I've read that type of thing too. The statements pertain to higher order crossovers where one or more L or C are connected to ground. The proponents seem to think that part of the signal is shunted to ground and therefore (since it is drained off) that path has less influence than the path with components in series.
In my view, that shows a lack of appreciation of the function of the circuit.
If you look at a second order filter, the L and C form a voltage divider with the the speaker attached at the center. In a second order low pass (woofer) the L connects to the amp and the speaker (in series, as it would be in a first order) But the L is loaded by both the speaker and the cap across the speaker,
Let's look at a classic voltage divider with two resistors. Both resistors influence the output of the voltage divider. No one would say that a faulty resistor on the ground side has no effect.
Similarly, the shunt cap in a woofer crossover HAS to influence the output.
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It is interesting to look at the value of the L and C in a second order crossover. You can use the tables at Parts Express. I'm sure you'll find that at the cross over freq, they both have the same impedance. You can do the math of Zl and Zc. I believe you will find their impedance (each) are 0.707 of the assumed load. Therefore, at the crossover freq, both the L and C have equal influence.
In the first order crossover the impedance of the single L or C is equal to the impedance of the load, BTW.
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I always love to point out something which one of our moderators called to our attention years ago. Suppose we have a second order crossover. If we disconnect the speaker, there is a series L-C across the input. These resonate to form an impedance notch down to zero ohms (or the internal resistance of the inductor) at some frequency. It should be the crossover freq IIRC..
Therefore it is potentially dangerous to the amp to run higher order (other than 1st) crossovers with a disconnected or blown out speaker. At some freq, the amp is seeing a dead short.
Gil