Imagine you have a 10V RMS AC 100VA transformer. We will assume nice sine waves and that the current drawn is 10A RMS.
We know that RMS measurements create an equivalency in the ability to heat something, and thus they allow us to measure power.
If we assume that current and voltage are exactly in sync, we can deliver 100W of energy. If voltage and current are 90 degrees out of phase, the load is still 100VA, but the power is zero watts. We will initially assume everything is nicely in phase and track where that 100W is spent.
Lets assume we have a perfect bridge rectifier and a perfect capacitor, and a perfect linear regulator.
The capacitors charge up to the peak voltage of 14.14V. Then the regulator reduces the voltage to 10VDC which is passed to a load.
Let's assume the load draws 7.07A.
What is the total power being delivered by the transformer? Well it is equivalent to the voltage on the caps multiplied by the current through the load. 14.14 x 7.07 = 99.97W. And remember that the peak power our transformer can supply is 100W.
In this case 71W goes to the load and 29W is dissipated by the regulator.
Assuming a linear regulator, no matter what the voltage output, the output current should not exceed 0.707 times the rated transformer current.
It's also interesting to note that the actual current drawn from the transformer may peak at 30 or 40 A as ALL the power is delivered in short high current pulses at the peak of each half cycle. The larger the filter capacitors, the shorter duration and higher current these peaks are.
Different transformer, rectifier, and filtering topologies have different multipliers, but bridge rectification and capacitive filtering is so common that the figures get etched into your mind. (a different set would be etched into your mind if you were supplying HV using a vacuum tube double diode, a centre tapped transformer, a small capacitor, and the field coil of your loudspeaker as an inductor).