# Can I convert to less AC volts to get more amps?

I was wondering if I have (for example) an input of 12V and 100A DC, instead of going to, let's say 120V and 10A AC, if it could be converted into 100V for more amps. (I already know about loss so let's just assume this is a magical lossless system.)

• @Chu I know, I am describing a transformer and that lossless is impossible but I mean, instead of going to 120V 10A AC, can I go to, let's say, 100V, 11A AC? (Or use extra amps to account for loss.) – Prowler1000 Aug 3 '16 at 0:17
• You can convert from any ac or dc voltage to any other ac or dc voltage and if it's lossless, power in = power out. But of course transformers only work on ac – Chu Aug 3 '16 at 0:22
• Also, the load defines the current demand. A 10V source, for example, will supply 1A to a 10 ohm resistor and 0.5A to a 20 ohm resistor. Ohm's law must be obeyed. – Chu Aug 3 '16 at 0:33
• @Chu Okay. So if less volts are converted, that means more amps are available. Right? – Prowler1000 Aug 3 '16 at 0:36
• Assuming no losses, yes. 120V x 10A is 1200W. This could be converted to 12V x 100A, or 24V x 50W, or any other similar combination. However, losses do enter into it, as do maximum current ratings, etc. Also, you can only use a transformer with AC. If you want to convert DC it becomes a more complicated problem. – bitsmack Aug 3 '16 at 1:09

Assuming no losses, yes. 120V x 10A is 1200W. This could be converted to 12V x 100A, or 24V x 50A, or any other similar combination. However, losses do enter into it, as do maximum current ratings, etc.

Also, you can only use a transformer with AC. If you want to convert DC it becomes a more complicated problem.

Energy must be conserved. You cannot get out more than you put in. Assuming you are having a constant power draw, the power in is going to equal the power out. Therefore,

[Amps in] x [Volts in] = [Amps out] x [Volts out]

So, if you are stepping up 12V 100A to 100V, you are going to get 12A (less with losses) out.

In principle yes, because not considering losses, for the same power, less voltage means more current.

The thing you have to consider though, is whether your load will absorbe the same power at a lower voltage. A purely resistive load will have proportional voltage and current as per Ohm's law, where some loads could have different behavior. If the load is regulated by e.g. a DC/DC converter you can expect the behavior you described.