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 BYJU'S Exam Prep
Updated on: September 13th, 2023
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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