How Long Will My Camcorder Battery Last?

[Jennifer2010]

Hey guys!

I have two camcorders.

Cam #1: Power Supply: DC 7.2V (battery)
Power Consumption: 7.4W

Cam #2: Power Supply: DC 7.4V (battery)
Power Consumption: Doesn't say :( - but should be the same roughly.

How long can a "portable powerpack" with the following specs keep these camcorders running before running out of power?:

Capacity: 1000VA / 500W
Output:
Nominal Voltage: 120V / 230V
Battery Capacity: 12V / 9Ah x 1pc (think it says 1pc cause it's a UPC for computers?)

Could someone please let me know the math on how to figure this out myself for future equations, too?

Thanks so much!

 

[sprayer_faust]

Hi.

The potential difference/voltage between the battery poles tells you how much energy (in Joules) is available per unit of charge (1 As). Short: Higher voltage means a single charge carrier can do more work for you.

A charge capacity of 9 Ah tells you, that the battery can drive a current of 9 A for an hour, or:
that it stores 9 A * 3600 s = 32400 As of charge. A unit of charge can do 12 J of work (12 V = 12 J/As) and you have 32400 units of charge. Consequently you have 12 V * 32400 As = 388800 J available.

Your camera consumes 7,4 W = 7,4 J/s on average. Which tells you that it should run for 388800 J * s / 7,4 J =52541 s = 14,6 h. Which is a lot, but then again you have a very big battery - capacity is on par with a car battery.

I do not understand these two lines though:

Capacity: 1000VA / 500W
Nominal Voltage: 120V / 230V

I am guessing that your 12V battery has a built in 12 V DC to 120 V or (switch) 230 V AC converter - since you said it is a PC reserve unit. So the second line represents possible output voltages and the first one maximum power output respectively (1000 W at 120 V and 500 W at 230 V). Be sure to use an adapter/converter (120 V to 7,4 V or 230 V to 7,4 V) or you will roast your equipment.

It would also be better (more efficient) to bypass the built-in converter and go straight from 12 V to 7,4 V, because converters are not perfect and each one degrades battery efficiency (battery time).