Increasing the voltage much beyond 5V to compensate for resistance in the switch and wire is difficult, because of how variable is the amount of current used by the Pi. In particular, the Pi draws between 0.68 and 3 amps which implies it provides a load between 1.67 and 7.35 ohms.bensimmo wrote: ↑Sat Mar 14, 2020 2:55 pmI have no idea, why one wouldn't use and extension lead before the PSU.

I'm just commenting that they would alter the voltage into the longer inline switched wire to compensate for the extra voltage drop. Maybe using a thicker gauge.so it would give 5.1V and a capability of delivering up to 3A at the usb-C end.

'they' being the manufacturers of the PSU.

Question 1. If the voltage of the source supply is stable, what is the maximum resistance of the delivery wire such that the voltage at the Pi end varies no more than plus or minus 10% over the range from minimum to maximum loads.

Question 2. Consider the design of a DC power supply with a feedback control that increases the output voltage linearly as current is increased. Calibrate this control as well as possible to provide a stable 5V potential at the end of a delivery wire that has an internal resistance of 1 ohm. Comment on the practicality of this design.

Question 3. Repeat question 2 but for the case when the wire provides 2 ohms of resistance. What happens if a 0.5 ohm wire is substituted without recalibrating the control in the power supply? Explain.