I have a 12 V DC input that I am stepping down to 2 V with 610 ohm resistor. When power is removed the light is staying on for a period of time and slowly dims.
I suspect that you mean when you switch off the mains power that the LED fades out.
simulate this circuit – Schematic created using CircuitLab
Figure 1. An old-fashioned power supply driving an LED.
How it works:
- SW1 switches the power on and off.
- XFMR1 is a transformer and steps the voltage down from mains to the low voltage required - probably about 9 V AC in your case.
- BR1 is a bridge rectifier consisting of four diodes. No matter what the polarity of the input voltage on the ~ ~ terminals the voltage is "rectified" and current always comes out on the + terminal. The bridge rectifier will output pulses of current at the peaks of the alternating voltage.
- C1 is the smoothing capacitor which stores some charge between pulses from BR1.
- R1 limits the current to the LED to a safe value.
- D1, the LED, will have a certain "forward voltage", Vf, at the current set by R1.
Would the resistor be acting like a battery?
Definitely not. The fading effect is caused by the charge stored in C1 slowly bleeding out through R1 - D1.