I was trying to measure the internal resistance of an LDO that will produce 2.85V. I used a Digital multimeter to do this, but I got a resistance of 810 Ohm and the LDO can support upto 450 mA. Will the internal resistance of the LDO can be that high?
If the input supply voltage to an "ideal" 5 volt regulator is 10 volts and the output load is 10 ohms (0.5 amp output current), then the regulator will "act" like a 10 ohm resistor in order to facilitate 5 volts at the output. It's just a simple potential divider.
If the load changes rapidly between 9 ohms and 11 ohms, then the regulator (due to feedback control), will keep the output voltage at 5 volts by rapidly altering its effective series resistance to counter the changing load impedance and current. So, when the load is at 9 ohms, the regulator will act like a 9 ohm resistor in order to keep the output at 5 volts.
If the input voltage reduces to (say) 8 volts and the load is 10 ohms then the effective series resistance of the regulator has to be 3 volts/0.5 amp = 6 ohm.
The effective series resistance of an ideal voltage regulator will change slowly or rapidly to counter the load and input voltage circumstances it faces.