If one measures resistance from L+ to ground OR R+ to ground with the cable on headphones your ohm meter will show the series resistance value of the headphone driver + cable. That is what I was getting at. The amp sees this total resistance.

Keep in mind, I said resistance (R) NOT impedance (Z). They really aren't the same thing, Z= R+jX. The reactance is really very negligible compared to resistance for a given headphone cable. We are dealing with a GHz signal over coax, etc... Also remember the driver is really going to have it's reactance change as it moves, and this change is way more dominant than the reactance of the cable.

You can model things however you want adding .1 ohms to the output impedance of the headphone amp, but by not including the driver impedance in your model you lose credibility and your assertion is factually flawed.

The amp sees the cable and headphones since they are connected to it (this is obvious). The headphone amp should be modeled as a voltage source. It typically does have a 120 ohm resistor on it's output for highgain loads (like the HD800). This is really a simple circuit as pictured below. Yes you lose some voltage across Rcable (resistance of the cable) so Rhp (resistance of one of the headphone drivers) does not see the full source voltage Vs, but sees Vheadphone instead.

__BUT this voltage change is very small as Rcable<<Rheadphone.__ **In other words it is simply a voltage divider: Vhp = Vs*(Rhp/(Rhp+Rc))** which is almost Vs since Rcable is so small. Replace R with Z if you want, still a voltage divider and definitely the only impedance we are worried about are that of the cable and driver.

PS: your source impedance (e.g. output impedance of the headphone amp) is off, although I would approach this circuit as shown above. The IEC standard is 120 ohms, and we are talking about a high impedance headphone (HD800), so 120 ohms makes sense here not 1 ohm.

DISCLAIMER: I know someone will critique my post again, and yes I am assuming Vs is an ideal source. This is perfectly ok to do here since changing headphone cables is NOT going to greatly impact the total impedance (headphone plus cable) that the headphone amp is seeing (e.g. 320.05 ohms versus 320.1 ohms, pretty sure Vs can keep up with that).

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Your attempt at a logical correlation between the quality of the HD800 stock cable and the quality of the engineers at Sennheiser is a false proposition. You assume quality is the only design parameter. Certainly they could design a better cable if they wanted to, but they have to consider other things like cost, construction time, supply chain availability, sourcing, etc.

**Secondly, the cable's resistance is not applied to the headphone's resistance, it is applied to the load from the source.** If the source is 1ohm and the cable is 0.1 ohm, you're essentially only getting 90.9% of the source signal. Drop that down to a 6ft 10AWG cable (i.e. 0.01 ohm) and you jump up to 99%.

Third, insertion loss is not uniform across the frequency spectrum, so you can't just "turn it up", as you say. It will affect all different kinds of variables from FR, to soundstage, to imaging clarity, bass impact and more. It's not insane nonsense, it's basic math and science.