1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measuring units

  1. Feb 14, 2007 #1

    I have a current through a simple circuit.
    What is the measuring unit of Log(I) where I is the intensity of the current measured in Ampers.

  2. jcsd
  3. Feb 14, 2007 #2


    User Avatar

    Staff: Mentor

    Well, it would usually be in some form of dB, where you are ratioing the measured I compared to some reference current (like 1uA or 1mA), or compared to some reference power if you know the impedance that the current is flowing through (like 1mW). Here are some typical examples:

    I in dBuA = 20 * log( I / 1uA)

    I in dbmA = 20 * log( I / 1mA )

    Power from I through 50 Ohms in dBm = 10 * log( I^2 * 50 / 1mW )

    If instead you are just plotting I on one axis of a graph and using a logarithmic axis, then you still label the units for that axis as Amps (or mA or whatever is appropriate), and the numbers you put on the decades of that axis just run like 1, 10, 100, etc.

    Does any of that answer your question?
  4. Feb 15, 2007 #3
    Not quite.

    The dB is out. I don't want to use the representation in dB. And as far as the dB is concerned I think it has no physical dimension:
    Log(Amps/Amps)=no phys units...now look at the problem below.

    I don't think that using the same units for I as for log(I) is correct. I want to represent for example Log(I)=f(U).

    for example,
    Because if I use the square function I would have the units Amps^2...The Log(I) should have the units of Log(Amps)?!...

    This is the place where maths meets physics:smile: What does a mathematical function to a physical unit?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook