Combination of Errors

If we do an experiment involving several measurements, we must know how the errors in all the measurements combine. For example, mass density is obtained by deviding mass by the volume of the substance. If we have errors in the measurement of mass and of the sizes or dimensions, we must know what the error will be in the density of the substance. To make such estimates, we should learn how errors combine in various mathematical operations. For this, we use the following procedure.

(a) Error of a sum or a difference Suppose two physical quantities A and B have measured values A ± ∆A, B ± ∆B respectively where ∆A and ∆B are their absolute errors. We wish to find the error ∆Z in the sum 

Z = A + B. 

We have by addition, 

Z ± ∆Z = (A ± ∆A) + (B ± ∆B). 

The maximum possible error in Z 

∆Z = ∆A + ∆B 

For the difference Z = A – B, we have 

 Z ± ∆ Z = (A ± ∆A) – (B ± ∆B) 

 = (A – B) ± ∆A ± ∆B 

or, ± ∆Z = ± ∆A ± ∆B 

The maximum value of the error ∆Z is again ∆A + ∆B. 

Hence the rule :

When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities.

Example 1

The temperatures of two bodies measured by a thermometer are $t_1 = 20^0C$ ± 0.5 $^0C$ and $t_2 = 50^0C$ ± 0.5 $^0C$. Calculate the temperature difference and the error theirin.

Answer:

t′ = $t_2–t_1$ = ($50^0C$ ±0.5 $^0C$)– ($20^0C$± 0.5 $^0C$) 

t′ = $30^0C$ ± 1 $^0C$

(b) Error of a product or a quotient 

Suppose Z = AB and the measured values of A and B are A ± ∆A and B ± ∆B. Then 

Z ± ∆Z = (A ± ∆A) (B ± ∆B) 

= AB ± B ∆A ± A ∆B ± ∆A ∆B. 

Dividing LHS by Z and RHS by AB we have, 

1±(∆Z/Z) = 1 ± (∆A/A) ± (∆B/B) ± (∆A/A)(∆B/B). 

Since ∆A and ∆B are small, we shall ignore their product. Hence the maximum relative error 

∆Z/ Z = (∆A/A) + (∆B/B). 

You can easily verify that this is true for division also.

Hence the rule: When two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.

Example 2

The resistance R = V/I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R.

Answer 

The percentage error in V is 5% and in I it is 2%. 

The total error in R would therefore be 5% + 2% = 7%.

Example 2

Two resistors of resistances $R_1$ = (100 ± 3) ohm and $R_2$ = (200 ± 4) ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation $R = R_1 + R_2$, and for (b) $\frac{1}{R'}=\frac{1}{R_1}+{\frac{1}{R_2}}$ and $\frac{\Delta R'}{R'^2}=\frac{\Delta R_1}{R_1^2} + {\frac{\Delta R_2}{R_2^2}}$

Answer 

(a) The equivalent resistance of series combination 

$R = R_1 + R_2$ = (100 ± 3) ohm + (200 ± 4) ohm 

      = 300 ± 7 ohm. 

(b) The equivalent resistance of parallel combination

$R'=\frac{R_1R_2}{R_1+R_2}=\frac{200}{3}$ = 66.7 ohm

Then, from $\frac{1}{R'}=\frac{1}{R_1}+{\frac{1}{R_2}}$

we get,

$\frac{\Delta R'}{R'^2}=\frac{\Delta R_1}{R_1^2} + {\frac{\Delta R_2}{R_2^2}}$

$\Delta R'=(R'^2)\frac{\Delta R_1}{R_1^2} + (R'^2){\frac{\Delta R_2}{R_2^2}}$

$\Delta R'=\left(\frac{66.7}{100}\right)^2 (3)+ \left(\frac{66.7}{100}\right)^2(4)$

$\Delta R'=1.8$

Then, R′ =  66.7 ± 1.8 ohm 

(Here, ∆R is expresed as 1.8 instead of 2 to keep in confirmity with the rules of significant figures.)

(c) Error in case of a measured quantity raised to a power Suppose $Z = A^2$,

Then, 

 ∆Z/Z = (∆A/A) + (∆A/A) = 2 (∆A/A). 

Hence, the relative error in $A^2$ is two times the error in A. 

In general, if Z = $A^p B^q/C^r$ 

Then, 

 ∆Z/Z = p (∆A/A) + q (∆B/B) + r (∆C/C). 

Hence the rule: The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity.

Example 1 

Find the relative error in Z, if $Z = A^4B^{1/3}/CD^{3/2}$.

Answer 

The relative error in Z is 

∆Z/Z = 4(∆A/A) +(1/3) (∆B/B) + (∆C/C) + (3/2) (∆D/D).

Example 2

The period of oscillation of a simple pendulum is $T=2 \pi \sqrt{\frac{L}{g}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g ?

Answer: $g=4 \pi^2L/T^2$

Here, $T = \frac{t}{n}$  and $\Delta T=\frac{\Delta t}{n}$. Therefore, $\frac{\Delta T}{T}=\frac{\Delta t}{t}$. 

The errors in both L and t are the least count errors. Therefore,

(∆g/g) = (∆L/L) + 2(∆T/T ) 

          = $\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)=0.027$

Thus, the percentage error in g is 

100 (∆g/g) = 100(∆L/L) + 2 × 100 (∆T/T) = 3%