Current tranformer, burden resistor calculation
If you have 2 Ap-p flowing through your primary, the secondary current would be: -

\$\dfrac{2 Ap-p}{50}\$ = 40 mAp-p. If you used a 1 ohm burden resistor you'd get 40 mVp-p which wouldn't be enough to drive a bridge rectifier (diodes need about 0.6 V across them to conduct).

What is worth a try is feeding the output of your C.T. strainght into a bridge and then putting the burden resistor on the output of the bridge. This may work fine but I have a doubt.

A C.T. is a step up transformer and if the voltage dropped on the primary single turn is 10 mVp-p (for 2 Ap-p flowing) then the output voltage will only be 500 mVp-p open circuit and not enough to drive a bridge.

If you wanted it to work, you could still use a 1 ohm (or whatever the manufacturer recommends in their data sheet for the burden resistor) and then use an op-amp precision rectifier to boost the 40 mVp-p AC voltage across the burden to some more usable DC value.

If you are using the signal from the C.T. to calculate power then you can't use a bridge or any rectifier because the AC waveforms of current and voltage need to be multiplied together and then averaged in your MCU.

Here's a brief reason why the burden resistor has to be about 1 ohm or less on the OP's CT: -



It's a little more complex than this in that the secondary copper loss and leakage inductance are also in series with the burden and this adds another small error - it's a linear error so it isn't too bad but it can account for several percent of the signal. The upshot of all this is make the burden as low as is practically possible - use op-amps to bring the signal back to something that is usable. 

Better CTs have higher turns ratios to avoid the fall-off in signal due to the magnetizing inductance not being thoroughly shunted by the relected impedance of the burden on the primary.