Question

Section BreakProblem 08.048 2.value: 5.00 pointsRequired information Problem 08.048.b Compute the expected error. The expected error is _____%. (Round the final answer to the nearest whole number.)

Answer 1

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%