Trading voltage for current

BradtheRad said:

1.

If you want 20V at 5A, then you must provide 5V at 20A (not counting losses due to inefficiency). In other words, to get 100W out, you must put 100W in.

Small batteries cannot provide that many watts.
I believe a string of 4 D batteries might be able to provide over 10 A at 5V.

2.

The Ehow page has a misstatement or two.
You can get more AC current through a capacitor by increasing its Farad value, but for that to happen, the AC source needs to be able to provide that much current.

As for DC, the capacitor blocks DC as soon as it has charged sufficiently.



It claims that a diode will increase current. This works in the sense that an AC waveform has a zero net average value..., however when you take one half of it (by rectifying it), then it gives you a waveform which can do things that AC cannot. You're taking either the positive or negative power and that is 50 percent of the original.

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