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The capacitor quickly absorbs inductive kick at turn-off. The amount of webers (A * L) is converted to a certain increase of volts in the capacitor, depending on its Farad value. Then it discharges through the resistor during the remainder of the cycle.
If capacitor value is too small, then it charges to a very high voltage suddenly possibly exceeding its safe rating.
If the resistor is too great, then it fails to discharge the capacitor enough. The capacitor may rise to a destructive volt level sooner or later.
If the resistor is too low, then the primary winding may continue to push current wastefully around a loop which includes the resistor and diode.
Resistor watt rating must be chosen to endure heat dissipation as it draws current from the capacitor.
To find a sensible RCD combination, it's a tradeoff as to what amount of energy comes from the transformer primary as it shuts off, versus what voltage you wish to accumulate on the capacitor, versus the watt rating of the resistor.
The formulas take energy from the transformer, and convert it to energy stored in the capacitor. I believe the formulas have units of joules on one side of the equation.
We might speculate that if you have 1 Henry carrying 1 Ampere, it is the same energy as 1 Farad charged to 1 Volt. However the calculation is not linear. It is not the same energy as 1uF charged to 1 million volts.
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