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

Answer:

$\Delta R_{max} = 0.46 ohm$

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

$\Delta R_1 = 0.005\times 64 = 0.32 ohm$

$\Delta R_2 = 0.005 \times 20 = 0.1 ohm$

$\Delta R_2 = 0.005 \times 8 = 0.04 ohm$

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

$\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3$

$\Delta R_{max} = 0.32 + 0.1 + 0.04$

$\Delta R_{max} = 0.46 ohm$