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.
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
Why is lightening always quoted in the amount of Volts...and not the...??? current finally discharged
it's hard to get a clamp meter around it :d
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.
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.
It's hard to get a clamp meter around it![]()
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.