LED7. TP1. Consider checking F116 (like the message says).

12vu.jpg

Check the voltage immediately after power on when the machine has been off for a while. If the voltage is in the expected range (~70V) watch the voltage and see if it rises over time.

You do not have "electron vision" so you cannot definitively say a fuse is blown or not by simple inspection.

You must measure with a multimeter to be sure a fuse is blown or not. Always isolate the board at the very least (disconnect all connectors) or isolate the fuse itself (lift one side out of the holder or clip). Only the multimeter has "electron vision".

This is normal. It is unregulated. There are two "+12V" rectification or regulation circuits on the board and unfortunately Williams decided to indicate a nominal voltage of "+12V". These are far too often confused by many people on this forum.

I have always wanted to use the emoticon. Very well done. You searched and you found the information. I was hoping you would do that. I see far too many people ask questions that have been answered MANY times over on the forum - i.e. they don't search. I pretty much stopped answering questions that have been asked many times before. Two decades ago I was a member of a forum that had a strict rule that you could not ask a question that had been previously asked and answered before. They strong encouraged using the search feature. I agree. Waste your own time - not somebody else's. I gave you a small hint and you took it. I am truly impressed.

Now go replace that open resistor. This problem is somewhat common on the board. The resistor is underrated for power dissipation and it cooks slowly over time. Replace it with a 15k or higher (even up to 27k) Ohm 0.5W resistor. Don't replace it with a 10k. It will likely slowly cook over time again and result in the same failure.

You have reached the limit of my knowledge. I only did high school physics which is basically V=IR. All I know is that R260 is the "bleed" resistor for the 50V supply (see schematic snippet below). The power dissipation is P=V^2/R and the nominal voltage is 70VDC.

P=V^2/R
P=70*70/10000
P=4900/10000
P=0.49

This means 70VDC dissipates 0.49W but the resistor is only 0.5W. The 0.5W rating is maximal and ideally (from what I have read) power ratings for the resistor should be double actual (computed) power dissipation.

r260.jpg

How does this cause the voltage to slowly rise over time? No clue. You'll need to consult someone like zaza or G-P-E who have vastly more electronics knowledge than me. I'm only guessing that it may have something to do with AC ripple and RMS but I have no clue and no evidence for that.