Съдържание

Условие Решение

USAPhO 2000, Полуфинал — A2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2000

2000 Semi-Final Exam · 20 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Work Part A first. You have 90 minutes to complete all four problems.
After you have completed Part A, you may take a break.
Then work Part B. You have 90 minutes to complete both problems.
Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

A1 – 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. It may not be used in graphing mode. Calculators may not be shared. You may not use any tables, books, or collections of formulas. You may use a ruler or straight edge.
Questions with the same point value are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8$ N/kg
Newton's gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Coulomb's constant	$k=1/4\pi\epsilon_0=9.0\times10^9$ N·m $^2$ /C $^2$
Biot-Savart constant	$k_m=\mu_0/4\pi=10^{-7}$ T·m/A
Speed of light in a vacuum	$c=3.0\times10^8$ m/s
Boltzmann's constant	$k_B=1.38\times10^{-23}$ J/K
Avogadro's number	$N_A=6.02\times10^{23}$ (mol) $^{-1}$
Ideal gas constant	$R=N_Ak_B=8.31$ J/mol·K
Elementary charge	$e=1.6\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.6\times10^{-19}$ J
Planck's constant	$h=6.63\times10^{-34}\text{ J}\cdot\text{s}=4.14\times10^{-15}\text{ eV}\cdot\text{s}$
Electron mass	$m=9.1\times10^{-31}\text{ kg}=0.51\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\simeq1+nx\quad\text{for }\lvert x\rvert\ll1$

Условие

The circuit diagrammed at the right consists of a 6.0-V power supply and eight 12.0- $\Omega$ resistors, labeled 1–8.

6.0-V supply and eight resistors labelled 1 through 8.

a. What is the equivalent resistance of the circuit? [5 т.]

b. What is the current in each resistor? Please use $I_j$ to denote the current in resistor $j$ . [5 т.]

c. Resistors 1, 5, and 7 are replaced with 4.0- $\mu$ F capacitors. What is the charge on each capacitor after the circuit has been connected for a very long time? Please use $Q_j$ to denote the charge on the capacitor in location $j$ . [5 т.]

d. The capacitor in location 5 is replaced with a second 6.0-V power supply (positive end closest to resistor 3 and negative end closest to resistor 6). What is the current through each resistor after the circuit has been connected for a very long time? Please use $I_j$ to denote the current in resistor $j$ . In this part 2, 3, 4, 6, and 8 denote 12.0- $\Omega$ resistors, 1 and 7 denote 4.0- $\mu$ F capacitors, and 5 denotes a 6.0-V power supply. [5 т.]

Решение

a. (5 т.)

Because of symmetry, $I_1=I_2$ and $I_4=I_5$ and $I_7=I_8$ .

No current flows through 3 or 6. $I_3=0$ and $I_6=0$ .

1, 4, and 7 are effectively in series with an equivalent resistance of $R_s=R+R+R=3R$ .

Similarly for the right branch containing 2, 5, and 8.

The left and right branches are in parallel so, the equivalent resistance is given by

\frac1{R_{eq}}=\frac1{3R}+\frac1{3R}=\frac2{3R}

R_{eq}=\frac{3R}{2}=\frac{3(12\,\Omega)}2=18\,\Omega.

b. (5 т.)

As stated in the argument in Part a. $I_3=0$ and $I_6=0$ . All the other currents are equal $I_1=I_2=I_4=I_5=I_7=I_8=I$ .

Applying Kirchhoff’s loop rule to the left loop

0=6.0\text{ V}-I(3R),\qquad I=\frac{6.0\text{ V}}{3R}=\frac{6.0\text{ V}}{36\,\Omega}=\frac16\text{ A}.

c. (5 т.)

When the circuit has been connected for a very long time, the capacitors are fully charged. No current flows through 1, 5, and 7. There is only one path for the current, through 2, 3, 4, 6, and 8. These five resistors are effectively in series with

V_R=(1/5)\,6.0\text{ V}=1.2\text{ V}

across each. Applying Kirchoff’s loop rule to the upper loop $0=V_R+V_R-V_{C1}$

Q_1=CV_{C1}=C\,2V_R=(4.0\,\mu\text{F})(2.4\text{ V})=9.6\,\mu\text{C}

Similarly for $Q_7=9.6\,\mu\text{C}$ .

For the middle loop we have $0=V_R+V_R+V_R-V_{C5}$

Which yields $Q_5=CV_{C5}=C\,3V_R=(4.0\,\mu\text{F})(3.6\text{ V})=14.4\,\mu\text{C}$

d. (5 т.)

When the circuit has been connected for a very long time, no current flows through 1 or 7. The same current flows through 3, 4 and 6.

0=6.0\text{ V}-3RI,\qquad I_3=I_4=I_6=\frac{6.0\text{ V}}{36\,\Omega}=\frac16\text{ A}.

There is 2.0 V across each of the resistors 3, 4, and 6 and capacitors 1 and 7. Hence the voltage across 2 and 8 is zero and $I_2=I_8=0$ .

Оригинал в Архива: Semifinal 2000.pdf · официални решения: Semifinal 2000 Solutions.pdf