How does the current flow in this full-wave filtered DC power supply

I am reading an introductory electronics book. At some point there is this power supply and I couldn't quite understand.
I understand that when the switch is open, the capacitor is getting charged until its full. So electrons concentrate on the bottom side of the capacitor (in the case of this diagram) and positive charges resides on the top side.
However, when the switch is closed, the book says, the capacitor is being discharged by the load. I assume the electrons flow from the bottom of the capacitor to the load through the ground and then go to the upper side of the diagram and go through the switch. (Does the current actually flow through the ground?) However, my another question is, if the current flow through the ground, why doesn't the electricity supplied by the transformer go to the load directly? It seems really confusing when the book suggests the capacitor is being charged and discharged while it looks much more convenient for current from the transformer to go to the load directly and the voltage in the capacitor stays unchanged.
Thanks.

What happens is that the current the diodes are supplying varies throught the mains cycle. If there was no load, once the capacitor has charged up there would be no current flow through the diodes as the capacitor has charged to the peak (max) voltage and the transformer only supplies this for a short time, the rest of the time the diodes would be trying to supply a lower voltage which they can not.
If a load is put on the capacitor, it supplies current when the diodes voltage is too low, but the caps voltage decreases, like a slope from say, + 12V to 10V. By which time the voltage on the transformer has risen and the diodes then pump up this 10V back to 12 V. It is only during this time that the diodes are actually conducting current into the load and capacitor. The voltage from the transformer then falls, the diodes stop conducting and the capacitor then provides power into the load by it self and its voltage falls again from +12 to +10V.
The fall from 12- > 10V i.e. 2V is called RIPPLE and is an important factor in a power supply. If the load was doubled (twice the current), the voltage on the capacitor would fall from +12 to +8, so the ripple has doubled to 4V (peak to peak). The diodes will start to supply current earlier (transformer = 8V) but will still stop conducting at 12V.
Frank
Frank

Hi Frank,

But doesn't the transformer connect to the load through ground? Why doesn't the transformer keeps supplying current to the load and ignore the capacitor?

'Ground' is an often confusingly-used and misused term.

The circuit ground in your schematic (indicated by the two standard symbols) is not necessarily connected to the real Earth 'ground'. Circuit ground and real ground are separate things.

All that the schematic indicates is that the two points with the ground symbols are connected to each other, on the PCB or by wiring inside the equipment.


Even if the circuit ground is connected to the Earth, the two ground points in the schematic are still connected by actual conductors in the equipment and so nothing will need to flow through the Earth.