## Monday, March 28, 2011

### Kerala Engineering Entrance [KEAM (Engineering)] 2007 Question on Percentage Error

In continuation of the previous post (dated 23rd March 2011) related to errors in measurements, I give you today the following question which appeared in KEAM (Engineering) 2007 question paper:

The value of two resistors are R1 = (6 ± 0.3) kΩ and R2 =(10 ± 0.2) kΩ. The percentage error in the equivalent resistance when they are connected in parallel is

(a) 5.125 % (b) 2 % (c) 3.125 %

(d) 7 % (e) 10.125 %

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 R2 and ∆(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 %.

As an extension of the above problem consider the following question:

In the above question what will be the percentage error in the heat produced per second when a current of (4 ± 0.1) ampere is passed through the parallel combination of R1 and R2?

(a) 3.5 %

(b) 7 %

(c) 10.125 %

(d) 15.125 %

(e) 20.25 %

The heat H produced is given by H = I2R, where I is the current.

The fractional error in the heat produced per second is

H/H = 2×(∆I/I) + ∆R/R

Percentage error in H is 2×(∆I/I) ×100 + (∆R/R) ×100

Therefore, percentage error in H = 2×(0.1/4)×100 + 10.125 = 15.125%