Short Answer: The resolution depends on the electrode spacing. It is half of the receiver dipole spacing.
Long Answer: Mathematically we cannot calculate the response of the ground at any better than one half of the dipole spacing. For example, if the dipole spacing for a particular measurement is 5 meters, the best possible (neglecting any noise) resolution is 2.5 meters.
Does that mean that my resolution will be 2.5m at any depth? No, it means that we have to look at the "size" of the receiver dipole. To get better depth penetration, the transmitter and receiver dipoles are separated until the point where no signal can be recorded accurately (this is usually 8 dipole separations (due to the inverse square law), in 2D, hence why we have an 8 channel SuperSting instrument).
At this point, we increase the receiver size to increase the signal and continue to separate the transmitter and receiver dipoles. However, increasing the receiver dipole size also decreases the resolution (remember that we can only resolve features in the ground at least half the size of the receiver dipole).
At depth, the dipole spacing is increased in order to increase the signal-to-noise ratio. The effect is that the resolution drops with depth. The target needs to be larger to be detectable at greater depths.
Example:
Let's say you have 56 electrodes spread out to 110m.
You would have n-1 spaces between the electrodes which = 55. Let's divide the 110m distance by the 55 spaces and we get a 2-meter electrode spacing.
Now according to the Nyquist-Shannon sampling theorem, the resistivity method is only sensitive to resistivity changes greater than half of the electrode spacing: half of 2m is 1m. That would be the smallest resistivity changes that you would be able to image.