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INPhO 2022, Теория — Задача 5

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Всички материали от INPhO 2022

Indian Olympiad Qualifier in Physics (IOQP) 2021-2022 — Part II: Indian National Physics Olympiad (INPhO) · 13 март 2022 г. · 12 т.

Указания

This booklet consists of 4 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.
Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Условие

A circuit consists of an emf source and five resistors with unknown resistances. When an ideal ammeter is connected between points 1 and 2, its reading is $I_A$ . If instead a resistor $R$ is connected to the same two points, the current through that resistor is $I_R$ . If instead an ideal voltmeter is connected between points 1 and 2, its reading is $V$ . Obtain $V$ in terms of $I_A$ , $R$ and $I_R$ only.

Circuit with three parallel horizontal branches between two vertical wires: top branch R1 and R3 with terminal 1 between them, middle branch R2 and R4 with terminal 2 between them, bottom branch R0 in series with a cell.

Отговори

Решение

The circuit redrawn with a resistance RM between terminals 1 and 2 and currents marked: I1 through R1, I1 − IM through R3, IM through RM, I2 through R2, I2 + IM through R4, I1 + I2 through R0, and the emf ε of the cell.

Resistances, currents and emf are shown in the diagram. Imagine also there is a resistance $R_M$ between terminals 1 and 2. We can set later $R_M = 0$ for an ideal ammeter and $I_M = 0$ for an ideal voltmeter placed between two terminals. Applying Krichoff’s law

\varepsilon = (I_1 + I_2)R_0 + I_1 R_1 + (I_1 - I_M)R_3 = (R_0 + R_1 + R_3)I_1 + I_2 R_0 - I_M R_3 \qquad (5.1)

= R_{013} I_1 + R_0 I_2 - R_3 I_M \qquad (5.2)

\varepsilon = (I_1 + I_2)R_0 + I_2 R_2 + (I_2 + I_M)R_4 = (R_0 + R_2 + R_4)I_2 + I_1 R_0 + I_M R_4 \qquad (5.3)

= R_{024} I_2 + R_0 I_1 + R_4 I_M \qquad (5.4)

Solving above equations

I_1 = \frac{\varepsilon R_{24} + I_M (R_0 R_{34} + R_3 R_{24})}{R_0 R_{1234} + R_{13} R_{24}} \quad \text{and} \quad I_2 = \frac{\varepsilon R_{13} - I_M (R_0 R_{34} + R_4 R_{13})}{R_0 R_{1234} + R_{13} R_{24}} \qquad (5.5)

Potential difference between terminal 1 and 2 is

V_{12} = I_2 R_2 - I_1 R_1 = \varepsilon A - I_M B \qquad (5.6)

where the coefficients $A$ and $B$ depends only on resistances in the circuit. When an ideal ammeter is placed between terminals $V_{12} = 0$ and $I_M = I_A$ .

\varepsilon A = I_A B

When a resistance $R$ is placed between terminals, $I_M = I_R$ and $V_{12} = I_R R$ . This yields

I_R R = \varepsilon A - I_R B = I_A B - I_R B \Rightarrow B = \frac{I_R R}{I_A - I_R} \qquad (5.7)

When an ideal voltmeter is placed between terminals, $I_M = 0$ . Hence $V_{12} = \varepsilon A - 0 = I_A B$ or

V_{12} = \frac{I_A I_R}{I_A - I_R} R

Оригинал в Архива: IOQP2022-PartII-Questions-en.pdf · официални решения: IOQP2022-PartII-Solutions-20220503.pdf