At 09:27 PM 7/11/2011, Shawn Wright wrote:
>I have two big 120ohm resistors, and a 470ohm in series with the
>LED. All three check out ok, yet the resistance between terminal 8 &
>11 is 540ohms, not the 160 it should be. Or is that also an old spec
>in Bentley?

 From the diagram for the '88, 97.126 tracks 52-4...it shows two
resistors in parallel with the LED/resistor pair.  If each of those
resistors is 120R, the effective resistance of that pair is 60R.

The diagram shows a diode in series with the two parallel resistors
and LED/resistor.  That will make ohmmeter readings a matter very
much dependent on the individual characteristics of the particular
ohmmeter in use.

Put +12 on the panel and see how much current you can draw from pin
11.  Roughly:

You're starting with 12-12.5 V.  Deduct .6 V for the series diode,
that leaves 11.5-12.

I = E / R.  Taking into account just the two 120R resistors, I =
11,5/60 = .192 A or 192 mA.  Add 15 mA or so for the LED/resistor
pair to get 200+ mA.

If you can only pull half that then one of the 120R is open.  If only
~15 mA then both are open.

Easiest way may be to unplug the fridge relay and alternator D+, then
plug in the panel and ground either of those leads through your milliammeter.

Power dissipated would be P = E^2 / R = 12^2 / 120 = 144/120 = 1.2 W
at each of the 120R resistors, so I'd take your measurement promptly
and disconnect.

To track which resistor is open, put a pea bulb of some sort instead
of your milliammeter, then use the voltmeter to track voltages on the
resistors.  The one(s) with the same voltage at both ends is the open one(s).

Yours,
David