I made a few measurements on an older OMC TRIM Gauge regarding the resistances. We have seen in ETEC literature that voltage for the TRIM circuit is supplied in the ICommand Wiring Harness by a 47Ohm resistor connected between the VIOLET and WHITEwithBROWNstripe conductors.
On the older TRIM gauge there are three terminals for the gauge: I, S, and G. There is an external resistor connected between I and S. I removed this resistor and measured it; the resistance was 84Ohm. With the external resistor removed, I measured the resistance between I and S as 96.6Ohm. When in parallel, the combination would have a resistance of about 45Ohm. This is apparently why there is a 47Ohm resistor used in the ICommand Wiring Harness; it duplicates the resistance that would be in the circuit if a conventional trim gauge were used. A value of 47Ohm is a standard value, and close enough to 45Ohms.
I also measured the resistance of the gauge between S and G: it was 165.5ohms. The external resistance of the TRIM SENDER is connected across these same terminals. The TRIM SENDER is typically a 0 to 100 ohm rheostat. When at 100ohms the combination with the internal resistor would be about 62Ohms.
I also measured the resistance from I to G (without the external TRIM SENDER); it was 210.1Ohms, or the sum of the resistances form I to S and S to G. This suggests that the resistances are in series. With the TRIM SENDER at 100ohms, the resistance from S to G would be about 62Ohms. This implies the total resistance I to G with the TRIM SENDER at 100Ohms would be about 45 + 62 = 107Ohms.
Next, I applied 13.2Volts to the gauge at I and connected a 100Ohm resistor between S and G to simulate the TRIM SENDER. I measured the current flow: 0.12Ampere flowed into I and returned at G. According to Ohm's Law, the resistance of the circuit was therefore 13.2/0.12 = 110Ohms. This is a good agreement with my measured resistances.
With the TRIM SENDER at 100Ohms, the TRIM gauge read near the "DN" end of the scale.
Next I put a 1.5Ohm resistor between S and G, to simulate the other end of the TRIM SENDER range, near 0Ohm. With 13.2Volts applied the TRIM gauge read "UP" and I measured 0.27Ampere of current. This is in good agreement with the expected current, which would be 13.2/46.5 =0.284Ampere. (My meter has only a 10Ampere scale, so there can be some error in these low current readings.)
This narrative is a hard way to describe things. Let me try to draw with ASCII. I will show the external connections to the TRIM gauge:
13.2 Volts>(I)84Ohm(S)100Ohm(G)Power Common
This produces 0.12A flow and the meter reads "DN."
Next:
13.2 Volts>(I)84Ohm(S)1.5Ohm(G)Power Common
This produces 0.27A flow and the meter reads "UP."
It is easy to see that the TRIM SENDER is close to 100Ohms when in the DN position and close to 0Ohms when in the UP position.
What is not apparent to me is which terminals of the gauge actually have the meter connected to them. To test this, I modified the circuit. I did not connect anything to (G). Instead, I just took the 100Ohm resistor lead from (S) and connected it to Power Common:
13.2 Volts>(I)84Ohm(S)(nothing)(G)

L100OhmPower Common
To my surprise, the TRIM gauge pegged hard to the UP position.
I had to scratch my head on that one. I had been expecting the meter to be connected between (S) and (G), and when I removed the connection from (G) there would be no current in that part of the circuit, which would imply that the meter should not move at all.
Where is the meter movement connected? What is going on inside the TRIM gauge?
Pondering what might be inside the TRIM gauge, I came up with the idea of the meter being arranged in a bridge circuit. The circuit seems to be like this:
The use of an external resistor across the I and S terminals always had me puzzled. But this is really a very good idea. It allows the manufacturer of the gaugewhich is very likely not Evinrude themselvesto make a universal gauge. The value of the resistor installed externally at I and S is related to the range of the TRIM SENSOR that is connected between S and G. The internal resistors values are selected to work well with a range of external resistors values. This allows the gauge to be used with various ranges of TRIM SENSOR resistances. For example, suppose another outboard engine manufacturer used a TRIM SENSOR that varied from 5 to 200Ohms. By changing the value of the external resistor connected between I and S, the same trim gauge could be used with that new range of values for TRIM SENSOR.
The bridge arrangement also explains very nicely why the meter reacted so unexpectedly when the ground connection to G was removed. The bridge balance was completely upset and much more current flowed through the meter. That is why the meter pegged so hard.
I would love to disassemble an old TRIM gauge to see if the internal circuitry is the same as my hypothesis. If anyone has an old TRIM gauge that no longer works, I
MORE ABOUT RESISTANCE MEASUREMENTS
The resistance of the coil of an ammeter is typically very low, perhaps only a fraction of an Ohm. When I measured the resistance from the three terminals (with the external 83.5Ohm resistor disconnected), I was probably measuring the coil resistance of the meter in those measurements.
If
RM = resistance of the meter
R1 = the internal resistance of the upper arm of the bridge
R2 = the internal resistance of the lower arm of the bridge
Then
I to S = R1 + RM = 95.8Ohms
S to G = RM + R2 = 166.2Ohm
I to G = R1 + R2 = 262Ohm
Solving those relationships leaves RM = 0, which may be close to the truth. The resistance of the meter coil is very low, perhaps only 0.1Ohm. It may be too low to see in the measured values unless very accurate resistance measurements are made.
Another clue that perhaps the meter is wired into a bridge circuit is the meter itself. It appears to be a Zerocenter meter, that is, a meter that shows current flow in two directions, with the rest position in the center. When there is no current flow in the meter the dial pointer drifts toward the center. would love to take it apart to see exactly what is in there. I do think the bridge circuit I describe (above) is very likely what will be found.