Съдържание

Условие Решение

USAPhO 2005, Полуфинал — A1

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2005

2005 Semi-Final Exam · 25 т.

2005 Semi-Final Exam — Instructions

AAPT / AIP — United States Physics Team 2005.

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. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s, or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Questions with the same point value are not necessarily of the same difficulty.
Do not discuss the contents of this exam with anyone until after the submission deadline.
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	$k=1/4\pi\epsilon_0=8.99\times10^9$ 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	$c=3.00\times10^8$ 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)
Stefan-Boltzmann constant	$\sigma=5.67\times10^{-8}$ J/(s·m $^2$ ·K $^4$ )
Elementary charge	$e=1.602\times10^{-19}$ C
1 electron volt	$1\text{ eV}=1.602\times10^{-19}$ 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	$m=9.109\times10^{-31}\text{ kg}=0.511\text{ MeV}/c^2$
Binomial expansion	$(1+x)^n\approx1+nx$ for $\lvert x\rvert\ll1$
Small angle approximations	$\sin\theta\approx\theta$ , $\cos\theta\approx1-\tfrac12\theta^2$

Условие

Seven-resistor network between terminalsA andB, with labelsR1 throughR7.

Each of the seven resistors in the network above has the same resistance $R$ .

a. If an ideal battery of emf $\mathcal E$ and no internal resistance is connected across the points labeled A and B, what is the current $I_j$ that flows in the $j^{th}$ resistor $R_j$ (for each of the seven resistors)? [10 т.]

b. What equivalent resistor could be used to replace the network so that the same current flowed from the battery? [5 т.]

c. If the resistor labeled $R_4$ was removed and replaced by a capacitor with capacitance $C$ , what would the charge be on the capacitor after the circuit had been connected to the battery for a very long time? [10 т.]

Решение

a. (10 т.)

Since $R_3$ and $R_6$ are in series, $I_3=I_6$ . Since $R_2$ and $R_5$ are in series, $I_2=I_5$ . By symmetry, the current through $R_3$ and $R_6$ must be the same as the current through $R_2$ and $R_5$ since the current through A = the current through B and this current splits into two branches — one with a resistance of $2R$ and the other with a resistance of $R$ . Therefore,

I_2=I_3=I_5=I_6\qquad (A1-1)

and $I_1=I_7$ . (A1-2)

By the junction rule,

I_1=I_2+I_4.\qquad (A1-3)

Applying the loop rule from A to B across the top of the circuit,

\epsilon=I_1R+2I_2R.\qquad (A1-4)

Applying the loop rule from A to B going through $R_1$ , $R_4$ , and $R_7$ yields

\epsilon=I_1R+I_4R+I_1R.\qquad (A1-5)

Substitute (A1-3) into (A1-4) and (A1-5).

\epsilon=3I_2R+I_4R\qquad (A1-6)

and $\epsilon=2I_2R+3I_4R$ . (A1-7)

Subtract (A1-7) from (A1-6) to obtain

I_2=2I_4.\qquad (A1-8)

Substitute into (A1-6) or (A1-7) to find that

I_4=\frac\epsilon{7R}.\qquad (A1-9)

Then, from (A1-8) and (A1-1),

I_2=I_3=I_5=I_6=\frac{2\epsilon}{7R}.\qquad (A1-10)

Using (A1-3) and (A1-2),

I_1=I_7=\frac{3\epsilon}{7R}.\qquad (A1-11)

b. (5 т.)

I_{\rm battery}=\frac\epsilon{R_{\rm equiv}}\qquad (A1-12)

and $I_{\rm battery}=I_1+I_2$ . (A1-13)

Therefore, by substituting (A1-10) and (A1-11) into (A1-12) and (A1-13),

\frac\epsilon{R_{\rm equiv}}=\frac{5\epsilon}{7R}.\qquad (A1-14)

Thus, $R_{\rm equiv}=7R/5$ . (A1-15)

c. (10 т.)

After a very long time, no current will flow through the branch of the circuit with the capacitor. Therefore, $R_1$ , $R_2$ , and $R_5$ are in series and $R_3$ , $R_6$ , and $R_7$ are in series. The current through the top branch and the current through the bottom branch will then be

I=\frac\epsilon{3R}.\qquad (A1-16)

The voltage drop across $R_1$ is $IR$ and the voltage drop across $R_3$ and $R_6$ is $2IR$ . Thus, the voltage drop across the capacitor is $2IR-IR=IR$ .

Using (A1-16),

V_{\rm cap}=\frac\epsilon3.\qquad (A1-17)

The charge on the capacitor is $Q=CV_{\rm cap}=C\epsilon/3$ . (A1-18)

Оригинал в Архива: Semifinal 2005.pdf · официални решения: Semifinal 2005 Solutions.pdf