Discuss Energy Dissipation in the Australia area at ElectriciansForums.net

M

makawhite

Hey guys

Sitting studying at the moment but can't get my head around calculating for energy dissipation. I was wondering if anyone could shed some "light" on the subjuct...

Lets say a current of 4.5A flows through a 3.5 Ohm resistor for 5.5 mins. What is the energy dissipation?

Thanks for your help

Mark
 
Last edited by a moderator:
Hi.

Well lets start at the beginning. First off find V, = IxR = 3.5x4.5 = 15.75V.

So power dissapation, P = Vsquared/R = (15.75x15.75)/3.5 = 70.87W.

So P = 71W.

Now there is a chance that isn't correct as I was in work all night glanding off 12 x 800mm single core swa's into a 3200A ACB and my swede is fried.:(
 
Last edited by a moderator:
Hi

I worked it out the same way Lenny did.

Basically...

V = IR. so V = 4.5 x 3.5 = 15.75W
P = IV. so P = 4.5 x 15.75 which = 70.875W

So to work out Energy Dissipation you take 70.875W x 330s ( 5.5 minutes x 60 to give seconds) and this = 23388.75 J
 
(Abit off the topic of enegery dissipation)

But hats off to Lenny! An all nighter glanding 800mm SWAs! Harsh! I glanded off 2 4 - core 120mm cables yesterday...Did my head in! Haha
 

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