Veteran electrician gave me a challenge

[Sea Nile]

So, I asked a 36 year veteran electrician to teach me something. He had me map out how to wire 4 or 5 different scenarios and I was doing very well, but this last one stumped me.

The task is this, how to wire (5) 3-way switch's so that the light will only come on when all the switch's are in the up position, or all the switch's are in the down position.

I said it wasn't possible and he drew out the wires to give me a hint.

He says it's possible, but I dont see it. I'm not asking anyone to solve it for me, I just wanted to share the challenge.

 

[wwhitney]

Have not figured it out yet, but the bottom right cable can be a 2-conductor cable, rather than a 3-conductor cable, as the light only has 2 connections. If you do something fancy like connect black and red to one terminal on the light, you could instead in the last switch box connect together whatever is connected to black, and whatever is connected to red, and put those just on the black, say, and not use the red.

Cheers, Wayne

 

[Sea Nile]

It's that last switch that throws a monkey wrench in it for me. If you put two conductors on one terminal then the last switch would always have a connection I think. Unless he's switching the neutral somehow. BTW, this challenge doesn't have to be code compliant, it just has to be wired in a way that satisfies the stated requirements.

 

[Eddie702]

I am too lazy to figure that out. It should be 2 3 way switches with 3 4 way switches between the two three ways

 

[hillbilly1]

The multiple wires are red herring’s. 12-2 to the first box, 12-3 to the next, the white goes straight thru, and is not connected to any of the switches. On the 12-3, the black goes to the top traveler, the red goes to the bottom traveler. The same on the second box.Then a 12-2 goes to the third box, the black comes from the common on the second box, then to the common on the third box. Another 12-3 comes from the third box, to fourth box repeating the same pattern of connections, 12-2 from the fourth box to the fifth. The black on the common in the fifth box, a jumper between the traveler screws on the fifth box, with a 12-2 going to the light (black also attached to the traveler terminals going to the light.) (12-2 is just being used as a reference, can be any gauge, or set of wires)

 

[wwhitney]

That's going to power the light whenever S1 = S2 (both up or down) and S3 = S4. So 8 different combinations. But the goal is to light it only when S1 = S2 = S3 = S4 = S5, which is 2 different combinations.

Cheers, Wayne

 

[hillbilly1]

It's that last switch that throws a monkey wrench in it for me. If you put two conductors on one terminal then the last switch would always have a connection I think. Unless he's switching the neutral somehow. BTW, this challenge doesn't have to be code compliant, it just has to be wired in a way that satisfies the stated requirements.

Your right, it should be six switches for that to work. Even switching the neutral on the last switch would not make it work.

 

[Buck Parrish]

It doesn't make any sense. It wouldn't be prudent. Next, he'll tell you to retrieve the wire stretcher out of the truck.

 

[hillbilly1]

It doesn't make any sense. It wouldn't be prudent. Next, he'll tell you to retrieve the wire stretcher out of the truck.

That or sky hooks!

 

[hillbilly1]

I think I figured it out, he is using a California three way at the end, while flipping one of the two three ways around to match the other throw positions. Sneaky!

 

[hbiss]

The first thing you have to do is learn to draw schematics so you can see how everything is wired! I hope that crayon drawing didn't come from that 36 year veteran!

-Hal

 

[GoldDigger]

Note that you are not using any 4-way in this problem. (And using them would make it harder to limit the number of ON states.)

 

[winnie]

I think I see the code compliant version, but have to draw it out.

If I am correct, then the answer is related to the puzzle about running a three way setup so that you switch loads at both locations (say a switch and a light at both house and garage) and provide an always on supply with only 4 conductors between the two locations.

Jon

 

[4x4dually]

I'm confused as to why anyone would give a challenge that is definitely not a real life scenario. What is there to learn from this? I'm so lost. I may have to figure this out just to see how absurd it really is. LOL

 

[4x4dually]

I give up. It's rheetartid. LOL

I can set the first two switches up as normal 3-way for the line side and the second set of two up as a 3-way for the neutral side and it will work. The 5th switch is the bastard.

 

[d0nut]

It does look like there are may more conductors shown than are actually needed. Also, the solution is the same for any odd number of 3-way switches, so if you can solve for one or three switches, you can solve for five.

Spoiler

 

[david luchini]

The task is this, how to wire (5) 3-way switch's so that the light will only come on when all the switch's are in the up position, or all the switch's are in the down position.

I said it wasn't possible and he drew out the wires to give me a hint.

He says it's possible, but I dont see it.

He's shown extra conductors to throw you off.

Edit: or what d0nut said.

 

[4x4dually]

I don't think the "spoiler" above meets the exact criteria of the challenge. The challenge says the light will ONLY be on when all switches are UP or DOWN.

If it were that the light will be on when all switches are up and all switches are down AND a few other combinations, then the "spoiler" holds true.

 

[4x4dually]

So, in the spoiler, if you switch both sw3 and sw4 together, the light will be off?

EDIT: Never mind. I just noticed the colors. Black wires and blue wires. That was dumb. I totally missed that. Carry on.

 

[d0nut]

So, in the spoiler, if you switch both sw3 and sw4 together, the light will be off?

EDIT: Never mind. I just noticed the colors. Black wires and blue wires. That was dumb. I totally missed that. Carry on.

No, you are correct. The switch pairs of S1 and S2 or S3 and S4 would need to be in the same position, not necessarily all up or all down. My proposed solution fails the test. I will have to think about it some more.