Discuss Shower overheating no loose connection in the UK Electrical Forum area at ElectriciansForums.net

No it won't. The shower is a fixed resistance load. If the voltage goes down, so do the amps.

But wouldn't the fixed resistance draw higher amps as the voltage is reduced - the same level of resistance at lower voltage would demand higher amperage to meet the resistance surely?

As ever, I accept I might be wrong - ready to learn...
 
But wouldn't the fixed resistance draw higher amps as the voltage is reduced - the same level of resistance at lower voltage would demand higher amperage to meet the resistance surely?

As ever, I accept I might be wrong - ready to learn...
No! It's a fixed resistance load, the elements resistance will not change. So if the voltage drops then the current will drop, and vice versa.

Ohms law, R (the element resistance) is the one that will not change. I = V/R
 
Exactly. Fixed resistance doesn't equal fixed current draw. With lower voltage it has the potential to carry more amps was my way of thinking :confused:
It's Ohm's law - I = V/R

R is fixed so as V gets lower, so must I.

You can think of it as the voltage pushing the current through the resistance. A lower voltage means it pushes less current through.

If you run some examples with say a resistance of 10ohms and voltages of 240 and 220, you'll see the current drop.

Edit: beaten to it!
 
Last edited:
No! It's a fixed resistance load, the elements resistance will not change. So if the voltage drops then the current will drop, and vice versa.

Ohms law, R (the element resistance) is the one that will not change. I = V/R

It's Ohm's law - I = V/R

R is fixed so as V gets lower, so must I.

You can think of it as the voltage pushing the current through the resistance. A lower voltage means it pushes less current through.

If you run some examples with say a resistance of 10ohms and voltages of 240 and 220, you'll see the current drop.

Edit: beaten to it!

Ok, I accept the weight of opinion and I'm off to google until I've got my head around this :cool:

On this basis then, lower voltage would mean lower amps so that the element would put out less heat?

My confusion comes from us once plugging an electric pool heater made in the UK, into a 220v system overseas and the cable quickly overheated. Why didn't the cable stay cool and the heater simply perform less well than it would at 240ish volts? It was a ~56a heater of a 63a supply.
 
Ok, I accept the weight of opinion and I'm off to google until I've got my head around this :cool:

On this basis then, lower voltage would mean lower amps so that the element would put out less heat?

My confusion comes from us once plugging an electric pool heater made in the UK, into a 220v system overseas and the cable quickly overheated. Why didn't the cable stay cool and the heater simply perform less well than it would at 240ish volts? It was a ~56a heater of a 63a supply.

Yep, that's right-lower voltage and lower current would mean lower wattage and less heat output.

I'm not sure about your pool heater - maybe the cable just wasn't able to handle such a high current or perhaps there were other factors involved - damage in transit? Was this for the aquarium...?
 
Yep, that's right-lower voltage and lower current would mean lower wattage and less heat output.

I'm not sure about your pool heater - maybe the cable just wasn't able to handle such a high current or perhaps there were other factors involved - damage in transit? Was this for the aquarium...?

Not for the aquarium, but for a temporary pool so same thing..

The heaters are in regular use and are tested and still perform well. It was just that one location where both heaters struggled.

They became electrically hotter than they should around the cable entry. I can't accurately say if they heated the water more/less efficiently than normal as the issue was detected, and both disconnected too soon to make any judgement on that.
 
if the element is OK, then overheating of the cable in the shower could be down to a high resistance termination, either between the cable and crimp or crimp and terminal.
 
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
 

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