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Paha777 [63]
3 years ago
6

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

esistor and measure about 100 mA. Since there are variations in your resistor values, the resistor you chose is not exactly 100 . First, you measure y1 = 102 mA. Then, you ask a friend to measure, and she gets y2 = 97 mA. a. What do you get for the resistance value if you use the two measurements separately (two values)? b. Set it up as a linear system with measurement y = [y1y2]T and a single resistance value x.
Physics
1 answer:
Butoxors [25]3 years ago
3 0

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current y_{1} = 102 mA = 0.102 A

According to ohm's law, V = IR

R = V/I

Here, I = y_{1}

R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms

For the second measurement, current y_{2} = 97 mA = 0.097 A

R = \frac{V}{y_{2} }

R = \frac{10}{0.097} \\R = 103 .09 ohms

b) y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}

y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]

y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right]

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

x = \left[\begin{array}{ccc}100\end{array}\right]

\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3}  \end{array}\right] =  \left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2}  \end{array}\right] =  \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5}  \end{array}\right]

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<h3>What is density of substance?</h3>

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