Chapter 2.14 - Combination of Errors - Solved Examples

In the previous section, we saw the combination of errors. In this section, we will see some solved examples

Solved example 2.22
The resistance R = ^V⁄_I where V = (100 ± 5)V and I = (10 ± 0.2)A. Find the percentage error in R
Solution:
1. We have the rule for division:
If Z = ^A⁄_B, then the 𝚫Z is given by the relation: 
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
2. Applying this rule, we can write:
Relative error in R = Relative error in V + Relative error in I
3. Relative error in V = $\mathbf\small{\frac{\Delta V}{V}=\frac{5}{100}=0.05}$  
• Relative error in I = $\mathbf\small{\frac{\Delta I}{I}=\frac{0.2}{10}=0.02}$
4. Substituting these values in (2), we get:
Relative error in R = 0.05 + 0.02 = 0.07
5. So percentage error in R = Relative error in R × 100 = (0.07 × 100) = 7%

Solved example 2.23
Two resistors of resistances R₁ = 100 ±3 ohm and R₂ = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination
Solution:
Part (a):
1. When two resistors R₁ and R₂ are connected in series, the equivalent resistance R is given by the relation: R = R₁ + R₂ (Details here)
2. We have the rule for addition:
If Z = A + B, then the error 𝚫Z in Z is (𝚫A + 𝚫B)
3. Applying this rule, we can write:
Absolute error in R = ±𝚫R = ±(𝚫R₁ + 𝚫R₂) = ±(3 + 4) = ±7 ohm
4. So the equivalent resistance of the series combination = (100 + 200) ± 7 = 300 ± 7 ohm  
Part (b):
1. When two resistors R₁ and R₂ are connected in parallel, the equivalent resistance R' is given by the relation: $\mathbf\small{\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}}$ (Details here)
2. Substituting the values, we get:
$\mathbf\small{\frac{1}{R'}=\frac{1}{100}+\frac{1}{200}}$
$\mathbf\small{\Rightarrow R'=\frac{200}{3}\; \text{ohm}}$
3. From the relation in (1), we get: (R')^-1 = (R₁)^-1 + (R₂)^-1
(i) Let Z' = (R')^-1
• Then we can write: $\mathbf\small{\frac{\Delta Z'}{Z'}=-1\left(\frac{\Delta R'}{R'} \right)}$
• Substituting for Z', we get: $\mathbf\small{\frac{\Delta Z'}{(R')^{-1}}=-1\left(\frac{\Delta R'}{R'} \right)}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{R'} \right)(R')^{-1}}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{R'} \right)\left(\frac{1}{R'} \right)}$
$\mathbf\small{\Rightarrow \Delta Z'=-1\left(\frac{\Delta R'}{(R')^2} \right)}$
(ii) Let Z₁ = (R₁)^-1
• Similar to the above, we will get:
$\mathbf\small{\Delta Z_1=-1\left(\frac{\Delta R_1}{(R_1)^2} \right)}$
(iii) Let Z₂ = (R₂)^-1
• Similar to the above, we will get:
$\mathbf\small{\Delta Z_2=-1\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
4. We succeeded in finding 𝚫Z'
    ♦ It is the absolute error in (R')^-1  
• We succeeded in finding 𝚫Z₁
    ♦ It is the absolute error in (R₁)^-1 
• We succeeded in finding 𝚫Z₂
    ♦ It is the absolute error in (R₂)^-1
5. Now consider the relation that we wrote in (3):
(R')^-1 = (R₁)^-1 + (R₂)^-1
• On the right side, we have an addition of two quantities. So the absolute errors can be added
We get:
$\mathbf\small{-1\left(\frac{\Delta R'}{(R')^2} \right)=-1\left(\frac{\Delta R_1}{(R_1)^2} \right)+-1\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
• Multiplying both sides by '-1', we get:
$\mathbf\small{\left(\frac{\Delta R'}{(R')^2} \right)=\left(\frac{\Delta R_1}{(R_1)^2} \right)+\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
6. Substituting the values, we get:
$\mathbf\small{\left(\frac{\Delta R'}{(200/3)^2} \right)=\left(\frac{3}{(100)^2} \right)+\left(\frac{4}{(200)^2} \right)}$
• Thus we get: 𝚫R' = 1.777 = 1.8 ohm 
7. This 𝚫R' is the absolute error in R'
• We can report the effective resistance in parallel connection as:
(R' ± 𝚫R') = (200/3 ± 1.78) = 66.7 ± 1.8 ohm

