Does Ohm's law work the same way for a potentiometer if one is to put
another resistor in parallel with the pot?
Suppose I have a pot with a maximum resistance ohm of 500OHM. But I want to
reduce this maximum to a lower value say around 85ohm. According to ohm's
law Resistance(total) = 1/( (1/Resistor1) + (1/Potentiometer)). Therefore
I could substitute Resistor1 with a 100Ohm resistor. And now I can vary the
Potentiometer resistance and I would get a range from 1Ohm to 83.3Ohm as
shown
You can also use the formula R = R1*R2/(R1+R2), which works fine if
either resistor is zero.
And if you want to know what resistor to put in parallel, you can use
Rp = 1/(1/R - 1/Rpot)
Eg. pot 500 ohms, you want 85 ohms, so 102.4 ohms.
Check: R = 102.4 * 500/(602.4) = 84.99 ohms
Now, the problem with what you are doing is this-- look at the
linearity of the resulting compound pot:
Pot % Rpot R Ideal pot value
0 0 0.00 0.00
5 25 20.09 4.25
10 50 33.60 8.50
15 75 43.29 12.75
20 100 50.59 17.00
25 125 56.29 21.25
30 150 60.86 25.50
35 175 64.60 29.75
40 200 67.72 34.00
45 225 70.37 38.25
50 250 72.64 42.50
55 275 74.62 46.75
60 300 76.34 51.00
65 325 77.87 55.25
70 350 79.22 59.50
75 375 80.44 63.75
80 400 81.53 68.00
85 425 82.52 72.25
90 450 83.42 76.50
95 475 84.24 80.75
100 500 84.99 85.00
As you can see the last 5% of pot rotation changes the parallel value
as much as half the rotation of an ideal pot. So the pot setting might
be finicky, unstable, noisy etc. if it happens to be down in that
range.
Best regards,
Spehro Pefhany