With a ideal 2x teleconverter, you will be 2 f-stops down from what the lens is set to.

Think about the basic physics and this should be clear. A 2x teleconverter makes the dimension of anything in the image 2x larger. Something that would result in a 1x1 mm square with the bare lens results in a 2x2 mm square with the teleconverter. That 2x2 mm square has 4 times the area of the original 1x1 mm square. That means the same light coming out of the lens is now spread over 1/4 the area, which means it will be 1/4 as bright. A factor of 4 less light is 2 f-stops, since each f-stop represents a factor of 2.

Any real teleconverter will loose a little light due to it being reflected off the front surfaces of the lens elements and the like. Therefore a 2x teleconverter will loose a little more than 2 f-stops of light. However, this extra loss is generally small enough to ignore. In any case, the automatic metering system in the camera should compensate for whatever the teleconverter does. You will see this as slower shutter speed, or requiring you to open the lens more for the same shutter speed.

Therefore, to answer your question, if the lens is set to f/1.8, then the effective aperture for the purpose of light level will be 2 f-stops higher. f/1.8 is not one of the standard f-stop values, but computing 2 f-stops higher is still easy. Each f-stop is sqrt(2) higher from the previous, so 2 f-stops is 2x higher, so 2 f-stops past f/1.8 is f/3.6. If the lens was set to f/4.0, then you'd get f/8.0, if set to f/11, you'd get f/22, etc.