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BPO 2025, Теория — Задача 4. N resistors and an ammeter

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Всички материали от BPO 2025

BPO7 Problems 26-30 June 2025, Ohrid · 26 юни 2025 г. · 25 т.

Условие

A circuit diagram is given in the figure. It contains a battery with voltage $E$ , $N$ resistors of equal value $R_1 = R_2 = \dots = R_N = R$ , and an ammeter with resistance $r$ .

Circuit: battery E with current I_E in series with resistors R_N … R_m+1, R_m … R_2, R_1 (current I_m); the ammeter A with resistance r (current I_Am) is connected in parallel with the group R_1 … R_m.

a) Obtain a formula for the current $I_{\mathrm{Am}}$ through the ammeter expressed in terms of the given quantities $E, R, r, N, m$ . [5 т.]

b) In a circuit with $N = 50$ and $E =$ 12 V a series of measurements of the ammeter current $I_{\mathrm{Am}}$ were made at 10 consecutive values of $m$ ( $m, m+1, m+2, \dots, m+9$ ). These values are given in the table below:

$k$	$I_{\mathrm{Am}}$ , mA
$m$	8.63
$m+1$	9.16
$m+2$	9.77
$m+3$	10.5
$m+4$	11.2
$m+5$	12.2
$m+6$	13.3
$m+7$	14.5
$m+8$	16.1
$m+9$	18.1

Using suitable variables represent the measurements on such a graph that the values of resistance $R$ and number $m$ can be obtained. Assume that $m \gg 1$ and $r \lt R$ . Draw the dependence on the graph paper provided with the answer sheets. All additionally calculated data must be present in a table on the answer sheets provided. Use as many columns in the table in your answer sheets as you need. Calculate $R$ and $m$ . Round the value of $R$ to a multiple of 5. [13 т.]

c) Calculate $r$ . Any additional calculated data should be presented in the table. Round the value of $r$ to a multiple of 5. [7 т.]

Отговори

Решение

Solution

From the circuit it is seen that $rI_{\mathrm{Am}}=mRI_m$ (1.0 p) (1)

The same circuit as in the statement: battery E, current I_E, resistors R_N … R_m+1, R_m … R_2, R_1 with current I_m, and the ammeter A with resistance r and current I_Am in parallel with R_1 … R_m.

and $I_E=I_{\mathrm{Am}}+I_m$ . (0.5 p) (2)

Therefore $I_m=I_{\mathrm{Am}}\frac{r}{mR}$ (3)

and $I_E=I_{\mathrm{Am}}\left(1+\frac{r}{mR}\right)$ . (1.0 p) (4)

Also $E=(N-m)RI_E+rI_{\mathrm{Am}}$ . (1.0 p) (5)

Substituting (4) in (5),

E=(N-m)RI_{\mathrm{Am}}\left(1+\frac{r}{mR}\right)+rI_{\mathrm{Am}}=I_{\mathrm{Am}}\left[(N-m)R\left(1+\frac{r}{mR}\right)+r\right].\qquad(6)

Finally, $I_{\mathrm{Am}}=\frac{E}{(N-m)R\left(1+\frac{r}{mR}\right)+r}$ . (1.5 p) (7)

b) (13 points) In a circuit with $N=50$ a series of measurements of the ammeter current $I_{Am}$ were made at 10 consecutive values of $m$ ( $m, m+1, m+2, \dots, m+9$ ). These values are given in the table below:

$k$	$I_{\mathrm{Am}}$ , mA	$\frac{E}{I_{\mathrm{Am}}}$ , Ω	$k=N-\frac{E}{I_{\mathrm{Am}}R}$	$r=\frac{m}{N}\left[\frac{E}{I_{\mathrm{Am}}}-(N-m)R\right]$ , Ω
$m$	8.63	1391	32.6	20.1
$m+1$	9.16	1310	33.6	20.5
$m+2$	9.77	1228	34.6	19.8
$m+3$	10.5	1143	35.7	16.5
$m+4$	11.2	1071	36.6	23.3
$m+5$	12.2	983.6	37.7	17.9
$m+6$	13.3	902.3	38.7	17.4
$m+7$	14.5	827.6	39.7	22.1
$m+8$	16.1	745.3	40.7	20.8
$m+9$	18.1	663.0	41.7	19.3

Using suitable variables represent the measurements on such a graph that the values of resistance $R$ and number $m$ can be obtained. Assume that $m \gg 1$ and $r \lt R$ . Draw the dependence on the graph paper provided. All additionally calculated data must be present in the empty columns of the table. Calculate $R$ and $m$ . Round the value of $R$ to a multiple of 5.

From (7) it follows that

\frac{E}{I_{\mathrm{Am}}}=(N-m)R\left(1+\frac{r}{mR}\right)+r=NR-mR+\frac{Nr}{m}.\qquad(8)

(1.0 p)

As $m \gg 1$ and $r \lt R$ , it is probable that $\frac{Nr}{m}\ll NR-mR$ and the dependence (8) can be nearly linear. (1.0 p)

Then from the slope of the graph of this dependence the resistance $R$ can be determined. The column with the values of $\frac{E}{I_{\mathrm{Am}}}$ is filled in the table. (2.0 p) We draw the graph $\frac{E}{I_{\mathrm{Am}}}=f(k)$ , where $k$ are arbitrary consecutive whole numbers. (5.0 p)

Graph of E/I_Am (Ohm, vertical axis 600–1500) against k (0–11): ten square data points on a falling straight line from about 1391 at k = 1 to 663 at k = 10, with the annotation R = -Δy/Δx = 80.6 Ohm.

Slope of the line is $-R$ . From the graph $R\approx 80.6\ \Omega$ . (1.0 p)

As the value must be multiple of 5, we round it to 80 Ω. (1.0 p)

Knowing already the value of $R$ , the value of $m$ can also be calculated. $m$ can be calculated from each measurement from the formula $m=N-\frac{E}{I_{\mathrm{Am}}R}$ . We fill the column in the table. (1.0 p)

From the values follows that $m=33$ . (1.0 p)

c) (7 points) Calculate $r$ . Any additional calculated data should be present in the empty columns of the table. Round the value of $r$ to a multiple of 5.

From (8) follows that $r=\frac{m}{N}\left[\frac{E}{I_{\mathrm{Am}}}-(N-m)R\right]$ . (1.0 p)

The value of $r$ can be calculated from each measurement. We fill the column in the table. (3.0 p)

The average value of $r$ is 19.8 Ω. (1.5 p)

As the value must be multiple of 5, we round it to 20 Ω. (1.5 p)

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