Learning Objectives:

• To calculate uncertainty in calculated result when dependent measurements have specified uncertainties.

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Solved Example: 88-1-01

The resistance of a certain size of copper wire is given as: $R = R_0 [1 + \alpha (T- 20)]$ where $R_0$ = 6 $\Omega$ $\pm$ 0.3$\%$ is the resistance at 20 $^\circ C$., $\alpha$ = 0.004 $^\circ C^{-1}$ $\pm$ 1 $\%$ is the temperature coefficient of resistance, and the temperature of the wire is T = 30 $\pm$ 1 $^\circ C$. The resistance of the wire and the uncertainty is:

A. 6.24 $\Omega$ and 0.49%

B. 6.24 $\Omega$ and 0.11%

C. 5.89 $\Omega$ and 0.49%

D. 5.89 $\Omega$ and 0.11%

To get nominal resistance, substitute the given values in the equation. R = 6.24 $\Omega$ \[\frac{\partial R }{\partial R_{0}} = 1.04, \frac{\partial R }{\partial \alpha} = 60\, \frac{\partial R }{\partial T} = 0.024\] \[w_{R_{0}} = (6)(0.003) = 0.018 \Omega\] \[w_{\alpha} = 4 \times 10^{-5}\ ^\circ C ^{-1}\] \[w_{T} = 1 ^\circ C\] $w_R$ = 0.0305 $\Omega$ which is 0.49$\%$

Correct Answer: A

Solved Example: 88-1-02

The electric power is calculated as: $P = EI$ where E and I are measured as: $E = 100 V \pm 2\ V$, $I = 10 A \pm 0.2\ A$. Calculate the error in calculation of power.

A. 2

B. 0.2

C. 1.484

D. 2.83

The nominal value of the power is 100 $\times$ 10 = 1000 W. The uncertainty in this value is calculated by applying the formula. The various terms are: \[\frac{\partial P}{\partial E} = I = 10, \frac{\partial P}{\partial I} = E = 100\] \[w_E=2\ V, w_I=0.2\ A\] Thus, the uncertainty in the electric power is: \[W_P= \sqrt{(10)^2(2)^2+(100)^2(0.2)^2} = 28.28\ W \mathrm{\ or\ } 2.83\%\]

Correct Answer: D