About Voltage Ratings

[xibalban]

I was wondering if the following makes any difference, technically.

Say, some load is rated at 100 W at 230 Volts (resistive load assumed)
By using Power formula:

I=P/V [Power factor=1 assumed]
So, I=100/230 = 435 mA

Let's connect a 12 V source [battery] across this same load:

I=P/V
So, I=100/12 = 8.3 A

Hence, if a 40 Ah, 12 V battery is connected across this load, it would last for approximately 5 hours:

Total hours, h = Ah/A
h = 40 Ah/8.3 A
So, h = 4.8 hours

I know that this setup would require thicker connecting wires due to higher amperage, but I have the following doubts:

Is it practical?
Can a 230 V rated load be connected to a 12 V source?
What happens if the load were to be inductive/capacitive?

Please shoot your comments fellas!

 

[godfreyl]

It doesn't work like that. If the load is truly resistive, the resistance wont change.

I=P/V [Power factor=1 assumed]
So, I=100/230 = 435 mA

Resistance = 230V / 435mA = 529Ω

When connected to a 12V battery, the current will be:
I = 12V / 529Ω = 22.7mA

The power will be:
P = 12V * 22.7mA = 272mW

Note that this calculation doesn't work for lightbulbs because their resistance changes with temperature. Even then, lower voltage will result in lower current.

 

[xibalban]

It doesn't work like that. If the load is truly resistive, the resistance wont change.

I=P/V [Power factor=1 assumed]
So, I=100/230 = 435 mA

Resistance = 230V / 435mA = 529Ω

When connected to a 12V battery, the current will be:
I = 12V / 529Ω = 22.7mA

The power will be:
P = 12V * 22.7mA = 272mW

Note that this calculation doesn't work for lightbulbs because their resistance changes with temperature. Even then, lower voltage will result in lower current.

Thanks for the answer. I wonder now, why aren't loads rated in resistance/impedence values instead of voltage/power ratings?