Squaring microamps

Thread Starter

ElectricMagician

Joined Jul 26, 2012
57
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:

Robartes

Joined Oct 1, 2014
57
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
The first.

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
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.
 

WBahn

Joined Mar 31, 2012
29,979
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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