UK,,
2007-07-17 04:29:56 · 4 answers · asked by ♣Kermit the Frog♣ 4 in Science & Mathematics ➔ Engineering
That's a maximum of 280 W, and 2.24 kW-hr. But, all the disk-drive or CD motors are not running all the time, nor are all the peripherals using their max. power, so that 280 W is probably more like 200 W, which is 1.6 kW-hr.
So, if your electricity costs are 15 c / kW-hr, then it will cost 24 c per day (that's an educated guess at 15 c/kW-hr for the cost of electric energy).
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2007-07-17 04:46:27 · answer #1 · answered by tlbs101 7 · 0⤊ 0⤋
It depends on your electricity suppliers' unit prices. But yeah, if your computer is using exactly 280w, that's 2.24KW or 2.24 units.
If your supplier charges you 12.99 pence per unit for the first 225 units in the quarter, then 7.99 pence per unit for the remaining, then this will either cost you...
29 pence (approx) earlier in the quarter, or
18 pence (approx) later in the quarter.
Hope this helps!
2007-07-17 12:56:50 · answer #2 · answered by Just some guy 6 · 0⤊ 0⤋
Well as the others said you can work out hourly consumption etc, but if it has an hibernate option then when not used it will decrease the comsumption. The others have worked out worst case scenarios but it will be less than this!
2007-07-18 10:56:08 · answer #3 · answered by big_kahuna_jaffa_cake 2 · 0⤊ 0⤋
let the cost per kWh be c (in pence)
0.280*8 = 2.24kWh
therefore your cost is 2.24c (pence)
Your meter or bill you tell you a value of c since it varies enormously with suppliers.
2007-07-17 11:33:31 · answer #4 · answered by SS4 7 · 1⤊ 0⤋