For two resistor R1 and R2 connected in parallel the relative error in their equivalent resistance is {where R1 = (10.0 +- 0.1) and R2 = (20.0 +- 0.4)} (a) 0.08 (b) 0.05 (c) 0.01 (d) 0.04

By Shivank Goel|Updated : August 17th, 2022

Given, that two resistors R1 and R2 are connected in parallel.

We have to find the relative error in equivalent resistance.

R1 = 10 ± 0.1

R2 = 20 ± 0.4

Since the resistors are in parallel, the relative error is given by

1/R = 1/R1 + 1/R2

1/R = 1/10 + 1/20

1/R = (20+10)/200

1/R = 30/200

R = 20/3

Differentiating both sides,

-1/R2 dR = -1/R12 dR1 - 1/R22 dR2

dR/R x 1/R = 1/R12 dR1 + 1/R22 dR2

dR/R x (3/20) = 1/(10)2 x(0.1) + 1/(20)2 x(0.2)

dR/R x (3/20) = 0.1/100 + 0.2/400

dR/R x (3/20) = 4(0.1)+0.2 / 400

dR/R x (3/20) = 0.4+0.2 / 400

dR/R x (3/20) = 0.6/400

dR/R x (3/20) = 0.3/200

dR/R = 0.3/200 x 20/3

dR/R = 6/600

dR/R = 1/100

dR/R = 0.01

Therefore, the relative error is 0.01.

Summary:

For two resistor R1 and R2 connected in parallel the relative error in their equivalent resistance is {where R1 = (10.0 +- 0.1) and R2 = (20.0 +- 0.4)} (a) 0.08 (b) 0.05 (c) 0.01 (d) 0.04

For two resistor R1 and R2 connected in parallel the relative error in their equivalent resistance is {where R1 = (10.0 +- 0.1) and R2 = (20.0 +- 0.4)}, 0.01

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