BPO 2023, Теория — Задача 4. Electric circuit

• A. 2000 Ω

  Averaged value printed as R₂ = 2002 Ω ≈ 2000 Ω.

• B. Plot U against I₁: a straight line U = E − R₁ I₁.

• C. E = 12.0 V, R₁ = 2000 Ω

• D. R₃ = U/I₃, η = U·I₃/(E·I₁); values in the table of the official solution.

• E. η ≈ 17%, R₃ ≈ 1350 Ω

• F. $\eta_{max} = 3 - \sqrt{8} \approx 0.172$

• G. $R_3 = \frac{R}{\sqrt{2}}$

• H. Theory: η = 17.2%, R₃ = 2000 Ω/√2 ≈ 1414 Ω — in excellent agreement with the experimental values.

BPU 2023 problem proposal

by Miroslav V. Abrashev

Sofia University, Faculty of Physics, 5 James Bourchier Blvd., 1164 Sofia, Bulgaria

Electric circuit (Solution)

A circuit diagram is given in the figure. It contains a battery with voltage $E$, two resistors of fixed value $R_1$ and $R_2$, one resistor $R_3$ whose value can be changed, one voltmeter and two ammeters. A series of measurements were made at different values of the resistor $R_3$, which are given in the table below. The notations of the measured electric currents are given in the figure. From the data presented, calculate (all additionally calculated data must be present in the empty columns of the table):

a) (2 points) The value of the resistance of resistor $R_2$ is calculated as $R_2 = \frac{U}{I_2} = \frac{U}{I_1 - I_3}$. (0.5 p) Calculated values are given on the table. (1.0 p) The averaged value is $R_2 = 2002\ \Omega \approx 2000\ \Omega$. (0.5 p)

b) (4 points) As $U = E - R_1.I_1$, the dependence $U(I_1)$ is straight line. (0.5 p) It crosses the ordinate at $E$ and has a slope $-R_1$. (0.5 p) The graph of the dependence is given below. (3.0 p)

c) (2 points) From the obtained graph we calculate the values of the voltage $E = 12.0$ V (1.0 p) and the resistance $R_1 = 2000\ \Omega$. (1.0 p)

d) (5 points) The value of the resistor is $R_3 = \frac{U}{I_3}$. (0.5 p) The so defined efficiency of the circuit is $\eta = \frac{U.I_3}{E.I_1}$. (0.5 p) The calculated values are listed in the table. (1.0 p) The graph of the dependence of the efficiency $\eta$ on the value of the resistance of the resistor $R_3$ is given below. (3.0 p)

e) (2 points) It is seen that the maximum of $\eta$ is somewhere in the interval (1200 Ω, 1500 Ω), (0.5 p) but as the dependence near the maximum is rather flat, it cannot be determined precisely from the presented data. (0.5 p) Good values are $\eta \approx$ 17%, (0.5 p) and $R_3 \approx 1350$ Ω. (0.5 p)

f) (5 points) The efficiency of the circuit is $\eta = \frac{U.I_3}{E.I_1} = \frac{{I_3}^2.R_3}{E.I_1}$. (0.5 p) In the simpler case $R_1 = R_2 = R$, $I_1 = \frac{E}{R + \frac{R.R_3}{R+R_3}}$. (0.5 p) As $R.I_2 = R_3.I_3$, $I_2 = I_3\frac{R_3}{R}$ . As $I_1 = I_2 + I_3$, $I_3 = I_1\frac{R}{R+R_3}$. (0.5 p) Substituting these results in the equation for the efficiency, $\eta = \frac{(I_1\frac{R}{R+R_3})^2.R_3}{E.I_1} = \frac{I_1.R^2.R_3}{E(R+R_3)^2} = \frac{R^2.R_3}{\left(R + \frac{R.R_3}{R+R_3}\right).(R+R_3)^2}$. (0.5 p) After simplifying the expression $= \frac{R.R_3}{2{R_3}^2 + 3R.R_3 + R^2}$ . (0.5 p) This expression can be written as a quadratic equation with variable $R_3$, $2\eta {R_3}^2 + (3\eta - 1)R.R_3 + \eta R^2 = 0$. (0.5 p) As for the maximal $\eta$ there is only one solution of $R_3$, then the discriminant of this quadratic equation must be zero: $D = (3\eta - 1)^2R^2 - 4.2\eta.\eta R^2 = 0$. (0.5 p) The new equation for $\eta$ is quadratic one: $\eta^2 - 6\eta + 1 = 0$. (0.5 p) This quadratic equation has roots $\eta = 3 \pm \sqrt{8}$. As $\eta < 1$, (0.5 p) only the root $3 - \sqrt{8} \approx 0.172$ is a solution. (0.5 p)

g) (4 points) Substituting the obtained value of $\eta_{max} = 3 - \sqrt{8}$ in the quadratic equation for $R_3$, $2\eta {R_3}^2 + (3\eta - 1)R.R_3 + \eta R^2 = 0$, we obtain $2(3 - \sqrt{8}){R_3}^2 + (3(3 - \sqrt{8}) - 1)R.R_3 + (3 - \sqrt{8})R^2 = 0$. (1.0 p) The discriminant of this quadratic equation is zero. (1.0 p) Then $R_3 = \frac{3\sqrt{8} - 8}{4(3 - \sqrt{8})}R$. (1.0 p) After simplification $R_3 = \frac{R}{\sqrt{2}}$. (1.0 p)

h) (1 point) The theoretical values of $\eta$ =17.2% (0.5 p) and $R_3 = \frac{2000\ \Omega}{\sqrt{2}} \approx 1414\ \Omega$ (0.5 p) are in excellent agreement with the experimental ones.