Help regarding Ac current limiting using resistor
I need to regulate it to 1to 4ma maximum output without having much
  voltage drop... Can I just use a resistor in series?


No. A resistor works linearly according to Ohm's Law. Resistance = Voltage / Current, so for a maximum (short circuit) current of 4 mA the resistance needs to be  12 / 0.004 = 3000&ohm;. However at 1 mA the resistor would drop 3000 * 0.001 = 3V, so output voltage would drop from 12V to 9V.

I you want a fairly stable 12V up to 4 mA and no more then you need a circuit that starts dropping the voltage only when current exceeds 4 mA. Doing this to AC current is further complicated by the fact that 4 mA AC is actually a sine wave varying from -5.66 mA to +5.66 mA. Most current limiting circuits also have significant voltage drop (typically 1~2 V minimum) which distorts the AC waveform as it crosses the zero voltage point, and might require a higher input voltage to make up for the loss. 

Here's an AC current regulation circuit that has minimal voltage drop below the current limit. A bridge rectifier formed by D1-D4 converts the AC voltage to DC so the current mirror (formed by Q1 and Q2) can work on both halves of the waveform. D5 and C1 half-wave rectify and smooth the AC input, producing ~+16V which is applied to R1 to produce a reference current of ~+5mA for the current mirror. The load (here represented by a variable resistor) is inserted on the AC side of the bridge rectifier.







simulate this circuit &ndash; Schematic created using CircuitLab

With matched transistors Q2 should sink the same current as Q1 if it is able to. If the load current is less than the reference current then Q2 saturates with a low voltage drop. The bridge rectifier uses Schottky diodes for low voltage drop, so the rms AC output voltage and current varies from ~11.6 V at 1 mA to 11.4 V at 3.8 mA. At 4 mA current limiting starts to kick in and voltage drops to 10.8 V. Short circuit current is ~5.6 mA rms.

To vary the current limit simply change the value of R3 according to the formula Resistance = (16 V - 0.6 V) / Current. The 0.6 V accounts for voltage drop across Q1. 

Q1 and Q2 should be in close thermal contact to ensure that they track well as temperature changes.