I am getting a platinum coated anode that will have a surface area of 26 cm2 (100 mm long x 8 mm diameter). For my application, the maximum recommended current density is 200 mA/cm. Based on that, how much current should be flowing through the anode? I calculate square root of
26 (5.1) times 0.2 which is 1.02 A, is this correct? Thanks.
I'm probably confused but, if it's 8mm in diameter, isn't it 25mm* in circumference? And, if it's 25mm, then 25mm x 100mm = 2500mm^2, convert to cm^2 and get 25cm^2. then multiply 200ma x 25cm = 5 amps. I'll let an adult correct me. I have never seen a platinum coated electrode.
You're at the same place I was originally when I did the calculation, which made me question whether or not I should take the square root of the area and then multiply that by current density. When I took the square, I got the 1.02A above, but much closer to your value when I didn't.
Platinum coated electrodes have many applications:
6 m
oot of
Lemme see- surface area of a cylinder = pi x D x H.
3.14159 x 8 = 25.13325.133 x 100 = 2513.3
0.0200 x 2513.3 = 502.7Divide by 100 (mm - cm) = 5.02 A.
add the end-cap, if relevant = pi x r^2 = 3.14159 x 16 = 50.27 = 0.
01AOr, close enough.
Peter Wieck Melrose Park, PA
I think the mistake your making is the 26 is already in cm^2, no need to take the square root. Just 26 x 0.2 = 5.2 amps.
OR, another way, your anode is 100mm x 8mm dia or 100mm x 25mm, convert to cm for 10cm x 2.5cm this equals 25cm^2. 25cm^2 x 0.2 amps = 5 amps.
Sometimes it would be nice for an OP to say, Hey, thanks for the help, or you're F'ing nuts and don't have clue. Just something, so we know they didn't get hit by a car!
Quite right, Thank you ALL!
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