Quote:
Originally Posted by astral highway
...in the instance described in the opening paragraph, Q is far less significant determinant than the relationship between inductance and capacitance. Simply, the smaller the capacitance for a given inductance, the greater by far the peak voltage.
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Absolutely! If the Q is infinite, the exponential term is just 1, and the equation simplifies to just Volts = Amps x Ohms.
The Ohms is just the reactance of the inductor (or capacitor) at resonance. You can easily prove this by working out the resonant frequency (F = 1 / (2 x pi x √(LC)) ), and putting the frequency into the formula for reactance (X = 2 x pi x f x L). Or do a bit of easy algebra, and you find things just drop out and you get X = √(L/C).
If Q is less than infinity, it's all multiplied by a very slowly decaying function. But unless Q is very low indeed, it hardly decays more than a couple of a percent in the first quarter cycle. So, Q has very little significance on the peak voltage, but inductance and capacitance matter a lot!