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BPO 2020, Теория — Задача 4. Electric resistances measurement

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Всички материали от BPO 2020

Balkan Physics Olympiad – BPO 2, online contest, 24 October 2020, Craiova, ROMANIA · 24 октомври 2020 г.

Условие

A. For measuring resistances, a student makes the electric circuit illustrated in Fig. 1. The switch k can be commuted in positions labeled with 1, respectively 2 in the diagram. $R$ is the unknown value of the resistance, V is the voltmeter and A is the amperemeter. The internal resistance of the voltmeter is $R_V$ and of the amperemeter is $R_A$ .

Circuit: a battery in series with ammeter A and resistor R, closed by a wire along the bottom; a voltmeter V is connected from the right end of R to a switch k, whose position 1 connects to the node between A and R and whose position 2 connects to the node between the battery and A. — Fig. 1

a) Derive, for each position of the switch (1, respectively 2), the measured value of the resistance ( $R_m$ ), as a function of $R$ , $R_V$ , and $R_A$ . [4 т.]

Because the amperemeter and the voltmeter are not ideal, the measured values $R_m$ of the resistance are different from the true value $R$ of the resistance.

b) Derive the expression of the relative variation $\alpha_R = \frac{R_m - R}{R}$ (as a measure of accuracy) for each position of the switch. [1 т.]

Depending on the value of the resistance $R$ , a certain position of the switch is preferred over the other.

c) Based on the results obtained above, derive the range of values for the resistance $R$ for which each position of the switch is preferred. Express the results in terms of $R_A$ and $R_V$ . Using the well-known relationship between $R_A$ and $R_V$ , prove that the (practical) approximate mathematical expression for the limit between the two resistance ranges is $\beta\sqrt{R_A R_V}$ and derive the numerical value for the coefficient $\beta$ . [6 т.]

B. Using some of the circuit elements illustrated in Fig. 1, the student is assembling their own ohmmeter. The purpose is to measure an unknown resistance $R_x$ of a resistor. The result is shown in Fig. 2. The e.m.f. of the generator is $E = 9.00\ \mathrm{V}$ (its internal resistance is neglected) and the internal resistance of the amperemeter is $R_A = 10.0\ \Omega$ . The amperemeter is an analog one, with the scale range of $I_{max} = 50\ \mathrm{mA}$ and with $N = 100$ divisions (Fig. 3). The current is indicated by a needle.

Ohmmeter circuit: inside a dashed box labelled ohmmeter, a source E, ammeter A and resistor R in series; the unknown resistor Rx is connected outside the box across its two terminals. — Fig. 2

Photograph of an analog milliammeter (mA) with a curved scale numbered 0, 10, 20, 30, 40, 50 and fine divisions; the needle rests at 0. — Fig. 3

The student chooses the resistor with the resistance $R$ such that, when their ohmmeter is short-circuited, the needle indicates the maximum value on the scale.

d) Derive the mathematical formula for $R$ and calculate its numerical value. [2 т.]

When the resistor with the unknown resistance is connected to the ohmmeter, the needle deviation is $n$ divisions.

e) Express $R_x$ as a function of $n, N, E$ , and $I_{max}$ . [1,5 т.]

The student measures the resistances of several resistors and wonders for which of them the precision is the best possible.

f) Derive the mathematical formula for the relative error of determining $R_x$ . Based on this result, derive the value of $n$ for which this relative error is minimum.

Note: consider that $\left(\frac{\Delta n}{n}\right)^2 \ll 1$ , where $\Delta n$ is precision with which the student reads the number $n$ of divisions on the amperemeter scale. [4 т.]

g) Obtain the expressions for $R_x$ and for the absolute error $\Delta R_x$ , using $n$ derived at question f) and calculate their numerical values. [3,5 т.]

