I have another switch puzzle (probably not very hard for you).

2 gang switch box has neutral splices in back of box and the following ungrounded configurations:

Switch A is 3W. 2 oranges (travelers) and on the common there are 2 black wires on one screw. One of the two jumper over to....

Switch B, single pole. One screw has two yellow wires (switch legs) the other has 2 black wires (1 being the jumper).

So is the solution the following?

pigtail the switch legs at Switch B.

Power in feeds both switches. So wirenut 4 wires (1 pigtail to Switch A, 1 pigtail to switch B and then power in and power out).


My guess how it got this way in the first place? The idiot previous owner hooked these up with fancy 10 dollar spec grade switches... but forgot to pick up some wirenuts at the store as well. I think the switches he was trying to replace were probably looped like you might see in an old installation. He couldn't reuse the loops so he snipped and doubled up on the screws.

 

If the switches have clamp-back terminals with two slots for conductors at each terminal, then they're rated for two conductors on each terminal. If they cost $10 each, they may have those.

Yep, that sounds like the standard solution.

 

Oh they indeed had the clamping backwire terminals... but the original installer preferred to put two hooks under a screw instead. Probably because they were drunk or blind, Ive seen there backwire clamps on other outlets.... the guy didn't pay much attention to the strip gauge if you know what I mean....

Even if they used clamp backwire terminals, surely that cant be preferable to a pigtail right? Then the rest of the circuit would be dependent of the device.

 

Six of one and a half-dozen of the other, IMO. Continuity to the rest of the circuit is dependent on the integrity of the connection in each terminal, but not on the device(s).