If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1 R3 . If the resistances are measured in ohms as R1 = 64 Ω, R2 = 20 Ω, and R3 = 8 Ω, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R. (Round your answer to three decimal places.)

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aristocles aristocles · Answer 1 · 2019-08-18T08:13:49+02:00

Answer:

[tex]\Delta R_{max} = 0.46 ohm[/tex]

Explanation:

Resistance is given of different values

now we have

[tex]R_1 = 64 ohm[/tex]

[tex]R_2 = 20 ohm[/tex]

[tex]R_3 = 8 ohm[/tex]

possible error in all resistance is 0.5 %

now we know that

[tex]\Delta R_1 = 0.005\times 64 = 0.32 ohm[/tex]

[tex]\Delta R_2 = 0.005 \times 20 = 0.1 ohm[/tex]

[tex]\Delta R_2 = 0.005 \times 8 = 0.04 ohm[/tex]

Now the maximum possible error when all resistance is connected in series

[tex]\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3[/tex]

[tex]\Delta R_{max} = 0.32 + 0.1 + 0.04[/tex]

[tex]\Delta R_{max} = 0.46 ohm[/tex]