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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.

Thanks

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- Thread starter cristian1500
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- #1

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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.

Thanks

- #2

berkeman

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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?

- #3

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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 What does a mathematical function to a physical unit?

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