Ohm’s Law Calculation

I've never been proficient with Ohm's Law regarding Resistance. Thanks for any help. I have a relay coil of 82 Ohms. System voltage is 13.5 Nominal. System voltage on #86, #85 to ground. .164 Amp. I can do this calculation. Now: I want to put a resistor in series with #85 and ground. Then insert system voltage between #85 and the resistor. B+ to B+, the relay opens. I pick 10 volts across the relay coil for reliable-enough closure. Hopefully the resistor doesn't have to be too large a wattage to sink system voltage to ground, nor the current in the "insertion" wire too high. Here's a visual: 13.5V--82 Ohms--10V--R=?--0V. What is the value of the second resistor "R=?" ?Even better, how is this calculation performed? I have two resistances in series, and know what the desired voltage in the middle of the two resistances is. I can't figure out how to calculate the value of the second resistor. Thanks.

If you want to go from 12V to 10V, create a voltage divider with two resistors. Voltage source positive to resistor 1. Voltage source ground to resistor 2. Resistor 1 to resistor 2. Voltage positive output where resistor 1 and resistor 2 connect. The resistance of the coil of the relay is not relevant to the voltage divider, but I'll use 82 ohms as the value of resistance 1 in my example. Voltage out = (voltage source x resistance 2) / (resistance 1 + resistance 2) Vout = (Vs x R2) / (R1 + R2) 10V = (13.5 x R2) / (82 + R2) R2 = (10 x 82) / (13.5 - 10) R2 = 234.2857142857143 R2 = 234.286 ohms the12volt • Support the12volt.com

Just re-read your post and a bit confused as to what it is you're trying to do. Are you wanting to use the relay coil as R1 to get 10V between #85 and R2, or are you wanting to send 10V to #85? the12volt • Support the12volt.com

Thanks for the tutorial. I'll mess with it and see if I can figure it out with your help. Thanks. What I'm trying to do may not work for a relay coil. It works for light bulbs. You may be familiar with the way GM did front side marker lights in the 80's and 90's. The marker bulb was a 194. One side was to the tail lights, and the other to the signal bulb. When the tail lights were on, the 194 grounded through the signal filament. When the tail lights were off and the turn signal was on, the 194 grounded through the tail light circuit. When both were on at the same time, the 194 went out. B+ on both sides = 0-Volts. My idea was to do the same thing with a relay coil, BUT not ground 85 through my "injection wire". I am trying to size the second resistor so that the relay coil sees 10V, so it has enough voltage to close. The second resistor will be between my "injection voltage" and ground so as not to present a direct short to ground on my "injection wire". As I mentioned, I use the B+ to B+ = 0-Volts strategy from time to time. It works pretty well with light bulbs. But may not with the relay coil. Thanks again for the interest.