Thomas said:

Thanks for the input. Would the circuit vary if 12V instead of 5V would be

used? Also, is that a 22 Ohm resistor in series with the pot? And again, I

really appreciate the help!

If everything was in proportion, then no, the circuit could be the

same. IOW, if the pull-in voltage of the 12V relay was (12/5)*4 =

9.6V, and the its DC resistance was (12/5)*400 = 960 ohms, then the

timing would follow a similar curve.

Yes, it is a 22 ohm resistor. With the pot at its maximum, the total

resistance in series with the big cap is therefore 90 ohms.

Dealing with your other post's query, a formula (which would be

complex, as we're dealing with exponential voltage over time) would be

of little practical value to you. As I said, your relay will not have

the specs I've assumed.

There are also other factors I didn't raise earlier, which will also

have a bearing on the practical results, and reinforce my suggestion

to use trial and error (assuming you want to use this ultra-simple

circuit). In practice I suspect the lid won't be as co-operative as in

my simplistic assumptions. IOW, it won't falsely open for 0.1 - 1

second and then stay closed again for a second or so before opening

again (giving the 4700uF capacitor plenty of time to discharge through

the relay's coil). It will probably be opening and closing to some

extent constantly - almost vibrating - with some jumps higher and

therefore lasting longer than others. That sort of behaviour presents

quite a complex signal at the input to the series resistor.