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BPO 2019, Теория — Задача 4

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Всички материали от BPO 2019

1st Balkan Physics Olympiad – 2019 BPO · 14 юли 2019 г. · 25 т.

Условие

An electric circuit consists of four resistors and two switches (see the figure below). Let’s denote the state of the switch with “0” if it is opened, and with “1” if it is closed (in the figure both switches are opened but this may not be true for the initial state of the circuit). The initial states of the switches are unknown. The resistance $R_{\mathrm{AB}}$ between the two ends A and B of the circuit is measured. Its initial resistance (state “ $\alpha$ ” of the circuit) is $R_{\mathrm{AB\alpha}} = 240\ \Omega$ . The switch K1’s state is reverted and the resistance of the circuit remains the same, $R_{\mathrm{AB\beta}} = 240\ \Omega$ . After that, the switch K2’s state is reverted and the resistance of the circuit becomes $R_{\mathrm{AB\gamma}} = 400\ \Omega$ . Finally, the switch K1’s state is once again reverted and the resistance of the circuit becomes $R_{\mathrm{AB\delta}} = 280\ \Omega$ .

Circuit between terminals A and B: upper branch R1 in series with R3 and open switch K2; lower branch R2 in series with R4; open switch K1 connects the midpoint between R1 and R3 to the midpoint between R2 and R4.

From the available experimental data, calculate the values of the resistances of the four resistors in the circuit and the initial state of each of the two switches. Fill your answer in the table below.

State of the circuit	State of K1	State of K2	$R_{\mathrm{ABi}}$ Ω
$\alpha$			240
$\beta$			240
$\gamma$			400
$\delta$			280

Resistor	Value, Ω
$R_1$
$R_2$
$R_3$
$R_4$

Initial states of K1, K2 [2 x 2,5 points] [5 т.]

Resistor values [4 x 5 points] [20 т.]

Отговори

Решение

As after the switching of the K1 the resistance of the circuit remains the same, the potentials on the both ends of K1 are equal (voltage on K1 is zero). This situation is only possible when the initial state of К2 is ”1”. Later, as the $R_{\mathrm{AB\gamma}} > R_{\mathrm{AB\delta}}$ , it can be concluded that the last state of K1 is ”1”. So, using these conclusions we can fill the state of the switches for all states of the circuit (see the table).

Now we can start calculations of the values of the resistors. As in the beginning both switches are closed and the opening of the K1 does not change the circuit resistance, it follows $U_{R_1} = U_{R_2}$ and $U_{R_3} = U_{R_4}$ . The relations of the currents are $I_{R_1} = I_{R_3}$ and $I_{R_2} = I_{R_4}$ . From these equations we can obtain that $\frac{R_2}{R_1} = \frac{R_4}{R_3} = x$ . To minimize the above expressions, we can note $R_1 = R$ , $R_2 = xR$ , let $R_3 = yR$ , then $R_4 = xyR$ .

For circuit state “ $\beta$ ”: $\frac{R(1+y)Rx(1+y)}{R(1+y)+Rx(1+y)} = 240\ \Omega$ . After simplification $\frac{Rx(1+y)}{1+x} = 240\ \Omega$ (1).

For circuit state “ $\gamma$ ”: $x(1 + y)R = 400\ \Omega$ . (2)

For circuit state “ $\delta$ ”: $\frac{xRR}{xR+R} + xyR = 280\ \Omega$ . (3)

Substituting (2) in (1), $\frac{400\ \Omega}{1+x} = 240\ \Omega$ , then $x = \frac{400}{240} - 1 = \frac{2}{3}$ . Subtracting (3) from (2), $x(1 + y)R - \frac{xRR}{xR+R} + xyR = 120\ \Omega$ . After simplification, $\frac{x^2R}{x+1} = 120\ \Omega$ . Substituting $x$ with the obtained value, $R = 450\ \Omega$ . Now substituting both $x$ and $R$ with their values in (2), $\frac{2}{3}(1 + y)450\ \Omega = 400\ \Omega$ , we obtain $y = \frac{1}{3}$ . So, the final values of the resistor are, $R_1 = 450\ \Omega$ , $R_2 = 300\ \Omega$ , $R_3 = 150\ \Omega$ , and $R_4 = 100\ \Omega$ .

The same circuit between A and B: R1 and R3 with switch K2 in the upper branch, R2 and R4 in the lower branch, switch K1 between the midpoints.

The answer is given in the table below.

State of the circuit	State of the K1	State of the K2	$R_{\mathrm{ABi}}$ , Ω
$\alpha$	1	1	240
$\beta$	0	1	240
$\gamma$	0	0	400
$\delta$	1	0	280

Resistor $R_j$	Value, Ω
$R_1$	450
$R_2$	300
$R_3$	150
$R_4$	100

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