Throwing stones into the water, look at the circles they form, otherwise this throwing will be an empty fun.

Consider the simple task of increasing the accuracy of a resistor using a parallel connection of two resistors, one of which we call the base one and denote r, and the second compensate and denote R. Just note that both resistors are constant and in no case variable. The reasons for this decision (do not use a variable resistor) will be left out of the discussion brackets, although this may be overall characteristics, cost indicators, etc. In the practice of developing such a solution is found repeatedly.

The procedure for setting the value is as follows: set the resistor r, measure its resistance, determine the nominal trim resistance necessary to create an equivalent resistance equal to the reference one, take the nearest nominal value and set it. Perhaps you had to do this with a real scheme.

And the question immediately arises - why does it work? After all, if r we measured and know absolutely exactly (within the limits of a measuring instrument error), then R we simply take from the cash register and its accuracy is determined by the accuracy of the resistor manufacturing. To answer this question we will involve a little mathematics.

Equivalent resistance formula for parallel-connected resistors r`0 = r * R / (r + R),
set the error R in the form of dR, we get r` = r * (R + dR) / (r + R + dR). Then dr` = r`-r`0 = ... = r * r * dR / (((r + R + dR) * (r + R)) is an absolute error, we obtain the relative error qr` = dr` / r `0 = dR * r / (R * (r + R + dR)),
considering dR = R * qR, we get qr` = qR * r / (r + R + dR) ~ qR * r / (r + R).
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It is easy to see that the relative error of the equivalent resistance is r / (r + R) times less than the relative error of the resistor R, and if you make r << R, then it will be significantly less. That is, if we initially strive for the situation r << R, then we can really create a very accurate resistor from those that are not too accurate.
In this case, to compensate for the deviation of the resistor r from rE, we need R = r * rE / (r-rE).

The question arises about the achievable accuracy, for which we consider the worst case. The starting point is the choice of the resistor r, which should be as close as possible to rE, since in this case, R is the maximum. Since for r <rE, we cannot compensate for this deviation (we will need R negative value and I do not really understand the physical meaning of this case), to ensure the absence of such a regime, we should choose r0 = rE / (1-qr).

Then the worst case is r = r0 * (1 + qr) and we get for R0 = rE * (1 + qr) / (2 * gr).

In this case, the ratio r0 / (R0 + r0) = 2 * qr / (1 + qr * qr) ~ 2 * qr, and the total error will be 2 * qr * qR.
If we take 5% resistors, the result will be 2 * 0.05 * 0.05 = 0.5%, which is not bad, and for 1% resistors 2 * 0.01 * 0.01 = 0.02%, which is most likely surpass the capabilities of our portable measuring device.

Unfortunately, such accuracy is unattainable in real conditions due to the fact that we cannot take the nominal values ​​obtained from the calculation for r0 and R0, since we are limited to the standard series of resistors E24 or E96. Let us try to take into account this fact, which will lead to the fact that for the resistor r0 we have to take the nearest higher nominal value of the series and in the worst case we get r0 = rE * (1 + qr) * (1 + qr) / (1-qr) ~ rE * (1 + 3 * qr). But for the resistor R, we can take the nearest nominal value of the series, which will lead to the worst case, R = R0 * (1 + 1.5 * qR).

That is, taking into account the standard series, the total error will be 4.5 * qr * qR, which for 1.1% of resistors will be 1.13% (which is not bad, but somehow less than we expected and generally easier to buy a 1% resistor), and for 1 % resistors - 0.05%, which is very cool, but we need a cash register trimmer with an accuracy of 1%. The truth, as always, lies in the middle (well, it has this property, always lie in the middle) as a base we take a 1% resistor (especially since we know its face value in advance), and we take the compensating one from the standard cash register of 5% (well, You definitely have) and we get an accuracy of 0.2%. If this accuracy is not enough for us, then we can either use a variable resistor as a compensating one, or put another 1 resistor in parallel to the first two of even larger nominal values ​​and get the required accuracy here. The calculation for this case is left to the inquisitive reader.

So in conclusion about where such an exotic scheme is used. First of all, these are current-measuring resistors (shunts), since in this case r must be, first of all, small, and the small face value trimmer is exotic, and secondly, powerful, and this is even more exotic. But the compensating resistor can be a trimmer, since all of the above refers to it to a much smaller extent. But the question of what is easier to do is to measure the nominal, take a resistor from the cash register and solder it, or measure the nominal and twist the trimmer until the result is achieved - it remains open. Of course, we did not consider single-turn resistors for such a circuit, even as a hypothesis, so it would take a long turn (i.e. 5-6 revolutions in the whole range).