On Sat, Sep 13, 2003 at 09:59:12AM +0100, Jon Mitchell wrote:
> Assuming the "murkin" (is that right ?) kettle requires the same
> amount of power to heat the water, then if you half the voltage
> (approximately correct) then the current is doubled.
Agreed, mostly ... (dredging up high school physics thru a fog of
stout... t=SHw*dT/E ) 1 Joule= 1 Watt-second, specific heat of water=
4200J/degreeC/litre so, 1 litre water @ 2400W, 10deg -> 100deg = 157.5
seconds, ignoring inefficiencies & heat loss to environment (I would
say this is where the non-linearity is!) -at half power,
would be double the time...
> But of course, it is the Current that kills you - not the Voltage.
with static electricity we are now back to Coulombs of charge (NOW my
head hurts...), which can be related to available energy, once
dissipated, there's no volts left- with the VdG genset the effective
source resistance is so high so, if you short it out with your body the
volts fall away, result is negligible AVAILABLE current, -unlike mains
power which can quite happily supply up to the rated current at the
rated voltage (almost) forever - as long as someone keeps paying the
bill!!!.
remember ohms law... V=IR -> I=V/R. for the same resistance, double the
voltage = double the current. But of course it's the resistance that is
important, sweaty hands means less resistance = more current. a serious
drinker retaining a lot of fluid could almost be un-personed by a car
battery.... link included for shock value <everybody groan here>...
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/shock.html
David(O)T..!
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