用電量點計 ?

V =電壓 A=電流 W=電功率

V 乘 A=W

1000W = 1KW = 1度電

火牛input 220v 1.5A / output dc 5A
220V X1.5A =330W 代表一小時的用電量

330W X24小時= 7920W =7.92KW =7.92度電

如火牛input 220v 1.5A / output dc 5A 只是印在火牛的標明
那在使用火牛時的用電量就不是 220V X 1.5A ,
因為input 220v 1.5A / output dc 5A 只是火牛的input 和 output最大使用電流.
input 電流可以是大於0至1.5A
真正電流要事乎output 所用電器用電量有多大.
而要知道真正使用的電流是要用電流錶量度input 電流的.

Q1)
Assumed pf = 1 (unity power factor) and transformer full load operate.

AC Power (P) = Volt * A * pf, (where pf is phase angle between V and A).
= 220 * 1.5 * 1
= 330 (Watt)

Total energy (kwh) per day is:
= (330/1000) * 24
= 7.92 kwh

Electricity charge ($):
= 7.92 * $0.9
= $7.128/day

Q2)
A - current unit (input current , output load current)
V - voltage unit (input voltage, output voltage)
W - power unit (input power = output power + loss)

In this case, I belive that you are not used the transformer/adaptor under full load (100% working) over 24 hours, therefore the actual charge may less than the apparent calculation.

Q1.
Assume this transformer is losses (IDEAL)

Input Voltage, Vin = 220V
Input Current, Iin = 1.5A

Therefore,
Input Power, Pin = Vin * Iin = 220*1.5 = 330VA

As mentioned above, it is an ideal transformer, Input REAL Power = 330W

Energy Comsumption = 330W * 24hr = 7.92kWh


Q2.
Generally speaking, transformer should shows following parameters:
i) Pri. Voltage, V1 (Unit: V)
ii) Sec. Voltage, V2 (Unit: V)
iii) Transformer Rating (Unit: VA)

2006-12-18 19:47:46 補充：
Cont. Q1If unit charge = HK$0.9/kWh,then, Total electricity charge = 7.92kWh * HK$0.9/kWh = Hk$7.128

W=V*A unit
A:電流 current ampere
V:電壓 voltage volt
W:功率 power watt