Two resistors are to be combined to form an equivalent resistance of 1000 Ω. The resistors are taken from available stock on hand as acquired over the years. Readily available are two common resistors rated at 500 ± 50 Ω and two common resistors rated at 2000 ± 100 Ω. (a) Calculate the uncertainties for both combinations. (b) What combination of resistors (series or parallel) would provide the smaller uncertainty in an equivalent 1000 Ω resistance?

Even though each of 2000 ohms resistor has twice the uncertainty of a 500 ohms resistor, the parallel circuit has half the uncertainty of the series circuit

Thus, the PARALLEL CIRCUIT would be more accurate

Explanation:

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codiepienagoya codiepienagoya · Answer 2 · 2021-12-29T22:50:29+01:00

Considering the [tex](500 \pm 50 \Omega)[/tex]impedance combination. Connection between series.

Considering the first pairing.

[tex]\to R_1 = 500+ 50= 550 \Omega \\\\ \to R_2= 550 \Omega\\\\[/tex]

Determine the corresponding resistance.

[tex]\to R_{eq} = R_1 +R_2 = 550 + 550 =1100 \Omega\\\\[/tex]

This uncertainty is [tex](1000+10\%)[/tex] in this case. Find the second pairing.

[tex]\to R_1 = 500 - 50 = 450 \ \Omega\\\\ \to R_2 = 500 - 50 = 450 \ \Omega\\\\[/tex]

Determine the corresponding resistance.

[tex]\to R_{eq}=R_1+R_2= 450 +450 = 900 \ \Omega\\\\[/tex]

Its ambiguity is [tex](1000 - 10\%)[/tex] in this case. The total uncertainty form [tex]1000 \ \Omega[/tex] of the resistors[tex](500 \pm 50 \Omega)[/tex] is [tex](1000\pm 10\%)[/tex].

A parallel connection of resistances is not considered in this case since it introduces a significant uncertainty in the needed resistance values.

Considering a combination of [tex](2000 \pm 5\% \Omega)[/tex] resistances. Connecting in parallel. Take the first pairing.

[tex]\to R_1 = 2000+ 5\% = 2000+100= 2100 \ \Omega\\\\\to R_2 = 2000+ 5\% = 2000+100= 2100 \ \Omega\\\\[/tex]

Determine the corresponding resistance.

[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} =\frac{1}{2100}+\frac{1}{2100}= \frac{1+1}{2100}=\frac{2}{2100}\\\\\frac{1}{R_{eq}}=\frac{2}{2100}\\\\R_{eq}=\frac{2100}{2}= 1050 \ \Omega\\\\[/tex]

Its uncertainty in this case is [tex](1000 \ \Omega + 5\%)[/tex]. Find the second pairing.

[tex]\to R_1 = 2000- 5\% = 2000-100= 1900 \ \Omega\\\\\to R_2 = 2000- 5\% = 2000-100= 1900 \ \Omega\\\\[/tex]

Determine the corresponding resistance.

[tex]\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2} =\frac{1}{1900}+\frac{1}{1900}= \frac{1+1}{1900}=\frac{2}{1900}\\\\\frac{1}{R_{eq}}=\frac{2}{1900}\\\\R_{eq}=\frac{1900}{2}= 950 \ \Omega\\\\[/tex]

The uncertainty is present here [tex](1000 \Omega -5\%)[/tex]. Its overall variance form [tex]1000 \ \Omega[/tex] of the resistors [tex](2000 \Omega + 5\%)[/tex] is [tex](1000 \ \Omega + 5\%)[/tex].

This parallel connection of resistance values is not addressed here since it introduces a significant uncertainty in the needed resistance values.

As just a result, the lowest uncertainty is now in the parallel combination of the resistances [tex](2000 \Omega + 5\%)[/tex].

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