Help with resistors

[fstalfire]

I'm swapping an LS motor into my '71 C10....my fuel gauge apparently is 3o ohm and my new electric fuel pump is 90ohm....a search here for nfo led me to the solution that a 45ohm resistor would fix my issue...my problem now s I've never used a resistor before and when i google search a 45 ohm resistor i find ones you use in circuit boards and no info on what i'm looking for or how to connect it to my wire harness
...so, whar type of resistor do i need for automotive application?
what wattage resistor do i need?
how do you connect it to your wire harness?

[pjmoreland]

If you connect one end of a 45-Ohm resistor to your sender wire and the other end of the resistor to ground, then your gauge will show full when the tank is full. The gauge will tend to show a higher level of fuel than there actually is in the tank for anything between full and empty. For example, the gauge will read half when your tank is actually at 1/4.

Anyway, if your fuel gauge works like a stock gauge, then the 12V that powers the gauge passes through the 30-Ohm resistance of the gauge winding and then it runs to the sender. That means the current flowing through the 30-Ohm gauge plus the 45-Ohm resistor you're adding is passing through a total of 30 + 45 = 75 Ohms. To calculate the current, use the equation V = I x R:

14.5V = I x 75
I = 0.193 Amps

To calculate the wattage for the 45-Ohm resistor use the equation P = I^2R:

P = 0.193^2 x 45 = 1.682 Watts

It's a good idea to go double on the rating of the resistor, so anything above 3 or 4 Watts will be good. 5W is a common size. Here's an example of a resistor:

https://www.digikey.com/en/products/...5F45R3/4297028

You'd have to solder wires to the leads and then cover it with shrink wrap. It's not exactly 45 Ohms, but it is close.

Another option would be to go with something like this:

https://www.autometer.com/fuel-bridg...er-gauges.html

[fstalfire]

Thank you pjmoreland for taking time to reply...you say i'll have to solder the ends to my wires, the only resistors I'm fnding look like a barrel with solid pin type ends....are those the ones i would use?

[pjmoreland]

Yep. You can just solder 18 or 20 gauge wires to the solid metal leads on the resistor. The resistor I linked in my previous post looks like this. You can trim down the leads to make them shorter, if you prefer.

[VetteVet]

If you have not already done so I would direct you to this thread.
http://67-72chevytrucks.com/vboard/s...d.php?t=640615

The first two pics show a stock gauge with a bakelite resistor on the back which is 45 ohms. The OP of that thread has some of those in stock. They vary in resistance and are there to protect the gauge windings. The additional resistance is added via the sending unit in the tank by a variable resistor to ground, on the float arm, from zero (empty) to 90 (full). This is how the stock units work.
I would also direct you to the LS swap forum for additional information.

I have not measure the sender resistance in any of the in-tank pumps that I have changed but they all had four wires in the harness. I assume two for the pump and two for the sender. I'm thinking two of those were for grounds for the pump and sender, and one was the gauge sender wire and the other was for 12 volts to the pump. It seems to me that if you wired a 45 ohm resistor between the sender wire and ground, the gauge would show a half tank at empty. If you wire the resistor in series with the sender wire and the variable resistor to ground, then you would read the 45 ohms plus the 30 ohms with a full tank, but you'd still see the 45 ohms on the gauge for a half tank with it empty.

I know there is a way to make the old 0-30 ohm senders work by adding a resistor but I don't remember how they did it. I will have to do some research on this.

[VetteVet]

This thread is old but it is pretty detailed. They did not mention the sender resistance except that all the gauges up to 1990 were 0-90 ohms so I assume the variable resistors matched that.
http://www.67-72chevytrucks.com/vboa...d.php?t=651699

[pjmoreland]

Originally Posted by VetteVet

It seems to me that if you wired a 45 ohm resistor between the sender wire and ground, the gauge would show a half tank at empty.

I believe the gauge would show empty at empty in this arrangement since the sender would be at 0 Ohms, which would effectively bypass the 45-Ohm resistor (two parallel resistors of 45 Ohms and 0 Ohms come out to a total of 0 Ohms). The gauge will generally read a higher level than is actually in the tank in a non-linear fashion.