If you want 30Vdc, the peak AC voltage has to be 30V + 2 diodes = 31.4Vpk. The RMS is this divided by sqrt(2) = 22.2V RMS. This takes no account of ripple. If you factor ripple into the equation then you can assume the 30V is in fact 30V +/-1V. This equates to half the supply being 15V +/-0.5V i.e. 1Vp-p ripple.

So now the peak AC voltage is 31V + 2 diodes = 32.4Vpk. The RMS is then 22.9V.

As this appears to be the "correct" answer I can see that when you added onto the 30V the peak ripple you used twice as much - remember the ripple is peak-to-peak and only half of that counts towards pushing up the peak AC voltage.

Take a simple example where the sums are trivial. I have a voltage that is on 50% of the time and off 50% of the time. It is 10V when it is on. The average voltage is thus 5V. If I connect a resistor of 1 ohm across it, it will dissipate 100W when it is on and 0W when it is off. The average power is thus 50W.

Now leave the voltage on all the time but make it 5V. Average voltage is still 5V, but the average power is only 25W. Oops.

Or suppose I have the voltage only on 10% of the time, but it is 50V. The average voltage is 5V again, but the power is 2500W when on, and 0W when off, so 250W average.

In reality to calculate power **in general** you have to integrate (instantaneous voltage) * (instantaneous current) over a period of the waveform to get the average (or from 0 to some time t as in your example to find the power over some interval).

**If** (and it's a big if) the load is a **fixed resistor R** you can say that v= i*R, so instantaneous power is i^2 * R and so then you can integrate i^2 over the period to get the "RMS current", and multiply by R later (since it's fixed it doesn't enter into the integral).

RMS current is not particularly useful if the load is something nonlinear like a diode. It can be useful in analyzing losses in something like a capacitor with a given ESR. The losses (and resulting heating effect which shortens the capacitor life) will be proportional to the RMS current, not the average.

## Best Answer

A resistive heating element gets hot when connected across 220V AC mains, even though the mean (average) voltage of the peak voltages (+311V and -311V) is 0 volts. Many loads don't care that the voltage reverses polarity with each half cycle. The 220VRMS value gives a more practical way to estimate the real power delivered to the load. For non-ohmic (inductive or capacitive) loads, there is a power factor correction, but as long as voltage and current are in phase, VRMS = IRMS * Resistance.