Shower overheating no loose connection

[telectrix]

Need to dust off the old books, I think I was getting confused with the amp thing as I would be calculating say the 9.5kw of the shower where of course ohms law throws that out the window because the 9.5kw is only relevant to the expected supply voltage so less voltage less kw less amps. Have I got there

spot on .example. a 10kW shower rated at 240V.

P = V x I, so
I = P/V = 10000/240 = 41.667
from this you can determine the resistance

R = V/I = 240/41.667 = 5.76 ohms.

now as the resistance is constant ( not allowing for temperature variations, then apply 230V to the constant resistance

I = V/R = 230/5.76 = 39.93 A

from that, P = VI = 9.184 kW

so the actual current draw and hence the power, is reduced.

can't you telli'm bored and got nowt better to do. it's pi$$ing down, so car wheel theft is not an option for this Scouser today..