Solved example 2.24
The values of two resistors are (5.0 ± 0.2) ㏀ and (10.0 ± 0.1) ㏀. What is the percentage error in the equivalent resistance when they are connected in parallel
Solution:
1. When two resistors R₁ and R₂ are connected in parallel, the effective resistance is given by:
$\mathbf\small{\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}}$
2. Substituting the values, we get:
$\mathbf\small{\frac{1}{R'}=\frac{1}{5}+\frac{1}{10}}$
$\mathbf\small{\Rightarrow R'=\frac{10}{3}}$ ㏀
3. Now we want the error when they are connected in parallel
• We can use the expression that we derived in the previous solved example 2.23
• We have: $\mathbf\small{\Rightarrow \left(\frac{\Delta R'}{(R')^2} \right)=\left(\frac{\Delta R_1}{(R_1)^2} \right)+\left(\frac{\Delta R_2}{(R_2)^2} \right)}$
4. Substituting the values, we get:
$\mathbf\small{\left(\frac{\Delta R'}{(10/3)^2} \right)=\left(\frac{0.2}{(5)^2} \right)+\left(\frac{0.1}{(10)^2} \right)}$
• Thus we get: 𝚫R' = 0.1 ㏀
5. So the relative error = $\mathbf\small{\frac{\Delta R'}{R'}=\frac{0.1}{10/3}=0.03}$
• Thus we get:
Percentage error = $\mathbf\small{\frac{\Delta R'}{R'}\times 100=0.03 \times 100 = 3 \text{%}}$

Solved example 2.25
Find the relative error in Z, if $\mathbf\small{Z=\frac{A^4 B^{\frac{1}{3}}}{C\,D^{\frac{3}{2}}}}$
Solution:
The relative error $\mathbf\small{\frac{\Delta Z}{Z}}$ is given by:
$\mathbf\small{\frac{\Delta Z}{Z}=4 \left(\frac{\Delta A}{A}\right)+\frac{1}{3}\left(\frac{\Delta B}{B}\right)+\left(\frac{\Delta C}{C}\right)+\frac{3}{2}\left(\frac{\Delta D}{D}\right)}$

Solved example 2.26
The error in measuring the diameter of a circle is 2%. What would be the error in measuring the radius of the circle?
Solution:
1. We have: r = 0.5d
• Here. 0.5 is a factor. It will not contain any error
2. Power of 'd' is '1'. So we get:
$\mathbf\small{\frac{\Delta r}{r}=1 \times\frac{\Delta d}{d}}$
3. That means, the relative error is the same for both r and d
• So percentage error will also be same for both r and d
• Thus the answer is 2%

Solved example 2.27
A force F is applied on a square plate of side L. The percentage error in the measurement of L is 2%. The percentage error in the measurement of F is 4%. What is the percentage error in the pressure?
Solution:
1. We have: $\mathbf\small{\text{Pressure}=\frac{\text{Force}}{\text{Area}}}$
• That is., $\mathbf\small{\text{P}=\frac{\text{F}}{\text{A}}}$
2. Area (A) of a square plate = L²
• So we get: $\mathbf\small{\text{P}=\frac{\text{F}}{{\text{L}}^2}}$
• So $\mathbf\small{\frac{\Delta P}{P}=\frac{\Delta F}{F}+2\left(\frac{\Delta L}{L}\right)}$
3. Given that percentage error in L = 2%. So $\mathbf\small{\frac{\Delta L}{L}}$ = 0.02
• Given that percentage error in F = 4%. So $\mathbf\small{\frac{\Delta L}{L}}$ = 0.04
4. Substituting the values in (2), we get: $\mathbf\small{\frac{\Delta P}{P}=0.04+2 \times 0.02=0.08}$
• So percentage error = $\mathbf\small{\frac{\Delta P}{P}\times 100=0.08 \times 100=8\text{%}}$

