4 Engineering Example 3

4.1 Error in power to a load resistance

Introduction

The power required by an electrical circuit depends upon its components. However, the specification of the rating of the individual components is subject to some uncertainity. This Example concerns the calculation of the error in the power required by a circuit shown in Figure 11 given a formula for the power, the values of the individual components and the percentage errors in them.

Problem in words

The power delivered to the load resistance $R_L$ for the circuit shown in Figure 11 is given by

$P=\frac{25R_L}{{\left(R+R_L\right)}^2}$

Figure 11 :

If $R=2000\Omega$ and $R_L=1000\Omega$ with a maximum possible error of 5 $\%$ in either, find $P$ and estimate the maximum error in $P.$

Mathematical statement of the problem

We can calculate $P$ by substituting $R=2000$ and $R_L=1000$ into $P=\frac{25R_L}{{\left(R+R_L\right)}^2}.$

We need to calculate the absolute errors in $R$ and $R_L$ and use these in the approximation $\delta P\approx \frac{P}{R}\delta R+\frac{P}{R_L}\delta R_L$ to calculate the error in $P.$

Mathematical analysis

At $R=2000$ and $R_L=1000$

$P=\frac{25\times 1000}{{\left(1000+2000\right)}^2}=\frac{25}{9000}=\frac{25}{9}\times 10^{-3}\approx 2.77\times 10^{-3}$ watts.

A 5 $\%$ error in $R$ gives ${\left|\delta R\right|}_{max}=\frac{5}{100}\times 2000=100$ and ${\left|\delta R_L\right|}_{max}=\frac{5}{100}\times 1000=50$

${\left|\delta P\right|}_{max}\approx \frac{P}{R}{\left|\delta R\right|}_{max}+\frac{P}{R_L}{\left|\delta R_L\right|}_{max}$

We need to calculate the values of the partial derivatives at $R=2000$ and $R_L=1000.$

$P=\frac{25R_L}{{\left(R+R_L\right)}^2}=25R_L{\left(R+R_L\right)}^{-2}$

$\frac{P}{R}=-50R_L{\left(R+R_L\right)}^{-3}$

$\frac{P}{R_L}=25{\left(R+R_L\right)}^{-2}-50R_L{\left(R+R_L\right)}^{-3}$

So $\frac{P}{R}\left(2000,1000\right)=-50\left(1000\right){\left(3000\right)}^{-3}=\frac{-50}{1000^2\times 27}=-\frac{50}{27}\times 10^{-6}$

$\frac{P}{R_L}\left(2000,1000\right)=25{\left(3000\right)}^{-2}-50\left(1000\right){\left(3000\right)}^{-3}=\left(\frac{25}{9}-\frac{50}{27}\right)\times 10^{-6}$

$=\left(\frac{75-50}{27}\right)\times 10^{-6}=\frac{25}{27}\times 10^{-6}$

Substituting these values into ${\left|\delta P\right|}_{max}\approx \frac{P}{R}{\left|\delta R\right|}_{max}+\frac{P}{R_L}{\left|\delta R_L\right|}_{max}$ we get:

${\left|\delta P\right|}_{max}=\frac{50}{27}\times 10^{-6}\times 100+\frac{25}{27}\times 10^{-6}\times 50=\left(\frac{5000}{27}+\frac{25\times 50}{27}\right)\times 10^{-6}\approx 2.315\times 10^{-4}$

Interpretation

At $R=2000$ and $R_L=1000$ , $P$ will be $2.77\times 10^{-3}$ W and, assuming 5 $\%$ errors in the values of the resistors, then the error in $P\approx \pm 2.315\times 10^{-4}$ W. This represents about 8.4% error. So the error in the power is greater than that in the individual components.

Exercises

1. The sides of a right-angled triangle enclosing the right-angle are measured as 6 m and 8 m. The maximum errors in each measurement are $\pm$ 0.1m. Find the maximum error in the calculated area.
2. In Exercise 1, the angle opposite the 8 m side is calculated from $tan\theta =8∕6$ as $\theta =53^{\circ }{8}^{\prime }$ . Calculate the approximate maximum error in that angle.
3. If $v=\sqrt{\frac{3x}{y}}$ find the maximum percentage error in $v$ due to errors of $1\%$ in $x$ and $3\%$ in $y$ .
4. If $n=\frac{1}{2L}\sqrt{\frac{E}{d}}$ and $L,E$ and $d$ can be measured correct to within $1\%$ , how accurate is the calculated value of $n$ ?
5. The area of a segment of a circle which subtends an angle $\theta$ is given by $A=\frac{1}{2}{r}^2\left(\theta -sin\theta \right)$ . The radius $r$ is measured with a percentage error of $+0.2\%$ and $\theta$ is measured as $45^0$ with an error of $=+0.1^{\circ }$ . Find the percentage error in the calculated area.

Answer