how to use CT to measure current
If the transformer is already putting out 1.08 V (that's with the specified load resistor, right?), then you don't need to do much more.  1.08V RMS is 3.1 V peak to peak.  That's enough range to go directly into a PIC A/D input without loosing much resolution.  However, the voltage needs to be centered about the middle of the A/D range.  Here is a simple way to do that:



T1 is the secondary of the current transformer.  R1 and R2 set the DC level, which will be in the middle of the A/D range.  C1 provides a little filtering against power supply noise.

You don't want to make C1 too large.  The parallel combination of R1 and R2 are a impedance effectively in series with the transformer.  If a sudden current spike occurs in the line being sensed, the output of the transformer could exceed the PIC's supply range.  Since the transformer output has low impedance, this can do bad things to the PIC, including blowing out the protection diodes, which then lets the voltage go higher and blow out other stuff.  The impedance in series with the transformer helps these rare cases not cause physical damage.  Since C1 lowers the impedance, you don't want to make it too big.  There are other better schemes for dealing with spikes, but this is a simple circuit that is easy to hook up.