Squaring microamps

Discussion in 'Homework Help' started by ElectricMagician, Feb 25, 2015.

  1. ElectricMagician

    Thread Starter Member

    Jul 26, 2012
    57
    0
    Hello,

    To calculate power due to a current in the uA range (say 25 uA) and a known resistance (say 50 Ohms), which is correct?

    P = (25 * 10^-6) ^2 * 50 Watts
    P = 25^2 * 10^-6 * 50 Watts

    This has confused me because I know that a square with sides of 1cm has an area of 1 cm2, not 1*10^-4 m2, but to do it this way does not seem mathematically accurate
    Thanks
     
    Last edited: Feb 25, 2015
  2. Robartes

    Member

    Oct 1, 2014
    57
    13
    The first.

    Sure it is. 1 cm * 1cm = 10^-2 m * 10^-2 m = 10 ^-4 m^2. 1 cm * 1cm is also 1 cm2.

    Edit: keyboard ate my ^2 after 10^-4 m. Added it again.
     
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  3. ElectricMagician

    Thread Starter Member

    Jul 26, 2012
    57
    0
    I think I get it now, the part I was missing is that
    1 cm2 = 1 x 10^-4 m2, not 1 x 10^-2 m2 as I assumed

    Thank you
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,792
    P = (25 * 10^-6) ^2 * 50 Watts

    Only if you believe that

    <br />
\(a \cdot b\)^2 \; = \; a^2 \cdot b<br />

    If all else fails, walk it through in full detail and TRACK YOUR UNITS!

    <br />
\text{<br />
P \; = \; I^2R<br />
P \; = \; \(25\mu A\)^2 50\Omega<br />
P \; = \; \(25 \strike{\mu A}\ \cdot \frac{1A}{10^{6} \strike{\mu A}}\)^2 50\Omega<br />
P \; = \; \(25 \times 10^{-6} A\)^2 50\Omega<br />
P \; = \; 625 \times 10^{-12} A\)^2 50\Omega<br />
P \; = \; 31250 \times 10^{-12} A^2 \Omega<br />
P \; = \; 31.25 \times 10^{-9} \strike{A^2 \Omega} \cdot \frac{1W}{1 \strike{A^2\Omega}}<br />
P \; = \; 31.25 \times 10^{-9} W<br />
P \; = \; 31.25 \times 10^{-9} \strike{W} \cdot \frac{10^{9}nW}{\strike{W}}<br />
P \; = \; 31.3 nW<br />
}<br />
     
    Roderick Young likes this.
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