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Scorpion4ik [409]
3 years ago
10

Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected e

rror is _____%. (Round the final answer to the nearest whole number.)
Engineering
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

a) 19 or select the closest answer

b) 5%

Explanation:

a)

from the voltage divide rule

V_{in} = V_0 * \frac{R_2}{R_2 + R_f}

\frac{V_0}{V_{in}} = \frac{R_2 + R_f}{R_2} => 1 + \frac{R_f}{R_2} = 20

\frac{R_f}{R_2} = 19

Select the nearest answer

b)

obtained gain = \frac{V_0}{V_{in}} = 1 + \frac{36}{2} = 19

Expected gain = \frac{V_0}{V_{in}} = 20

∴ error = |\frac{19 - 20}{20}| × 100

            = 1/20 × 100            

            = 5%

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\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

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