The other three switch puzzle: how to wire them all to a singe light

On Sat, 04 Nov 2006 12:08:17 +0000, Nick Atty


Arrggh!
s/on/off/

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Don't you just hate those questions designed to find out if your too technical for the job. You see the question and you think it's unusually complex for the type of people who might apply.
I remember as part of the interview test for a job as an arcade technician there was a question that said "What is the formula for capacitive reactance.) OK, I know it's Xc=1/(2pieFC) but I just put a question mark in the box. I still came across as too technical for the job despite acknowledging that it was primarily board swapping.
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Get a SPDT relay. Hook the common terminal to your output, and the two input wires to the "connected-when-coil-energized" and "connected-when-coil-not energized" terminals. Also run the wire from the "connected-when-coil-energized" terminal to the coil. Ground the other side of the coil.
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Matthew Russotto wrote:

First, the relay isn't in the original bill of materials. Second, you haven't explained well enough for me to( understand how each of the switches can change the state of the lamp.
Jerry
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The relay wiring was in parallel to the existing wires, so the lamp state switching didn't change at all. But others have posted better solutions sans relay anyway.
I l don't see a solution for three SPDT switches controlling a lamp, though two SPDT plus one DPDT can do it easily.
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Matthew Russotto wrote:
...

You and me both!
Jerry
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heinrich snipped-for-privacy@yahoo.com wrote:

For "three way" read "single pole, double throw."
A part of a larger proof that there is no solution. The missing parts are mumbled about in <these>.
<need proof that putting switches on both sides of light does not help>
Consider putting all the switches on one side. So you need to connect or break the connection between two wires.
(rest)----1 (switches go here) 2----(rest)
Let the three connections on a SPDT switch be A, B, and X. A and B are the sides, X the centre. The switch connects A to X, or it connects B to X, but not both. And you can connect wires to A, B, and X.
Consider the case that no X is connected to 1. Something has to connect to 1, or there is no circuit. Say S1 has its A connected. When S1 connects X1 to B1, there still has to be a circuit, so with no X connected to 1 we must have A1, A2, and A3 all connected. (Insert the phrase "without loss of generality" here as required.) Note that we can't have both A1 and B1 connected to 1, because then the switch S1 would not do anything, and could never turn the light off.
Consider now the subcase that no X is connected to 2. By the same reasoning, we must have B1, B2, and B3 all connected to 2.
Now we can easily see this cannot work. We cannot connect any of the X's to 1 or 2, so they can only be connected to each other. But, if we connect X1 to X2, say, then when S1 connects X1 to A1, and S2 connects X2 to B2, the light is on regardless of the position of S3.
So with no X connected to 1 or 2 there is no possible solution.
So we must have atleast one X. Say X1 to 1. If no X to 2, then we must have B1, B2, and B3 all connected to 2. But then when S1 connects X1 to B1, the light is on regardless of the other two switches.
So we can't have exactly one X connected to 1 and 2. Consider two.
A A ---1-X1-< >-X2-2--- B B
This is familiar from the two SPDT switch case. Just connect A to A and B to B and you've got the two switch problem solved. But if you do that you find that the third switch can't turn things off in some cases. So you can't just connect A to A and B to B. Indeed, you can't connect either, as then the light could be on regardless of S3 in some cases. Nor can A1 be connected to B2. Nor can A1 be connected to B1, as then S1 would be shorted, and so not be able to do anything. <I get tired here> Socks
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