# 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
20,057
5,644
P = (25 * 10^-6) ^2 * 50 Watts

Only if you believe that

$
$$a \cdot b$$^2 \; = \; a^2 \cdot b
$

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

$
\text{
P \; = \; I^2R
P \; = \; $$25\mu A$$^2 50\Omega
P \; = \; $$25 \strike{\mu A}\ \cdot \frac{1A}{10^{6} \strike{\mu A}}$$^2 50\Omega
P \; = \; $$25 \times 10^{-6} A$$^2 50\Omega
P \; = \; 625 \times 10^{-12} A\)^2 50\Omega
P \; = \; 31250 \times 10^{-12} A^2 \Omega
P \; = \; 31.25 \times 10^{-9} \strike{A^2 \Omega} \cdot \frac{1W}{1 \strike{A^2\Omega}}
P \; = \; 31.25 \times 10^{-9} W
P \; = \; 31.25 \times 10^{-9} \strike{W} \cdot \frac{10^{9}nW}{\strike{W}}
P \; = \; 31.3 nW
}
$

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