Question

determine the current, i, (in amps) given the following: v = 220 v ±2.2% and r = 830ω ± 1.3% round based on the applicable rules of accuracy/precision. i = \\(\frac{v}{r}\\) question 8 determine the area of a rectangle with the dimensions: l = 26.3 ft ±0.420% and w = 16.0 ft ±1.44% round based on the applicable rules of accuracy/precision. area = l × w =

Explanation:

Step1: Calculate the nominal current

Use Ohm's law $I=\frac{V}{R}$. Substitute $V = 220$ V and $R=830\Omega$. So, $I=\frac{220}{830}\approx 0.265$ A.

Step2: Calculate the relative - error in current

For a quantity $y=\frac{a}{b}$, the relative - error formula is $\frac{\Delta y}{y}\times100\%=\frac{\Delta a}{a}\times100\%+\frac{\Delta b}{b}\times100\%$. Here, $a = V$, $b = R$, $\frac{\Delta V}{V}\times100\% = 2.2\%$ and $\frac{\Delta R}{R}\times100\%=1.3\%$. Then the relative - error in $I$ is $\frac{\Delta I}{I}\times100\%=2.2\% + 1.3\%=3.5\%$.

for area calculation:

Step1: Calculate the nominal area

The area of a rectangle is given by $A=l\times w$. Substitute $l = 20.3$ ft and $w = 16.0$ ft. So, $A=20.3\times16.0 = 324.8$ square - feet.

Step2: Calculate the relative - error in area

For a quantity $y = a\times b$, the relative - error formula is $\frac{\Delta y}{y}\times100\%=\frac{\Delta a}{a}\times100\%+\frac{\Delta b}{b}\times100\%$. Here, $a = l$, $b = w$, $\frac{\Delta l}{l}\times100\%=0.420\%$ and $\frac{\Delta w}{w}\times100\% = 1.44\%$. Then the relative - error in $A$ is $\frac{\Delta A}{A}\times100\%=0.420\%+1.44\% = 1.86\%$.

Answer:

$0.265\pm3.5\%$