Question

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.)

Answer 1

`\Delta R_(max) = 0.46 ohm`

Step-by-step explanation:

Resistance is given of different values

now we have

`R_1 = 64 ohm`

`R_2 = 20 ohm`

`R_3 = 8 ohm`

possible error in all resistance is 0.5 %

now we know that

`\Delta R_1 = 0.005* 64 = 0.32 ohm`

`\Delta R_2 = 0.005 * 20 = 0.1 ohm`

`\Delta R_2 = 0.005 * 8 = 0.04 ohm`

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

`\Delta R_(max) = \Delta R_1 + \Delta R_2 + \Delta R_3`

`\Delta R_(max) = 0.32 + 0.1 + 0.04`

`\Delta R_(max) = 0.46 ohm`