1)Sinewave
2)Well at least 20 volts or 20 A.

Which? 20 V or 20 A. Do you understand the difference?

Amplitude, or RMS? Do you understand the difference?

Assuming RMS, you are asking for 400 W (more, since no conversion circuit is going to be 100% efficient). To get an idea of what you are asking for, do a search for automotive power inverters that can deliver 400 W.

Don't expect your set-up to last long. A typical 12 V automotive battery would be effectively dead in somewhere around an hour.

And don't even think of trying to get that kind of current out of most 9 V or 1.5 V batteries. Even if you could, they'd be dead in minutes or less.

Now, there ARE low voltage batteries that can deliver these kinds of currents for extended periods of time -- for instance, the 2 V batteries used to power submarines -- , but you probably can't afford them.

So take a step back and describe what you are trying to do. What is the end goal? Why do you want such a circuit? Then perhaps we can help you find a reasonable way to solve the underlying problem.

 

1)10 sec
2)Yeah,I do.Current is the electric charge that flows through a circuit.Voltage is the difference between the 2 points - and +.Load is the electrical charge.
3)500 mAh
4)Yeah.20 volts mean voltage and 20 A means electric current.I just realised that voltage doesn't change in a battery and that the current is the only one that changes.

500 mAh is NOT the current (which is what ScottWang asked for). 500 mAh is the "capacity" of the battery and is an indirect way of stating the total amount of charge it can deliver over its life. Ideally, it would mean that it could deliver a constant 1 mA for 500 hours, 500 mA for 1 hour, or 50 A for 0.01 hours (which is 0.6 seconds). Assuming this ideal case, to get 20 A at 20 V with 100% efficiency you would need to pull pretty close to that 50 A value, so your 9 V battery would last about a second. In reality, the actual current that a 9 V battery can deliver, even into a dead short, it limited to values well below that because of the internal resistance.

If you are just now realizing that the voltage in a battery doesn't change (at least in an ideal world), then you probably need to spend quite a bit of time studying the basics before you attempt making a circuit like you are describing.

But again, what is the actual problem you are trying to solve? What is it you need this AC signal for?