1.3 Working with errors  (Page 3/3)

$$C_1=1.2\pm 0.1\mu F$$

$$C_2=2.3\pm 0.2\mu F$$

Two capacitors are connected in parallel. What is the equivalent capacity of two given capacitors? Indicate error in percentage.

Solution : The equivalent capacitance of two capacitors in parallel is given by the sum of the capacitance of individual capacitors :

$$C=C_1+C_2=1.2+2.3=3.5\mu F$$

For addition of two quantities, the absolute error in the equivalent capacitance is given by :

$$\Delta C=\Delta C_1+\Delta C_2$$

$$\Rightarrow \Delta C=0.1+0.2=0.3\mu F$$

Percentage error in the equivalent capacitance is given by :

$$\Rightarrow \Delta Cp=\frac{\Delta C}{C}X100$$

$$\Rightarrow \Delta Cp=\frac{0.3}{3.5}X100=8.6$$

Hence, equivalent capacitance, “C”, with percentage error is :

$$\Rightarrow C=3.5\mu F\pm 8.6\%$$

Errors in product or division

Error in the product and division of two measured quantities can be similarly worked. For brevity, we shall work out the implication of error for the operation of product only. We shall simply extend the result obtained for the product to division. Let the two measured quantities be :

$$a=a\pm \Delta a$$

$$b=b\pm \Delta b$$

Let “Δc” be the absolute error in the product. Then, product of the two quantities is :

$$\left(c\pm \Delta c\right)=\left(a\pm \Delta a\right)\left(b\pm \Delta b\right)$$

$$\Rightarrow c\left(1\pm \frac{\Delta c}{c}\right)=ab\left(1\pm \frac{\Delta a}{a}\right)\left(1\pm \frac{\Delta b}{b}\right)$$

But, c = ab. Hence, expanding terms, we have :

$$\Rightarrow 1\pm \frac{\Delta c}{c}=1\pm \frac{\Delta a}{a}\pm \frac{\Delta b}{b}\pm \frac{\Delta a\Delta b}{ab}$$

$$\Rightarrow \pm \frac{\Delta c}{c}=\pm \frac{\Delta a}{a}\pm \frac{\Delta b}{b}$$

The terms “ $\frac{\Delta a\Delta b}{ab}$ ” is negligibly small and as such can be discarded :

$$\Rightarrow \pm \frac{\Delta c}{c}=\pm \left(\frac{\Delta a}{a}+\frac{\Delta b}{b}\right)$$

We see here that the relative error in the product of two quantities is equal to the sum of the relative errors in the individual quantities. This manifestation of individual errors in product is also true for division. Hence, we can broaden our observation that the relative error in the product or division of two quantities is equal to the sum of the relative errors in the individual quantities.

Example

Problem 2: The mass and the volume of a uniform body is given as :

$$m=9.5\pm 0.1kg$$

$$V=3.1\pm 0.2m^3$$

Determine density of the body with error limits.

Solution : Let us first calculate the value of the density, maintaining the decimal place in the result. It is given by :

$$\rho =\frac{m}{v}=\frac{9.5}{3.1}=3.1kg/m^3$$

We have rounded the result for precision of density is limited to the minimum of precision of the quantities involved. The relative error in the division of mass by volume is :

$$\frac{\Delta \rho }{\rho }=\left(\frac{\Delta m}{m}+\frac{\Delta V}{V}\right)$$

$$\Rightarrow \frac{\Delta \rho }{\rho }=\left(\frac{0.1}{9.5}+\frac{0.2}{3.1}\right)$$

$$\Rightarrow \frac{\Delta \rho }{\rho }=0.01+0.06=0.07$$

The absolute error in the density is :

$$\Rightarrow \Delta \rho =0.07\rho =0.07X3.1=0.2$$

Hence, density is given as :

$$\Rightarrow \rho =3.1\pm 0.2kg/m^3$$

Errors in raising a quantity to a power

In order to estimate error involving raising of a quantity to some power, we consider two measured quantities. Let us consider that they are related as :

$$x=\frac{a^n}{b^m}$$

Taking logarithm on either side of the equation, we have :

$$\Rightarrow \ln x=n\ln a-m\ln b$$

Differentiating on both sides, we have :

$$\Rightarrow \frac{dx}{x}=n\frac{da}{a}-m\frac{db}{b}$$

We can exchange each of the differential term with corresponding relative error terms.

$$\Rightarrow \pm \frac{\Delta x}{c}=\pm n\frac{\Delta a}{a}-m\left(\pm \frac{\Delta b}{b}\right)$$

Again there are four possible combinations of values for the relative error. The maximum being,

$$\Rightarrow \frac{\Delta x}{c}=n\frac{\Delta a}{a}+m\frac{\Delta b}{b}$$

Thus, relative error in the result is :

$$\Rightarrow \Delta x_r=\left(n\Delta a_r+m\Delta b_r\right)$$

We see here that error in the result is equal to power times the relative error, irrespective of whether the power is positive or negative. It leads to an important deduction that quantities with greater powers in an expression should be measured with highest accuracy to minimize error in the "derived" quantity.

Example

Problem 3: The torque required to produce a twist in solid bar is given by :

$$\tau =\frac{\Pi \eta r^4}{2L}$$

If percentage error in the measurement of η, r and L are 1%, 4% and 1% respectively. Find the percentage error in the value of torque.

Solution : We know that relative error in the resultant quantity i.e torque is :

$$\Delta t_r=\left(n_1\Delta {\eta }_r+n_2\Delta r_r+n_3\Delta L_r\right)$$

where $n_1$ , $n_2$ and $n_3$ are the powers of three quantities η, r and L respectively. The constants of the equations are not considered as they are not measured. Now percentage is obtained by just multiplying relative error by 100, we can write,

$$\Rightarrow \Delta t{}_p=\left(n_1\Delta {\eta }_p+n_2\Delta r_p+n_3\Delta L_p\right)$$

Here, $n_1=\mathrm{1,}{n}_2=4\mathrm{and}{n}_3=1$ . Note that position of a quantity either in numerator or denominator does not make difference in error combination. Putting values,

$$\Rightarrow \Delta t{}_p=\left(1X1+4X4+1X1\right)=18$$

This means that the maximum error estimate in torque would be 18%. This is quite a large amount of uncertainty. Note the role played by the error in “r”. Most of error (16 %) is due to 4% error in this quantity as it is raised to a power of 4. This result substantiates the observation that quantity with highest power should be measured with most accuracy.