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In Reply to: No, I want to see how a PSU can be quoted in joules. posted by cheap-Jack on April 17, 2007 at 11:10:53:
Energy in a CapacitorWe know that:
1. Energy = charge × voltage
2. Q = CV.
This second relationship tells us that the charge – voltage graph is a straight line.
The capacitor is charged with charge Q to a voltage V. Suppose we discharged the capacitor by a tiny amount of charge, dQ. The resulting tiny energy loss (dW) can be worked out from the first equation:
dW = V × dQ
If we discharge the capacitor completely, we can see that:
Energy loss = ½ QV
By substitution of Q = CV, we can go on to write:
E = ½ CV2
Follow Ups:
Hi.Assuming your dW=V.dQ is correct (relevant to the electrical defination of a joule), how many joules are you talking about a given PSU say 350V with its final cap being 8uF ?
There are tons of similar formula around, my question is how to apply them in the joule calculation to assess a PSU.
E joules = 1/2 C V 2350 volts, 8 μF
8 μF = 8 * 10 -6 F
E = 1/2 * 8 * 10 -6 * 350 * 350E = (1/2 * 8 * 350 * 350) * 10 -6
E = 490,000 * 10 -6
E = 4.9 * 10 +5 * 10 -6
E = 4.9 * 10 -1
E = .49 joules
Cheers, John
Hi.This makes the didfference btween actually knowing it or just hearsay about it.
a clue:
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