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Measure of Volts / Amps

  1. Apr 19, 2005 #1

    I was reading Asimov on Physics the other day, and found that a watt is a measure of gm cm^2/s^3. That got me thinking, if voltage times current equals watts, what would voltage and current be a measure of? I have been trying to figure it out for a while, and came up with: current= gm/s and voltage= cm^2/s^2. That is the only way I could make current work, but voltage just doesn't seem right to me.
  2. jcsd
  3. Apr 20, 2005 #2


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    Watts can be expressed in various systems e.g. cgs (cm gm sec), mks (m kg sec). To convert Watts to cgs you may want look at a table of conversions in physics or engineering texts
    or a verifiable online ref... for example ---> http://users.aol.com/tspquinn/units.html [Broken]

    1W = (1 J/s)
    1 J = 10^7 ergs
    1 erg = 1 dyne cm
    1 dyne = 1 gm cm/s^2

    So now calculate watts in cgs.

    1W = (1 J/s)*(10^7erg/J)*(1dyne cm/erg)*[(1 gm cm/s^2)/dyne]
    cancel J, erg,dyne... leaving you 1W (cgs) = 10^7 gm cm^2/s^3

    Which makes sense because converting this to mks
    (10^7 gm cm^2/s^3)(1m/100cm)(1m/100cm)(1kg/1000gm) =
    (10^7 gm cm^2/s^3)[1m^2 kg/(10^4 cm^2 10^3 gm)] =
    cancel gm, cm^2 ...
    10^7 kg m^2/ ((10^4*10^3) s^3) so 1W (mks) = 1 kg m^2 /s^3

    So cgs and mks systems differ by constant factor 10^7,
    if i choose to stay in mks (table including voltage an current)
    e.g. ref ---> http://www.csee.umbc.edu/help/theory/units.shtml [Broken]

    I can go through a similar derivation for voltage and current but you get the idea, to save time I obtain the following:

    1V = 1 (kg m^2)/( s^2 c)
    1A = 1 c/s

    1W(mks) = (kg m^2 c)/ (s^2 s c) , cancel c (c - Coulombs)
    = 1 kg m^2/ s^3
    Last edited by a moderator: May 2, 2017
  4. May 6, 2005 #3
    Thanks, but I still have one question. How can amperage - measured by the number of electrons that go by a certain point in a certain amount of time - be measured using cm/s? With mass I could see... 1.24 quintillion electrons weigh a certain amount, but how do 1.24 quintillion electrons have a length?
  5. May 6, 2005 #4


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    I didn't read the preceding post, but there is no way that amperes can be expressed in cm/s -- those are units of velocity! In fact, the ampere is fundamental unit in the SI system, just like the metre, kilogram, and second.


    It is not a derived unit, so it cannot be expressed in terms of these other three. Current was a physical quantity that demanded a new unit of measure. I won't try to state the official def'n of the ampere, because I'll get it wrong. You're better off looking it up. I think it has something to do with "the current required to produce _______ amount of force between two infinitely long, parallel, current - carrying wires." The ampere has, in turn, been used to define the coulomb (the amount of charge passing a point in one second when there is 1 A of current). I'm not sure whether current has a different unit in the cgs system, which you seem to be using, but I don't think so. I remember reading also that there is no British system of electrical units. Thankfully.

    One volt is one joule per coulomb (J/C), which makes sense because voltage is actually potential difference. E.g. the potential difference between two terminals can be interpreted as the potential energy per unit charge gained or lost as charge flows between the terminals. So it makes sense that the voltage drop across a resistor for example, times the current flowing through that resistor, gives you the power dissipated. For you have joules lost per coulomb times coulombs going by per second = joules lost per second.
    Mathematically: J/C * C/s = J/s = W

    But you wanted to see everything in base units:

    [tex] J = \frac{kg\cdot m^2}{s^2} [/tex]

    [tex] V = \frac{J}{C} = \frac{J}{A\cdot s} = \frac{kg\cdot m^2}{A\cdot s^3} [/tex]

    [tex] W = V\cdot A = \frac{kg\cdot m^2}{s^3} [/tex]
  6. May 6, 2005 #5
    You misread. The units given were c/s not CM. An amp is a coulomb/second or C/s.
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