I think the essential nature of the question has been missed. The power is coming from a rectified AC supply so the capacitor has to do more than just a little extra filtering of a DC supply, it has to supply the bulk of the current to the circuit during the "valleys" of the AC waveform.

As a first approximation, say the 14VAC supply is RMS, so the peaks are 14*sqrt(2) = 19.8V, minus two diode drops for the bridge (0.7V each) gives 18.4V. Say 18V to be conservative. So this input cap charges to 18V during the peaks and can't drop to any lower than 7V so the 7805 doesn't drop out.

This means the cap's voltage goes from 18V to 7V over an 8.33ms period (one cycle of the rectified waveform). Not exactly, but it's conservative. Given the current draw of 30mA, with a little extra 5mA for the 7805 we run the calculation C=i*dt/dV=0.035*0.00833/(18-7)=26.5 microfarads.

I would start with a 33uF capacitor.

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