Solved example 2.28
The radius of a ball is (5.2 ± 0.2) cm. What is the percentage error in the volume of the ball?
Solution:
1. Relative error in the radius = $\mathbf\small{\frac{\Delta r}{r}=\frac{0.2}{5.2}=0.0385}$
2. We have: Volume of the ball = $\mathbf\small{\frac{4}{3}\pi r^3}$
$\mathbf\small{\frac{4}{3}\pi}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta V}{V}=3\left(\frac{\Delta r}{r}\right)=3\times 0.0385=0.1154}$
3. So percentage error = 0.1154 × 100 = 11.54%

Solved example 2.28
The time period (T) of a pendulum is given by $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}}$. If l and g are measured with an error of 1% and 2% respectively, what is the percentage error in T?
Solution:
1. We have: $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}=2\pi\frac{l^{\frac{1}{2}}}{g^{\frac{1}{2}}}}$
$\mathbf\small{2\pi}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}\right)+\frac{1}{2}\left(\frac{\Delta g}{g}\right)}$
2. Substituting the values, we get: $\mathbf\small{\frac{\Delta T}{T}=\frac{1}{2}\left(0.01\right)+\frac{1}{2}\left(0.02\right)=0.015}$
• So percentage error = 0.015 × 100 = 1.5%

Solved example 2.29
The period of oscillation of a simple pendulum is $\mathbf\small{T=2\pi \sqrt{\frac{L}{g}}}$. 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 ?
Solution:
1. Given that: Measured value of L is 20.0 cm known to 1 mm accuracy  
• This is a least count error. The true value of L may be higher by 1 mm or lower by 1 mm
• 1 mm = 0.1 cm
• So we can write: L = (20 ± 0.1) cm
2. Given that: Time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution
• Time for 100 oscillations = 90 s
• The resolution of the watch is 1 s. That watch cannot measure time durations which are less than 1 s 
• That means least count of that watch is 1 s
• The 90 s may be 91 s or 89 s
• So we can write: The time for 100 oscillations is (90 ± 1) s 
• So time for 1 oscillation (T) = $\mathbf\small{\frac{90\pm 1}{100}=(0.9\pm0.01)\;\text{s}}$
3. Thus we get:
• Absolute error (𝚫L) in L = 0.1 cm
• Absolute error (𝚫T) in T = 0.01 s
4. Given that: The period of oscillation of a simple pendulum is $\mathbf\small{T=2\pi \sqrt{\frac{L}{g}}}$
• Squaring both side, we get: $\mathbf\small{T^2=4\pi^2\left(\frac{l}{g}\right)}$
• Rearranging the expression, we get: $\mathbf\small{g=4\pi^2\left(\frac{l}{T^2}\right)}$
5. $\mathbf\small{4\pi^2}$ is a factor. It will not contain any error
• So we get: $\mathbf\small{\frac{\Delta g}{g}=\left(\frac{\Delta l}{l}\right)+2\left(\frac{\Delta T}{T}\right)}$
6. Substituting the values, we get:
$\mathbf\small{\frac{\Delta g}{g}=\left(\frac{0.1}{20}\right)+2\left(\frac{0.01}{0.9}\right)}$ = 0.02722
7. So percentage error = $\mathbf\small{\frac{\Delta g}{g}\times 100}$ = 0.02722 × 100 = 2.722% = 3%

---

In the next section, we will see dimensions of physical quantities

PREVIOUS           CONTENTS          NEXT



Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved