Съдържание

BPO 2022, Теория — Задача 1. Resistor circuit

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от BPO 2022

Problem 1: Resistor circuit · Problem 2: The descent of a skier · Problem 3: “The bead on the spring” · Problem 4: Galileo Galilei – “The minimum descent time” · 25 т.

Условие

A constant voltage $U_0$ is applied to the ab ends of the circuit in figure. The resistors have resistances $R_1=R_2=R_5=R$ , $R_3=2R$ , $R_4=4R$ . The voltmeter and the ammeter are considered ideal.

A. (9 points) With the switch S open determine:

Circuit between terminals a and b: upper branch R1 and R2 in series, lower branch R3 and R4 in series; from the node between R1 and R2 a switch S in series with an ammeter A and resistor R5 goes down to the node between R3 and R4; a voltmeter V connects the node between R1 and R2 to the lower branch between R3 and R4.

A1. The equivalent resistance, $R_{ab}$ , between the ab ends of the circuit. [4 т.]

A2. The values measured by the ammeter and the voltmeter. [3 т.]

A3. The power dissipated on the resistor $R_4$ . [2 т.]

B. (13 points) With the switch S closed determine:

B1. The equivalent resistance, $R_{ab}$ , between the ab ends of the circuit. [9 т.]

B2. The values measured by the ammeter and the voltmeter. [2 т.]

B3. The power dissipated on the resistor $R_4$ . [2 т.]

C. The switch S is closed and opened periodically, so that in a time period T, 2T/3 of the time is closed and T/3 of the time is opened. Determine the average power dissipated on the resistor $R_4$ . [3 т.]

Assume that all current values are reached instantaneously.

Отговори

Решение

A1. (4 points)

R_s = R_A + R_B \qquad \textbf{(1p)}

R_p = \frac{R_A R_B}{R_A + R_B} \qquad \textbf{(1p)}

R_{12} = R_1 + R_2 = 2R \qquad \textbf{(0.5p)}

R_{34} = R_3 + R_4 = 6R \qquad \textbf{(0.5p)}

R_{ab} = \frac{R_{12}R_{34}}{R_{12} + R_{34}} = 1{,}5R \qquad \textbf{(1p)}

A2. (3points)

I_A = 0 \qquad \textbf{(1p)}

I_{12} = \frac{U_0}{R_{12}} = \frac{U_0}{2R} \qquad \textbf{(0.5p)}

I_{34} = \frac{U_0}{R_{34}} = \frac{U_0}{6R} \qquad \textbf{(0.5p)}

U_V = \left|I_{34}R_3 - I_{12}R_1\right| = \frac{U_0}{6} \qquad \textbf{(1p)}

A3. (2 points)

P_4 = I_{34}^2 R_4 \qquad \textbf{(1p)}

P_4 = \frac{U_0^2}{9R} \qquad \textbf{(1p)}

**B1. (9 points)**\_

I_1R - I_5R - 2I_3R = 0 \qquad \textbf{(1p)}

The circuit redrawn with switch closed: current I enters at a and splits into I1 through R (upper left) and I3 through 2R (lower left); I2 flows through R (upper right), I4 through 4R (lower right); the middle vertical resistor R carries I5 upwards from the lower node to the upper node; terminal b on the right.

I_2R - 4I_4R + I_5R = 0 \qquad \textbf{(1p)}

I_1 + I_5 = I_2 \qquad \textbf{(1p)}

I_3 = I_4 + I_5 \qquad \textbf{(1p)}

I_1R + I_2R = U_0 \qquad \textbf{(1p)}

I_1 = \frac{16}{34}\frac{U_0}{R} \qquad \textbf{(1p)}

I_3 = \frac{7}{34}\frac{U_0}{R} \qquad \textbf{(1p)}

\left(I_1 + I_3\right)R_{ab} = \frac{23}{34}\frac{U_0}{R} \qquad \textbf{(1p)}

R_{ab} = \frac{34}{23}R \qquad \textbf{(1p)}

B2. (2 point)

I_5 = \frac{1}{17}\frac{U_0}{R} \qquad \textbf{(1p)}

U_5 = \frac{1}{17}U_0 \qquad \textbf{(1p)}

B3. (2 points)

I_4 = \frac{5}{34}\frac{U_0}{R} \qquad \textbf{(1p)}

P_4 = 4I_4^2R = \frac{25U_0^2}{289R} \qquad \textbf{(1p)}

C. (3 point)

P_{med} = \frac{L}{T} \qquad \textbf{(1p)}

L = P_1\Delta t_1 + P_2\Delta t_2 \qquad \textbf{(1p)}

P_{aver} \approx 0{,}09\frac{U_0^2}{R} \qquad \textbf{(1p)}

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