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  1. Quote Originally Posted by Rogue View Post
    After a cup of tea and a bit of a read, I have a quick question in relation to 0.5CV^2. Where you give 0.5 * 0.005 * 70^2 = 6, I end up with 12.2. In fact using the time honoured technique of "randomly changing numbers until they worked", I only got 6j by changing the 70v to 16v?

    Trying to follow your calculation and sticking with the 5 second discharge, I ended up with this:
    0.5 * 0.005 * 70^2 = 12.25j
    12.25 / 5 = 2.45W
    70^2/2.45 = 2000ohm, so I would need a 2k ohm @ 3W resistor to discharge over 5 seconds.

    I'm rather hoping you made a typo because the alternative is that I'm more stupid than I thought!

    Errrmmm... looks like I made a boo boo... might have multiplied by 0.5 twice in error... your calculation is correct... now I'll go and edit my post..

    BTW 2k2 @ 3w will be fine
    Last edited by irving2008; 13-09-2012 at 09:36 PM.

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