In continuation of the previous post (dated 23^{rd} 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 R_{1} = (6 ± 0.3) kΩ and R_{2} =(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 *R*_{1} and *R*_{2} are connected in parallel the equivalent resistance* R* is given by

*R* = *R*_{1}*R*_{2}/(*R*_{1 }+ *R*_{2})

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

∆*R*/*R* = (∆*R*_{1}/*R*_{1}) + (∆*R*_{2}/*R*_{2}) + ∆(*R*_{1} + *R*_{2})/(*R*_{1 }+ *R*_{2})

[Here ∆*R*_{1} is the error in *R*_{1}, ∆*R*_{2} is the error in *R*_{2} and ∆(*R*_{1} + *R*_{2}) is the error in (*R*_{1} + *R*_{2}). 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 *R*_{1} + percentage error in *R*_{2} + percentage error in (*R*_{1 }+ *R*_{2}) = (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 *R*_{1 }and *R*_{2}?

(a) 3.5 %

(b) 7 %

(c) 10.125 %

(d) 15.125 %

(e) 20.25 %

The heat *H* produced is given by *H* = *I*^{2}*R*, 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%

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