Power Circuit (Commodore 16)
Power enters the C16 through the DC jack on the right of the console, and should be powered by 9.5V DC supporting up to 0.8A.
Ground and VCC pass through the line filter inductor L3, through the power switch, and then through fuse F1 before making its way to the 5V LDO regulator VR1.
Use a multimeter to verify you have 9.5V (or slightly higher) coming in the DC jack to the right side of the L3 inductor (ground top, VCC bottom).
Then check you have 9.5V on the right side of L3 (ground top, VCC bottom) or measure in continuity from bottom left to bottom right, and top left to top right (with the power off) to check the inductor is good.
If all is good, your input power should flow from the DC jack, through the L3 inductor, through the power switch when turned on, and to the start of the fuse on the right side.
After that, check your fuse for continuity or measure for voltage on the left side of the fuse.
If the L3 inductor is blown, replace it or bridge the top and bottom pins across.
If the fuse is blown, replace it with a 2A or close fuse, or temporarily bridge it.
After the fuse the input VCC goes into the VR1 LDO regulator (a standard 7805).
Nothing special here. If its working, you will have 9.5V, and 5V out.
If the regulator is faulty, or there is a short on the output or input it won't work and you need to find the cause.
The regulator also has an external diode. Some old LDO's had a need for external diodes to prevent when VOUT is greater than VIN such as when power is turned off and the charge in the capacitors on the output side have voltage. This diode is D11.
There is also a power resistor between the in and out of the LDO. The reason for this I'm not sure, but it certainly isn't needed and just wastes massive amounts of power.
My best guess is it is to discharge the capacitors on the output side of the regulator after power off, in an attempt at preventing reverse current damage. But if so, this is not the way to do it, and I would personally just remove this component.