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Homework Help: Two conventions

  1. Nov 2, 2006 #1
    "explain the difference between 'conventional current' and 'electron flow'. why were two conventions developed?

    my answer: 'Conventional current' describes the current flowing out of the positive terminal into the negative terminal. 'Electron flow' just describes the net movement of the negative charge

    im not sure what the second part of the question is about. what do they mean by 'conventions?'

  2. jcsd
  3. Nov 2, 2006 #2
    You should familiarize yourself with Google. This was on page 1.

  4. Nov 2, 2006 #3


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    Staff: Mentor

    I didn't follow the google link, but physicsgal, the "conventional" positive current flow is just in the opposite direction from the electron flow. That's all they're getting at.
  5. Nov 2, 2006 #4
    When Benjamin Franklin was first studying electricity, he discovered there were two types of charge that an object could have. He decided to call them positive and negative. (We now know that a positive charge comes from a deficiency in electrons and a negative charge comes from an excess of electrons, which is why we call electrons negatively charged and protons positively charge).

    Franklin then stated that it was the positive charge that moves in an electric current. Many years later, subatomic particles (protons, neutrons, and electrons) were discovered and scientists realized that it was actually the much smaller electrons that moved causing an electric current.

    Thus, the "conventional method" is Franklin's thinking: positive to negative. In reality we know that the reverse is true: electrons flow from negative to positive.
  6. Nov 2, 2006 #5
    thanks for the tips :smile: it makes sense now.

    here's what i wrote.

  7. Nov 2, 2006 #6
    another quick question:

    "a tv set has a rater power of 60.0W and is plugged into a 110 V household outlet"
    a) what is the value of the current flowing through the tv?

    I = P/V
    = 60W/110V
    = 0.54545 Amp

    b) How much energy does the tv use in 1 hour?
    so 1 hr = 3600 seconds
    Q = It
    = (0.54545amps)(3600s)
    = 1963.65C

    E = QV
    = (1963.65)(110V)
    = 216,001.5J

    is that pretty much accurate?

  8. Nov 2, 2006 #7


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    Staff: Mentor

    Looks correct, although I would do the 2nd one just as power * time directly. BTW, keep in mind that AC mains voltage is given in RMS, so that's why you can just multiply it by the RMS current (which is what you are given). Of course, that assumes that the voltage and current are in phase (negligible reactance in the load), which is a simplification.
  9. Nov 2, 2006 #8
    60W * 3600s = 216,000. thanks!

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