So you are saying that what we see in that circuit is false? That the input energy from the PSU does not cause the resistor to consume energy?
What I'm saying is that you can not connect just one end of a resistor to one terminal of the battery and expect a current except for the one needed to charge the newly created capacitor that is made up by the other battery terminal and the resistor with air being the dielectric.
It is about how current flows in to a capacitor when being charged and not through a capacitor.
Say you have an ideal voltage source 12V so there is no internal series resistance considered for this ideal voltage source.
Then say you have two wires each 0.5Ohms connected to this 12V supply and then you short the ends of those wires you will get 12V/ (0.5+0.5) =12A of current. This 12V * 12A will result in 144W energy wasted as heat so each second 144Ws (144J same thing) or 40mWh again same thing.
So 144Ws delivered by the source and 144Ws end as heat nothing stored.
Then say you connect a 5Ohm resistor and energy will go through wires and through resistor toal 6Ohms so 2A * 12V = 24W in one second it will be 24Ws delivered by the source and 24Ws all end as heat dissipated by wires and by resistor as energy travels through them.
Now say you leave the wires and instead of a resistor you connect a capacitor say it is 1000uF = 1mF and say internal DC ESR is basically zero to keep things simple.
a) How much energy will power supply deliver in one second after the discharged capacitor was connected ?
b) How much energy will end up as heat dissipated by wires?
c) How much energy is stored in the capacitor after this second?
d) How much energy is delivered by the source in the next second? so after first second ended and another second from that point.
Answer correctly to all this questions and you should understand what happens. Answer wrong and I may be able to help you understand what part you got wrong.