19 May '14 10:501 edit

So I was showing the CEO of our company a simple but redrawn schematic of a three resistor problem, which, when analyses correctly shows it to be just three resistors in parallel.

Now I have shown that problem to potential candidates for a position here and nobody incoming has passed.

I showed it to our CEO and I was surprised he just did the problem by making an assumption without actually going through a circuit analysis and said he remembers parallel resistors and used a formula, R1*R2*R3 divided by R1+R2+R3.

I didn't want to let him down, he's the CEO....

That formula poops out after 2 resistor problems, the best you can get is R1*R1 divide by R1+R2 and for those problems it works every time.

The way we were taught to do three and up parallel resistors is the formula R1 invert +R2 inv+ R3 inv and invert THAT result, which works every time for any number of resistors in parallel.

What I noticed about the R1*R2*R3 version is using 10 ohms apiece for the three resistors, the answer should be 3.333.... ohms, 3 and a third ohms.

But using the triple multiplication formula you get for 10 ohms, 1000 divide by 30 which happens to be exactly 10 times the real answer.

I thought I was on to something there where I could just go divide the whole thing by the resistor value, an added step and it works with 10 ohms but if I do the same with 7 ohms it comes out as 343 (7 cubed) divide by 21 (7X3) and that comes out as 16 and a third. So now divide THAT by 7 and you get 2 and 1/3. No help. It does appear SOMETHING is going on here, 7/2.3333 is 3. Not sure what that means but there it is.

So the question is, what is going on that the double formula works and the triple doesn't and of course quad and quints too.

So another question, is there a reasonably simple modification of the triple and up formula that will work in all cases as good as the inversion formula, is there some trick you can add to the triple and up formulae that would make it work to solve the simple case of parallel resistors of any number and any resistance, except zero of course. Stupid ZeroðŸ™‚

Now I have shown that problem to potential candidates for a position here and nobody incoming has passed.

I showed it to our CEO and I was surprised he just did the problem by making an assumption without actually going through a circuit analysis and said he remembers parallel resistors and used a formula, R1*R2*R3 divided by R1+R2+R3.

I didn't want to let him down, he's the CEO....

That formula poops out after 2 resistor problems, the best you can get is R1*R1 divide by R1+R2 and for those problems it works every time.

The way we were taught to do three and up parallel resistors is the formula R1 invert +R2 inv+ R3 inv and invert THAT result, which works every time for any number of resistors in parallel.

What I noticed about the R1*R2*R3 version is using 10 ohms apiece for the three resistors, the answer should be 3.333.... ohms, 3 and a third ohms.

But using the triple multiplication formula you get for 10 ohms, 1000 divide by 30 which happens to be exactly 10 times the real answer.

I thought I was on to something there where I could just go divide the whole thing by the resistor value, an added step and it works with 10 ohms but if I do the same with 7 ohms it comes out as 343 (7 cubed) divide by 21 (7X3) and that comes out as 16 and a third. So now divide THAT by 7 and you get 2 and 1/3. No help. It does appear SOMETHING is going on here, 7/2.3333 is 3. Not sure what that means but there it is.

So the question is, what is going on that the double formula works and the triple doesn't and of course quad and quints too.

So another question, is there a reasonably simple modification of the triple and up formula that will work in all cases as good as the inversion formula, is there some trick you can add to the triple and up formulae that would make it work to solve the simple case of parallel resistors of any number and any resistance, except zero of course. Stupid ZeroðŸ™‚