Does a higher gauge electrical wire impact voltage level?

When the power transformer’s secondary winding voltage is lower feeding the rectifier, due to a quick AC mains VD event, and the electrolytic capacitors voltage is higher, the rectifier will not conduct and the caps do not get recharged for that "(millisecond pulse)" in time.

@jea48  Yes, but this happens all the time. The capacitor's job is to integrate the incoming peak voltages over time. When idling and fully charged, the window of time when the rectifier is conducting is already pretty narrow.  It only conducts on the peaks of the 60Hz waveform.  The capacitors charge, and then, until the next cycle, continue to discharge.  Hence the natural ripple seen there.  If the mains VAC is now a little lower, it may not charge at all, or for a shorter period of time.

So the output of the power supply is in constant flux, especially if you consider how it discharges during musical playback of signals much faster than 60 Hz.  Kind of amazing it works at all. :)

With respect to variations in volts, aren’t all the capacitors in a amplifier designed to mitigate variations in power supply. Ie. Because capacitors store up energy waiting for demands to deal with frequency variations?

The caps are supposed to smooth out bumps in the power supply to a point, but...!

Most amplifiers are "unregulated." Meaning long term (several seconds) changes in the AC line will make it to the amplifier voltage rails. So if your DC rails are +- 50V with 120VAC input, they will vary in proportion to input. They could be 45V at 108V for instance (picking values that are 10% for ease of math).

Also, noise can jump across the power supply caps due to the inherent lack of perfection in the caps. Series resistance and inductance can reduce how perfectly they cut noise out.

Line level devices, like your preamp, CD player, DAC, etc. however are almost always fully regulated.  Meaning so long as the incoming VAC stays above a certain point, the voltage rails the circuits depend on stay locked at their designed voltage, often 5, 12 or 15 Volts.  In these cases, even wider incoming VAC variations won't really change what the working voltage the circuits see. 

OP : Yes, wire gauge can (in theory) impact voltage at the load.   A perfect wire has no voltage difference, and would have the same voltage at the load as at the source, whether we are talking AC power or speakers.

In AC power transmission in a home, 1-2 V per leg loss is normal.  Meaning, 2-4 V drop at the appliance relative to the panel.  A higher gauge wire than required can reduce this. 

As I mentioned before, the voltage loss (i.e. drop) is proportional to the current.  It is also proportional to the wire length.  For the same wiring, a high current device on the end of a long line will suffer more drop vs. a low current or short run.

For speaker cables it is extremely rare to expect voltage drops more than a couple of tenths of volts with anything equal to or better than 14 gauge cables.

@jasonbourne52

Not sure how.

V(dropped) = I(current) * R.

Under load, thicker gauge wiring has lower R for the same length and current.

Hey Emerging, not if there's no current on the circuit. :)  What you are talking about is commonly called "voltage drop."  Meaning, how much voltage drops on one end of a conductor compared to the other end, at the source.

Voltage drop is proportional to current in this formula:

V = I (current) * R (resistance)

That's the formula, but I is in Amps. So, no current, no drop. If the voltage at an outlet with no other current on the circuit is going up and down then it's caused at the source.  If the voltage goes up and down based on devices you turn on, then it's caused by wiring between you and the transformer.  In this case, higher gauge wiring would have lower drop because R would be lower.