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No you use less amps.

I=V/R

Resistance will remain constant so for an appliance with a resistance of 18 ohms

240/18 = 13.33A

230/18 = 12.77A

That's correct,

Now if from that you work out the power you get 3.199 Kw at 240v @ 2.937 Kw @ 230v.
 
So now they have lowered the domestic voltage down to 230 in some areas, this equates to more amps used, so my meter turns quicker and my electric bill gets bigger!

No it doesn't. Your electric meter doesn't measure amps, it measures kilowatt hours. If you use a one kilowatt heater for one hour, you use one kilowatt hour. Current and voltage are irrelevant to that. Additionally, while the standard is 230V +10%, -6%, (since 1995), you will find that your voltage still sits around 240V.
 
Last edited by a moderator:
No it doesn't. Your electric meter doesn't measure amps, it measures kilowatt hours. If you use a one kilowatt heater for one hour, you use one kilowatt hour. Current and voltage are irrelevant to that. Additionally, while the standard is 230V +10%, -6%, (since 1995), you will find that your voltage still sits around 240V.


Somebody needs to rewrite the equation P = I x V then . Power (the rate of energy dissipation) is a combination of current and voltage. Energy use, as you correctly say is, a combination of power and time.
 
No it doesn't. Your electric meter doesn't measure amps, it measures kilowatt hours. If you use a one kilowatt heater for one hour, you use one kilowatt hour. Current and voltage are irrelevant to that. Additionally, while the standard is 230V +10%, -6%, (since 1995), you will find that your voltage still sits around 240V.

How do you work that out? Current and Voltage are used to calculate Watts

Power (Watts) = Current x Voltage

*EDIT* pushrod pipped me at the post :D
 
Last edited by a moderator:
Somebody needs to rewrite the equation P = I x V then . Power (the rate of energy dissipation) is a combination of current and voltage. Energy use, as you correctly say is, a combination of power and time.


I think I put my point across badly.Perhaps I need to rephrase. Even if the faulty supposition that a decrease in voltage would lead to an increase in current, the number of units used would be the same or arguably less. The kilowatt hour meter registers the product of voltage, current and time. Does that make my point clearer?
 

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