Calculating the value of Rparallel

[lowell]

So resistors in parallel are the inverse of the sum of the reciprocals:

Rp=1/(1/x+1/x)

Can someone provide a bit of algebra here for me.

Rp=205k

Using standard resistor values how would I got about finding my parallel resistor values.
IE:

205k=1/(1/510k + 1/510k) this obviously will work but I'm out of 510k's.

I've wanted to know this for years and am officially now trying to find an answer.

I have 470k, 560k and most other values. Take your pick.

[NorCalTuna]

You're looking for 205k? easy, just parallel a 220k with a 3m, most builders have those parts around, 204.96k... Parallel Resistance Calculator R1 + R2 = equivalent resistor R equiv total resistor finder - sengpielaudio Sengpiel Berlin or just series two 100ks, then you could have two levels of gain available with a switch depending on where on the "totem pole" you pull your feed from.... ps, I have that link on my toolbar as its faster than math...

[Enzo]

Why make life difficult? None of these circuits are NASA rocket science. How about a standard 220k?

Got a bunch of 220k in a drawer? measure them, you could easily find one that measures pretty close to 205k.

If you are ordering parts soon,, look in Mouser. 1% resistors are not but a couple cents more expensive than 5% ones, and I am sure you can get DARN close to 205k in the 1% pile.

And my calculator doesn;t agree with your math. Two 510k in parallel is 255k in my book.

A pair of 410k in parallel makes 205k, but good luck finding 410k in your parts bin.

How about four 820k in parallel?


Really, just get the parts you need. If you need to dummy up a circuit in the mean time with an odd value like that, use a pot. A plate load for some 12AX7? Use a little trim pot - like a 250k trimmer.

[loudthud]

Rt = R total

The standard formula, 1/Rt = (1/R1)+(1/R2) can be rearranged to be:

1/R2 = (1/Rt) - (1/R1)

If you have a calculator with a {1/X} key, ( brackets {} denote a calculaor key) it gets pretty easy.
Rt {1/x} {-} R1 {1/x} {=} {1/x} answer is displayed.

[lowell]

You da man LT!!

[loudthud]

I was interviewing for a job one time and the guy ask me if I used the sum over the product rule or the reciprocal (sometimes called conductance) rule to calculate parallel resistors. I said I used the reciprocal rule because you can use it to find the unknown in the situation above. I got the job

[tubeswell]

So resistors in parallel are the inverse of the sum of the reciprocals:

Rp=1/(1/x+1/x)

Can someone provide a bit of algebra here for me.

Rp=205k

Using standard resistor values how would I got about finding my parallel resistor values.
IE:

205k=1/(1/510k + 1/510k) this obviously will work but I'm out of 510k's.

I've wanted to know this for years and am officially now trying to find an answer.

I have 470k, 560k and most other values. Take your pick.

1/470k + 1/560k = 1/255k

If you want 205k, try 1/390k + 1/470k = 1/213k, or 1/330k + 1/470k = 1/194k etc

[lowell]

Tubeswell thanks but I'm really sick of the trial and error method. time to really know the math so I don't have to fumble around for 5 minutes to find a value. LTs equation is gonna "fix" this! Yay!

[tubeswell]

Hi lowell - There was no 'guesswork'. I was merely doing a couple of examples using standard resistor values in the reciprocal equation to solve the resistance (hence the '+' sign and the '=' sign) - to put it another way 470k||560k = 255k etc.

(Perhaps I should've said 1/330k + 1/560k = 1/207k)

[lowell]

Ah gotcha. Thanks!

[Enzo]

The problem here is that you want some specific value to result, but you are only looking at combinations of standard values. You can calculate all you want to eliminate trial and error, but what are you going to do when you find the combination for 330k or 470k or 560k is something odd like 238k? You are right back where you started.

Or say you do find a combination of four or five standard values from your drawer. Then you have to shove five resistors in parallel into a circuit. WHat a wad.

[R.G.]

Or say you do find a combination of four or five standard values from your drawer. Then you have to shove five resistors in parallel into a circuit. WHat a wad.

No problem! 8-)

Just do the math by the method of casting out nines.

Well, OK, subtract the conductances of standard values from the intended conductance.

[J M Fahey]

Yes, if you calculate in Siemens instead of Ohms, *this* particular problem becomes somewhat easier.
And series resistors become slightly more complicated.
Oh well.

[R.G.]

Yes, if you calculate in Siemens instead of Ohms, *this* particular problem becomes somewhat easier.
And series resistors become slightly more complicated.
Oh well.

I like to flip back and forth between Siemens and Ohms, to suit the problem. It confuses the interviewer. 8-) It doesn't confuse me because capacitances already require me to think in terms of AC conductance and AC impedance, so I may as well do it to the resistors too.

[jmaf]

I got dizzy reading this thread.

I think mentally I do what loudthud said. I have a 220k and add a 1M, it'll lower 220k by approx. 1/5, because 220 / 1M is approx 1/5, which is somewhere around 40k so you end up with something near 180K.