During the experiments, the student did not place the amperemeter in front of them, but to their right, such that the angle between the reading direction and the perpendicular to the scale is $\theta$ . The length of the amperemeter scale is $L$ and the distance between the needle and the scale’s plane is $h$ .

h) Obtain the expression for the systematic error caused by the incorrect reading of the current. [1,5 т.]

i) Calculate the maximum value of the angle $\theta$ such that reading error to be less than one division on the amperemeter scale. The length of the amperemeter scale is $L = 10\ \mathrm{cm}$ and the distance between the needle and the scale’s plane is $h = 2.0\ \mathrm{mm}$ . [1,5 т.]

Отговори

Решение

a) Derive, for each position of the switch (1, respectively 2), the measured value of the resistance ( $R_m$ ), as a function of $R$ , $R_V$ , and $R_A$ . 4 p.

The measured value of the resistance is given by

R_m = \frac{U}{I},

1 p

where $U$ and $I$ are the indications of the voltmeter and of the amperemeter, respectively.

Switch in position 1:

I = \frac{U}{R} + \frac{U}{R_V},

1 p

such that

\boxed{R_m = \frac{U}{I} = \frac{RR_V}{R + R_V}.}

0.5 p

Switch in position 2:

U = I(R + R_A),

\boxed{R_m = R + R_A.}

0.5 p

b) Derive the expression of the relative variation $\alpha_R = \frac{R_m - R}{R}$ (as a measure of accuracy) for each position of the switch. 1 p.

Switch in position 1:

\alpha_{R1} = \frac{R_m}{R} - 1 = \frac{R_V}{R + R_V} - 1 = -\frac{R}{R + R_V}.

0.5 p

For this position of the switch the measured value of the resistance is smaller than the real value.

Switch in position 2:

\alpha_{R2} = \frac{R + R_A}{R} - 1 = \frac{R_A}{R}.

0.5 p

For this position of the switch the measured value of the resistance is greater than the real value.

Position 1 of the switch is preferred when

|\alpha_{R1}| < \alpha_{R2}.

1 p

Then

\frac{R}{R + R_V} < \frac{R_A}{R},

0.5 p

R^2 - R_A R - R_A R_V < 0.

0.5 p

The solution of this inequality is

R \in (R_-, R_+) \cap [0, +\infty).

0.5 p

Since

R_\pm = \frac{1}{2}\left(R_A \pm \sqrt{R_A^2 + 4R_A R_V}\right),

0.5 p

if follows that $R_- < 0$ , so

R < R_+ = \frac{1}{2}\left(R_A + \sqrt{R_A^2 + 4R_A R_V}\right).

0.5 p

Conclusion:

When $R < \frac{1}{2}\left(R_A + \sqrt{R_A^2 + 4R_A R_V}\right)$ the position 1 of the switch is preferred 0.5 p

and

when $R > \frac{1}{2}\left(R_A + \sqrt{R_A^2 + 4R_A R_V}\right)$ the position 2 is preferred.

Since

R_A \ll R_V,

0.5 p

the limit value of the resistance is

R_+ = \frac{1}{2}\left(R_A + \sqrt{R_A^2 + 4R_A R_V}\right) = \frac{1}{2}\left(R_A + R_A\sqrt{1 + 4\frac{R_V}{R_A}}\right) =

= \frac{R_A}{2}\left(1 + \sqrt{1 + 4\frac{R_V}{R_A}}\right) \cong \frac{R_A}{2}\left(1 + \sqrt{4\frac{R_V}{R_A}}\right) \cong \frac{R_A}{2} \cdot 2\sqrt{\frac{R_V}{R_A}} =

= \sqrt{R_A R_V}.

It follows that 1 p

\beta = 1.

0.5 p

The student chooses the resistor with the resistance $R$ such that, when their ohmmeter is short-circuited, the needle indicates the maximum value on the scale.

d) Derive the mathematical formula for $R$ and calculate its numerical value. 2 p.

E = I_{max}(R + R_A),

1 p

R = \frac{E}{I_{max}} - R_A.

0.5 p

Its numerical value is:

R = 170\ \Omega.

