Voltage doesn't "leak away".The voltage from the power supply measures 5.28 volts but on the PCB there is no voltage. It’s as though it leaks away.
If there was a perfect short-circuit (0 Ω) across the power supply, it would have 0V on its output, since V = I * R. And it would not like that, because it would be trying to supply infinite current. (Though a switched-mode PSU can do a much better job of protecting itself against that situation than a transformer and dissipative regulator.) If there was a partial short-circuit (very low resistance) then there would be a corresponding reduction in voltage. The ferrite core of the switching transformer can only store so many J of energy, and the oscillator only makes so many switching cycles every second, which limits the number of Joules that can be supplied in each second; and a Joule per second is a Watt. If the supply is good for 3A at 5V, that means it can supply 15W. If that was actually its maximum, and something was trying to pull 6A of current, the voltage would drop as low as 2.5V. The voltage might be more if the power supply had a little headroom beyond 15W; but it is fair to say, if there is still 5V coming from the power supply, then there isn't a short-circuit on its output.
So that means there must be a discontinuity somewhere, either in the +5V line or in the ground line.
With everything unpowered, set your meter to its lowest ohms range and measure between the PSU ground and the ground pin of any IC on the motherboard; and then measure between the PSU +5V output and the +5V pin of any IC on the motherboard. Remember that with no continuity, the meter will indicate an overload somehow; usually "1" in the leftmost position or "OL". You're looking for something different, ideally close to 0.
If in doubt, post pictures.
EDIT: Cross-posted with Mark.
Statistics: Posted by julie_m — Sat Mar 15, 2025 12:25 pm