represent a "sink" for electrical energy. In this case the back EMF voltage source is a real thing. It comes from the physics of a motor.
  7. Oct 9, 2016 #6


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    A resistor does indeed convert ALL of the power dissipated in it to heat.

    I suspect your confusion might be down to the difference between...

    a) Power dissipated in a resistor
    b) Power transmitted through a resistor to a load.

    Power Dissipated.png

    The power dissipated in the resistor = I*VR

    The power transmitted through the resistor to the load = I*VL


    The voltage drop across the resistor is..
    VR = VS-VL = I*R
  8. Oct 9, 2016 #7

    jim hardy

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    remember equivalence of heat and mechanical work(force x distance), both are measures of energy: 1 BTU = 778 ft-lbs.
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