Hi Chris.

Imagine you have a perfect voltage source, of E output across which you place your divider and from the mid-point of the divider you "feed" your circuit, then the equivalent circuit to yours, consists of a perfect voltage source of output voltage E*r1/(r1+r2). There is an "source impedance" of r1*r2/(r1+r2) between you and the perfect source.

The use of a voltage divider always has the implicit assumption that the load applied to it is very much larger (ideally infinite) than the source impedance, otherwise it will load the output and what you will get at the divider is NOT what you calculate. The "right" answer is based on a knowledge of what you are using the divider for ! If you were driving a transistor, then the lower values would be better, since the input of base of the transistor is quite low impedance. If you are using the divider to create a reference voltage for an ADC, then the higher impedance divider will work.

Didn't you do A level physics ?

Steve