Current Transformers and Energy Harvesting
If your load (aka burden) is 10 mA RMS and your primary current is 20 amps RMS then you have a turns ratio of 1:2000. If the load voltage is (say) 5 volts RMS then the load power is 50 mW.

That 50 mW is taken from the primary. What sort of impedance takes 50 mW from 20 amps? V = 50 mW / 20 amps  = 2.5 mV RMS. Not much of a volt drop. The effective impedance seen in the primary winding is 2.5 mV / 20 amps = 125 micro ohms.


  Is it 10mA more being drawn from the mains, on top of the 20A?


What you will see is a slight reduction in AC load voltage (2.5 mV) and this naturally results in a slight decrease in the 20 amps taken by the load (assuming it is a linear load).