Съдържание

IPhO 1971, Теория и експеримент — Задача 3. Question 3.

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 1971

V International Physics Olympiad, 1971. Sofia, Bulgaria

Внимание: Бележка към темата

Решенията са архивна адаптация на Victor Ivanov, както е посочено в оригинала. Формули (1.13), (1.17) и (4.6), както и повтореното означение P₂ в числените резултати на задача 2, са запазени според източника. Добавеният множител g в краткия отговор на задача 1 е отделно обозначена редакционна поправка; текстът на архивното решение остава непроменен.

Source

The problems and the solutions are adapted by Victor Ivanov University of Sofia, Faculty of Physics, 5 James Bourchier Blvd., 1164 Sofia, Bulgaria

Reference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High School Students”, eds. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).

Theoretical problems

Solutions — source

Solutions to the problems of the 5-th International Physics Olympiad, 1971, Sofia, Bulgaria

The problems and the solutions are adapted by Victor Ivanov Sofia State University, Faculty of Physics, 5 James Bourcier Blvd., 1164 Sofia, Bulgaria

Reference: O. F. Kabardin, V. A. Orlov, in “International Physics Olympiads for High School Students”, eds. V. G. Razumovski, Moscow, Nauka, 1985. (In Russian).

Theoretical problems

Условие

Four batteries of EMF $E_1 = 4\ \mathrm{V}$ , $E_2 = 8\ \mathrm{V}$ , $E_3 = 12\ \mathrm{V}$ , and $E_4 = 16\ \mathrm{V}$ , four capacitors with the same capacitance $\mathrm{C}_1 = C_2 = C_3 = C_4 = 1\ \mu\mathrm{F}$ , and four equivalent resistors are connected in the circuit shown in Fig. 3. The internal resistance of the batteries is negligible.

• Calculate the total energy $W$ accumulated on the capacitors when a steady state of the system is established.

• The points H and B are short connected. Find the charge on the capacitor $C_2$ in the new steady state.

Circuit drawn as a cube with vertices A, B, C, D, E, F, G, H: capacitors C1 (D–C), C2 (C–G), C3 (A–E), C4 (E–F); batteries E1 (A–D), E2 (B–C), E3 (H–E), E4 (F–G); four resistors joined near H and B in the middle of the cube. — Fig. 3

Отговори

Решение

A circuit equivalent to the given one is shown in Fig. 3. In a steady state (the capacitors are completely charged already) the same current $I$ flows through all the resistors in the closed circuit ABFGHDA. From the Kirchhoff’s second rule we obtain:

(3.1)

I = \frac{E_4 - E_1}{4R}.

Next we apply this rule for the circuit ABCDA:

(3.2) $V_1 + IR = E_2 - E_1$ ,

where $V_1$ is the potential difference across the capacitor $C_1$ . By using the expression (3.1) for $I$ , and the equation (3.2) we obtain:

(3.3)

V_1 = E_2 - E_1 - \frac{E_4 - E_1}{4} = 1\,\mathrm{V}.

Similarly, we obtain the potential differences $V_2$ and $V_4$ across the capacitors $C_2$ and $C_4$ by considering circuits BFGCB and FGHEF:

(3.4)

V_2 = E_4 - E_2 - \frac{E_4 - E_1}{4} = 5\,\mathrm{V},

(3.5)

V_4 = E_4 - E_3 - \frac{E_4 - E_1}{4} = 1\,\mathrm{V}.

Finally, the voltage $V_3$ across $C_3$ is found by applying the Kirchhoff’s rule for the outermost circuit EHDAH:

(3.6)

V_3 = E_3 - E_1 - \frac{E_4 - E_1}{4} = 5\,\mathrm{V}.

The total energy of the capacitors is expressed by the formula:

(3.7)

W = \frac{C}{2}\left(V_1^2 + V_2^2 + V_3^2 + V_4^2\right) = 26\ \mu\mathrm{J}.

Equivalent planar circuit: top branch A–R–B–R–F–C4–E with C3 on an outer wire above; vertical batteries E1 (A–D), E2 (B–C), E4 (F–G), E3 (E–H); bottom branch D–C1–C–C2–G–R–H; an outer wire with resistor R joins D and H at the bottom. — Fig. 3

When points B and H are short connected the same electric current $I'$ flows through the resistors in the BFGH circuit. It can be calculated, again by means of the Kirchhoff’s rule, that:

(3.8)

I' = \frac{E_4}{2R}.

The new steady-state voltage on $C_2$ is found by considering the BFGCB circuit:

(3.9) $V_2' + I'R = E_4 - E_2$

or finally:

(3.10)

V_2' = \frac{E_4}{2} - E_2 = 0\,\mathrm{V}.

Therefore the charge $q_2'$ on $C_2$ in the new steady state is zero.

Оригинал в Архива: ipho71_problems.pdf · решение от архива: ipho71_solutions.pdf · допълнителни данни: 1971_Sofia_Problems.pdf · допълнителни данни: 1971_Sofia_Solutions.pdf