On October 31, 2017 at 22:07, Fins said...
The lock and keypad have rj12 connections, and as I understand need six conductors to function. If I could figure out how to put an RJ12 on an 18awg wire, I would do so.
You seem to misunderstand something basic about resistance and wire! Let me list some details to illustrate what to do.
Let's say 24 ga wire has 0.025 ohms per foot. That's about what it has.* Let's say 18 ga wire has 0.005 ohms per foot, because that's about what it has.
If you have a ten foot run of 24 ga wire, its resistance is about a 0.5 ohm. If you had a ten foot run of 18 ga wire, its resistance would be about 0.1 ohm -- 20% of the resistance of the 24 ga wire. But you can't figure out how to put an RJ12 on an 18awg wire.
Don't. Instead make two 6 inch pigtails of 24 ga wire coming out of RJ12s. That's a foot of wire. Splice this into 9 feet of 18 gauge wires. The resistance becomes
1 foot of 24 ga = 0.025 ohms
9 feet of 18 ga = 0.045 ohms
The resistance of this combination of wires is 0.025 + 0.045 ohms = 0.07 ohms. You have to double that as the circuit has two wires, so the resistance of the circuit wiring is 0.14 ohms.
We started with a 24 ga wire with 0.5 ohms of resistance. We now have a cable of 0.14 ohms resistance.
Two things can be seen:
Those pigtails make it possible to drastically lower the resistance, and
Any system of batteries and parts that's susceptible to failure with resistances WAY under one ohm has design problems!
Tell your relative you have to call on a professional since you can't figure it out. Get the freakin' locksmith back out there.
You will suffer momentary shame because of being powerless to help your relative, and
he will never call you to waste your time on something for him for free again. How sad!
*Retrieved from AWG Characteristics file that's been in my "Tech Info" directory for ten years