Code Question of the Day with Charlie Trout

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Code Question Of The Day

Monday, June 27, 2011

Subject: Code Question of the Day for Monday, June 06, 2011 - Joules from 18KV.

What the submitter didn't tell you is the magnitude of the capacitor that gets charged. The energy, watt-seconds or joules is the product of the Capacitance "C" and the Voltage "E", squared. The rest of the information is just incidental and describes the result of the discharge. It has to obey the RC time constant discharge rate, which also depends on the internal and external series resistances. A rule of thumb is that a capacitor is essentially fully discharged after three times the RC value, which is in units of seconds. There may be some energy lost due to the internal resistance of the device as well so the actual capacitor could be larger then the calculation would yield.

Hope this helps.

Incidentally, how does this relate to anything in the NEC??

Nick Abbatiello



Hey Nick, I’m impressed. Not only by your answer, but by the clear concise brief manner in which you presented the information. You’re right – it really has little to do with the NEC but I am inclined at times to allow some leeway where an electrical question is asked without specific relation to the NEC. You have many times come up with good answers, generally lacking in specific Code references, but brevity was not your forte and since I would never edit any answers, I could not use your answers. Keep them short and to the point


Source: Charlie Trout

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