0.5 p

When the resistor with the unknown resistance is connected to the ohmmeter, the needle deviation is $n$ divisions.

e) Express $R_x$ as a function of $n, N, E$ , and $I_{max}$ . 1.5 p.

The current is

I = n\frac{I_{max}}{N}

0.5 p

R_x = \frac{E}{I} - R - R_A = \frac{E}{I} - \frac{E}{I_{max}} = \left(\frac{N}{n} - 1\right)\frac{E}{I_{max}}.

1 p

The student measures the resistances of several resistors and wonders for which of them the precision is the best possible.

f) Derive the mathematical formula for the relative error of determining $R_x$ . Based on this result, derive the value of $n$ for which this relative error is minimum. Note: consider that $\left(\frac{\Delta n}{n}\right)^2 \ll 1$ , where $\Delta n$ is precision with which the student reads the number $n$ of divisions on the amperemeter scale. 4 p.

\varepsilon_{R_x} = \frac{\Delta R_x}{R_x} = \frac{(R_x + \Delta R_x) - R_x}{R_x},

1 p

where

R_x + \Delta R_x = \left(\frac{N}{n - \Delta n} - 1\right)\frac{E}{I_{max}}.

1 p

Consequently

\varepsilon_{R_x} = \frac{\left(\frac{N}{n - \Delta n} - 1\right)\frac{E}{I_{max}} - \left(\frac{N}{n} - 1\right)\frac{E}{I_{max}}}{\left(\frac{N}{n} - 1\right)\frac{E}{I_{max}}} = \frac{\left(\frac{N}{n - \Delta n} - 1\right) - \left(\frac{N}{n} - 1\right)}{\frac{N}{n} - 1} =

= \frac{\frac{N}{n - \Delta n} - \frac{N}{n}}{\frac{N}{n} - 1} = \frac{N\Delta n}{(n - \Delta n)(N - n)} = \frac{N\Delta n(n + \Delta n)}{[n^2 - (\Delta n)^2](N - n)} =

= \frac{n^2 N\frac{\Delta n}{n}\left(1 + \frac{\Delta n}{n}\right)}{n^2\left[1 - \left(\frac{\Delta n}{n}\right)^2\right](N - n)},

\boxed{\varepsilon_{R_x} = \frac{N}{(N - n)}\frac{\Delta n}{n}.}

1 p

This error is minimal when the function

f(n) = n(N - n)

reaches its maximal value, which happens for

\boxed{n = \frac{N}{2}.}

0.5 p

The numerical value of $n$ is

\boxed{n = 50,}

0.5 p

which means that the most precise measurement is obtained when the needle is pointing exactly at the middle of the scale.

g) Obtain the expressions for $R_x$ and for the absolute error $\Delta R_x$ , using $n$ derived at question f) and calculate their numerical values. 3.5 p.

For $n = \frac{N}{2}$ , the requested value is

R_x = \frac{E}{I_{max}} = 180\ \Omega.

1 p

The absolute error is

\Delta R_x = R_x \varepsilon_{R_x} = 4R_x\frac{\Delta n}{N}.

0.5 p

Since the precision of the instrument is

\Delta n = 1\ \text{division},

0.5 p

then

\Delta R_x = \frac{4R_x}{N} = 7.2\ \Omega \cong 7\ \Omega.

0.5 p

In conclusion,

R_x = 180\ \pm 7\ (\Omega).

1 p

h) Obtain the expression for the systematic error caused by the incorrect reading of the current. 1.5 p.

Compared to the real indication, the student sees the tip of the needle deviated to the right with the amount

\Delta x = h \times tan\theta,

0.5 p

so, the error is

\Delta I = \frac{I_{max}}{L}\Delta x = I_{max}\frac{h}{L}tan\theta.

1 p

\Delta I < \frac{I_{max}}{N},

0.5 p

then

tan\theta < \frac{L}{Nh}.

0.5 p

Numerically,

tan\theta < 0.5,

which gives

\theta_{max} = tan^{-1}0.5 = 27^{\circ}.

0.5 p

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf