In what way do passive circuit elements change the functional form of the voltage?

Question

I heard capacitors affect the valleys and mounds of voltage sine curves, so that you get DC from AC. It's related to Graetz bridge, flipping signs of sine waves and seemingly afterwards smoothing sharp points, see pictures below.

How can that be derived and interpreted from electronics? What is the math behind capacitors in a parallel configuration of an AC source $U_\text{ext}$ leveling to give flater the output voltage?

For example the images suggest the capacitor provides the current at the right time, but I don't know why. If charges when the current is high, but how to compute the current at the vertex at the parallel capacitor and the partition to the three paths. The picture suggest the process is periodic and so I guess there might be a closed form for the time behaviour of the relevant functions.

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My approach:

$I_{U_\text{ext}}=I_C+I_R$

$U_R=I_R\ R$

$U_C=\frac{1}{C}\int I_C \mathrm dt$

$U_\text{ext}\overset{?}{=}I_{U_\text{ext}}\cdot \frac{1}{1/Z_C^{???}+1/R}$

$U_\text{ext}\overset{?}{=}\pm U_C\overset{?}{=}\pm U_R$

$U_R=\ ???$

Answer 1

The point is that the DC comes from the rectification (the diode/bridge is forcing current one way). The capacitor just behaves like a capacitor does normally, nothing special. When the potential is above the potential $Q/C$ then the capacitor will charge. When the potential drops below this value, it will discharge. By analyzing the flow of current across the load you can see that this extra source of current has a smoothing effect on potential (and current, for that matter). How much the capacitor is able to smooth transients depends on it's time constant relative to the time-scale of the transients.