About Voltage Ratings

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!

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.

godfreyl said:

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.

Click to expand...

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