The two halves of the crossover are always in parallel either at the speaker end in single wired or the amp end in bi-wired so the nominal impedance will always be maintained. The cable resistance would be halved in a bi-wired installation but should be negligeqble.
Since a 'biwirable' crossover is electrically complementary, the nominal impedance should be expected to stay the same. 8ohms single-wired should be expected to remain 8ohms in each half after splitting.
The fact it may not is down to speaker designers and their own cavalier assumptions about voltage sources...
TLS - no it doesn't... bi/tri-wirable crossovers split out 0v returns also.
In any case all voltages are differential
That means both halves would have common ground and that would blow up some amplifiers because they cannot be runned in parallel!
If you mean you cannot put 4 ohms on some 8 ohm amps - thats not what happens.
Not exactly.
In a simple first order ''normal polarity'' crossover, the positive signal is split. One end of the positive signal goes to the capacitor then goes to the + terminal on the tweeter. The other end of the positive signal goes to the inductor then goes to the + terminal on the woofer. Both have a common ground (negative) that is connected to the negative side of the binding post.
If you add a second set of binding post, the negative from the tweeter and the negative from the woofer is separated.
Tis true..but they meet at the amp. The drivers are therefore star earthed and star driven. You can only try it and see. If anything it should increase the differences between the drive units.