Съдържание

USAPhO 2016, Полуфинал — Задача 2. Question A2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2016

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

The native worked key contains printed mathematical inconsistencies at A1(b), B1 and B2(a)(iii)/(c)(ii); the source formulas and adjacent reasoning are retained as printed. The A3 key graphs omit the requested event-letter labels and do not draw the stopped-rod world-line changes; no new answer labels or lines have been invented. Exact originals and detailed source-defect evidence accompany this transcription.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 4 parts: the cover sheet (page 2), Part A (pages 3-6), Part B (pages 8-9), and several answer sheets for one of the questions in Part A (pages 11-12). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2016.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN - Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. - After you have completed Part A you may take a break. - Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. - Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. - Start each question on a new sheet of paper. Put your AAPT ID number, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

AAPT ID # 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. - In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 15, 2016.

Possibly Useful Information. You may use this sheet for both parts of the exam.

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Условие

A student designs a simple integrated circuit device that has two inputs, $V_a$ and $V_b$ , and two outputs, $V_o$ and $V_g$ . The inputs are effectively connected internally to a single resistor with effectively infinite resistance. The outputs are effectively connected internally to a perfect source of emf $\mathcal E$ . The integrated circuit is configured so that $\mathcal E=G(V_a-V_b)$ , where $G$ is a very large number somewhere between $10^7$ and $10^9$ . The circuits below are chosen so that the precise value of $G$ is unimportant. On the left is an internal schematic for the device; on the right is the symbol that is used in circuit diagrams.

Both complete native internal circuit and device symbol: Va/Vb/Vo/Vg, input resistor, emf source polarity and a/b/o/g terminal labels.

a. Consider the following circuit. $R_1=8.2\ \mathrm{k\Omega}$ and $R_2=560\ \Omega$ are two resistors. Terminal $g$ and the negative side of $V_{\mathrm{in}}$ are connected to ground, so both are at a potential of $0$ volts. Determine the ratio $V_{\mathrm{out}}/V_{\mathrm{in}}$ .

Full native feedback-amplifier circuit, Vin/Vout and a/b/o/g terminals, R1/R2, crossing-wire bridge, ground and all wires.

b. Consider the following circuit. All four resistors have identical resistance $R$ . Determine $V_{\mathrm{out}}$ in terms of any or all of $V_1$ , $V_2$ , and $R$ .

Full native four-resistor circuit with all four R labels, V1/V2/Vout and a/b/o/g terminals, source polarities, ground and feedback wires.

c. Consider the following circuit. The circuit has a capacitor $C$ and a resistor $R$ with time constant $RC=\tau$ . The source on the left provides variable, but bounded voltage. Assume $V_{\mathrm{in}}$ is a function of time. Determine $V_{\mathrm{out}}$ as a function of $V_{\mathrm{in}}$ , and any or all of time $t$ and $\tau$ .

Complete native RC circuit: Vin/Vout, C/R, a/b/o/g terminals, source circle, ground and all feedback/connection wires.

Решение

The key idea is that if $\mathcal E$ is finite, then $V_a\approx V_b$ , since $G$ is so large. If we work exactly, then the answers will contain terms like $(V_b-V_a)/G$ which are negligible. Thus we can find the same answers by just setting $V_a=V_b$ .

a.

For this first part, we will not assume $V_a=V_b$ . Since terminal $g$ is grounded, $V_g=0$ and $V_a=V_{\mathrm{in}}$ , so $V_{\mathrm{out}}=G(V_{\mathrm{in}}-V_b)$ . No current runs between $a$ and $b$ , so any current through $R_1$ also flows through $R_2$ . Then Ohm's law gives

\frac{V_b}{R_2}=\frac{V_{\mathrm{out}}}{R_1+R_2}\quad\Rightarrow\quad V_{\mathrm{out}}=G\left(V_{\mathrm{in}}-V_{\mathrm{out}}\frac{R_2}{R_1+R_2}\right)

and solving for $V_{\mathrm{out}}$ gives

V_{\mathrm{out}}=V_{\mathrm{in}}\frac1{\frac1G+\frac{R_2}{R_1+R_2}}.

But since $G\gg R_1/R_2$ , we can neglect the $1/G$ term, giving

\frac{V_{\mathrm{out}}}{V_{\mathrm{in}}}\approx\frac{R_1+R_2}{R_2}.

This circuit is an amplifier with feedback.

b.

For this part, we will assume $V_a=V_b$ . Again $V_g=0$ , and if current $I$ flows through the bottom resistor (below the $b$ and $g$ terminals) then $V_2=2V_b$ , since the voltage drop across the bottom two resistors must be equal. Similarly, the voltage drop across the top two resistors is equal, so $V_1+V_{\mathrm{out}}=2V_a$ . Then

V_{\mathrm{out}}=2V_a-V_1=V_2-V_1.

This circuit is a subtractor.

c.

We again set $V_a=V_b=0$ . Then the capacitor charge and current satisfy

Q=CV_{\mathrm{in}},\quad \dot Q=-\frac{V_{\mathrm{out}}}R

where the second result follows from Ohm's law. Then

\frac{V_{\mathrm{out}}}R=-C\frac{dV_{\mathrm{in}}}{dt}\quad\Rightarrow\quad V_{\mathrm{out}}=-\tau\frac{dV_{\mathrm{in}}}{dt}.

This circuit is a differentiator.

Оригинал в Архива: USAPhO-2016.pdf · официални решения: USAPhO-2016-Solutions.pdf