Question

The value of two resistance are R1(6+/- 0.3) k ohm R2(10+/-0.2) k ohm. Find the % error in the equivalent resistance when they are connected in parallel.

Answer 1

When resistances R1 and R2 are connected in parallel the equivalent resistance R is given by

R = R1R2/(R1 + R2)

The fractional error in R is ∆R/R, which you can find by taking logarithm and differentiating:

∆R/R = (∆R1/R1) + (∆R2/R2) + ∆(R1 + R2)/(R1 + R2)

[Here ∆R1 is the error in R1, ∆R2 is the error in R2and ∆(R1 + R2) is the error in (R1 + R2). On differentiation the third term is negative but we have taken the sign of the third term as positive since we have to consider the worst case in which the maximum possible error is obtained].

Therefore, ∆R/R = (0.3/6) + (0.2/10) + (0.5/16)

Percentage error in R = percentage error in R1 + percentage error in R2 + percentage error in (R1 +R2) = (0.3/6)×100 + (0.2/10)×100 + (0.5/16)×100 = 5 % + 2 % + 3.125 % = 10.125 %.

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