Electrical – Elevated voltage after a filtering capacitor

The answers already here are correct. I just wanted to add that since you are using a Full-Wave Rectifier (4 diodes), your ripple voltage is determined by:

$$ V_{ripple}=\frac{V_{peak}}{2fCR}$$

or equivalently, $$ V_{ripple}=\frac{I_{load}}{2fC}$$

Now, the thing here is that you want to keep the ripple voltage 'small.' A good number is within 10% of \$V_{peak}\$. From your post, I see \$V_{peak}=13.8{\mathrm V}\$

If you do the math, in order to keep the ripple voltage, say, 10% of \$V_{peak}\$, the maximum current you can draw from the rectifier is : $$ I_{load}=2(1.38{\mathrm V})(50{\mathrm{Hz}})(4700\mu{\mathrm F})$$ $$ I_{load}=0.65{\mathrm A}$$

That is the maximum current you want to draw if you want keep the ripple voltage at a maximum of \$V_{ripple}=1.38{\mathrm V}\$ or 10% of the peak voltage.

I saw in one of your comments that you were concerned about damaging components. You won't damage them as long as they are rated to handle the voltage/current they are being provided with. The capacitor should be fine (since it is 50V), but they may be a lot of current running through the circuit at startup (when the capacitor is fully discharged). So you want to make sure your diodes are rated to handle that much current, and also check the reverse voltage ratings for those diodes.

The reverse voltage across those diodes are theoretically the same as the peak voltage for a full wave rectifier with four diodes (in your case the diodes should handle more than the 13.8 volts you are getting at the output).

1. The sites formula is correct, but only under ideal conditions. The AC input must be a clean sine-wave. Some UPS's and DC-AC inverters put out a choppy sine wave that would make the 1.414 ratio of RMS value to peak value not true.

2. This peak voltage assumes no load, whether a single diode is used or a bridge rectifier, plus capacitor of sufficient value to remove any AC ripple. Even a tiny load of .1% of capacity will drop the voltage by the amount the diodes dropped. So subtract .7 volts or 1.4 volts from the expected peak, and the numbers should match better.

3. With heavier loads a bridge or full-wave rectifier will provide the most current. At high current levels >10 amps the Vdrop across each diode can be 1 volt. The output voltage will drop as the load increases until a full safe load is reached.

4. By now the peaking effect is gone and the DC voltage is more like the AC-RMS value. The RMS-to-Peak formula is correct but only under ideal conditions, and as you might have guessed by now, real-world conditions dominate once you apply a load or use a inverter/UPS for AC power, making the RMS to Peak formula useless, especially under heavy loads.

5. @ElliotAlderson made an important observation;

You are missing the point of the formula if you think it is wrong. The formula tells you the maximum dc voltage you can achieve from a given rms ac voltage. That value is very important when it is time to specify the voltage ratings of your components, such as the filter capacitor. The formula is very useful if you understand what it means.