amperage - Batteries & Output Current Limiting

I am currently working on a project that uses a 9.8V 1000mAh battery to power a solenoid, and as I understand it (please correct me if I got this wrong.) a battery capable of delivering 1000mAh is also capable of delivering 1Ah, or 2A for 1/2 hour, 4A for 1/4 hour, etc.

So, assuming I got the above correct, I need to know how to limit the battery output current to 1.0A (My circuit would get really hot otherwise.)

Answer

Your first assumption "battery capable of delivering 1000mAh is also capable of delivering 1Ah, or 2A for 1/2 hour, 4A for 1/4 hour" is incorrect. (That has been answered here already, I will have a look for it.)

The battery capacity vs discharge is far from linear, and the mAh rating is quoted against a low discharge rate (~0.1*capacity).

Secondly your circuit will use as much current as it needs. Trying to limit the current is likely to stop it working. To use less current, redesign the circuit. (You might want to write a new question about that, if you need help.)

Answer: So, if the current must be 1A, and the voltage is 9.8V, wind a coil with a resistance of 9.8ohms.

Edit: Clarified that the circuit is only the solenoid, and it is made by OP.

To use less current, either reduce the voltage, or increase the resistance of the coil.

Increase the resistance of the coil by using more turns, or thinner wire. Thinner wire will have a higher resistance, but you might not have any to hand. So more turns is easier to do.

Edit:

The relationship between resistance (R), current (I) and voltage (V) is Ohm's Law, and is: V = I x R

The battery provides a fixed voltage, in this case 9.8V

So I x R is fixed, it equals 9.8

Let's assume R is 2ohms
So I x 2ohm = 9.8V
I = 9.8V / 2ohm
I = 4.9A

If the resistance increases, the only possible way to ensure I x R stays the same, is to reduce I. Double the resistance, then I must half. increase R by 10, and that changes I by 1/10, or multiples I by 0.1.

Let's do it, R is 10 x bigger, so R become 2ohm x 10 = 20ohm
I = 9.8V / 20ohm

I = 0.49A

Further statements:
"If I run the same current through a wire with higher resistance ..."
Well how could you cause that to happen?

We know that the voltage (the force, if you like), the battery can exert is fixed at 9.8V. So if the resistance is increased, then the current must be decreased. Why? That is how the universe works.

So if we take the initial calcuation, 2ohm resistor allows 4.8A to flow
Power = V x I
Power = 9.8V x 4.9A
Power = 48.02W

Let's double the resitance to 4ohm

9.8V = 4ohm * new-I
new-I = 9.8 / 4
new-I = 2.45A

new-Power = 9.8V * 2.45A
new-Power = 24.01W - so a wire with twice the resistance will heat up more slowly