Power Consumption while screen brightness is minimum

[bmvakili]

The title is the question.

How to calculate the regular MacBook power consumption if it is left on lowest screen brightness for about 12 hours, plugged in, with just a browser running Gmail?

Thanks in advance for your input.

The following is the System Profiler->Power Information that pertains to this topic:


Battery Information:
Model Information:
Serial Number: DP-xxxxxxx-xxxx-xxxx
Manufacturer: DP
Device name: xxxxxxx
Pack Lot Code: 0002
PCB Lot Code: 0000
Firmware Version: 0110
Hardware Revision: 0500
Cell Revision: 0102
Charge Information:
Charge remaining (mAh): 4831
Fully charged: Yes
Charging: No
Full charge capacity (mAh): 4831
Health Information:
Cycle count: 188
Condition: Good
Battery Installed: Yes
Amperage (mA): 147
Voltage (mV): 12604

System Power Settings:

AC Power:
System Sleep Timer (Minutes): 0
Disk Sleep Timer (Minutes): 10
Display Sleep Timer (Minutes): 20
Automatic Restart On Power Loss: No
Wake On AC Change: No
Wake On Clamshell Open: Yes
Wake On LAN: Yes
Display Sleep Uses Dim: Yes
Battery Power:
System Sleep Timer (Minutes): 5
Disk Sleep Timer (Minutes): 10
Display Sleep Timer (Minutes): 1
Wake On AC Change: No
Wake On Clamshell Open: Yes
Display Sleep Uses Dim: Yes
Reduce Brightness: Yes

Hardware Configuration:
UPS Installed: No

AC Charger Information:
Connected: Yes
Charging: No

 

[alphaod]

MiniBatteryLogger show the current amperage draw. I don't know: is that what you want?

 

[Abstract]

Do you want P = IV (Power = current x voltage)???

If you do, grab a calculator and be my guest. "Amperage" is the same thing as "current". Also, divide your answer by 1000000 (1 million) so that you calculate Power with the correct units.

 

[alphaod]

Also, divide your answer by 1000000 (1 million) so that you calculate Power with the correct units.

Shouldn't it be divide by 1000, since milli- = one-thousandth?

 

[Abstract]

But in his 1st post, both Voltage and current are reported in "milli"-units. Multiply them together, and he'd have to divide by 10^6 to adjust for it.