Solving for current, Ohm's law is I = V/R. You only deliver .7 amps of current into a 4.285 ohm load.
If you connect a 3 volt battery to a 1,000 ohm resistor, then only 0.003 amps of current will flow, which is 9 milliwatts.
Connect a 3 volt battery to a .01 ohm load (a dead short, almost) and 300 amps of current will flow, (900 watts!) very briefly, until the battery voltage sags under load.
1: Or any voltage source in general, barring some trickery that you can't do with a simple electrochemical battery.
> Connect a 3 volt battery to a .01 ohm load (a dead short, almost) and 300 amps of current will flow
Except it won't, of course, because all real batteries have internal resistance. (How much depends on the battery chemistry and size.) Probably the 700mA rating for the AA battery is into a dead short.
He might be referring to the "hydraulic analogy" of electricity: A wire is a water pipe, voltage is water pressure, amperage is total water flow. A battery is a pump pushing water through the pipes, voltage is how hard the pump pushes, and amperage is limited by how wide the pipes are, and thus how much resistance they impose on the flow of water.
The basic idea was that you can't put a 5kW drive on a 2.5kW motor: it would shove too much power into the motor and break it. This is not quite at all how those systems work.
E.g., if you have a resistive load and an inductive power source, the power source will keep upping its voltage until the load accepts the current that's being shoved down its throat.
Solving for current, Ohm's law is I = V/R. You only deliver .7 amps of current into a 4.285 ohm load.
If you connect a 3 volt battery to a 1,000 ohm resistor, then only 0.003 amps of current will flow, which is 9 milliwatts.
Connect a 3 volt battery to a .01 ohm load (a dead short, almost) and 300 amps of current will flow, (900 watts!) very briefly, until the battery voltage sags under load.
1: Or any voltage source in general, barring some trickery that you can't do with a simple electrochemical battery.