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Условие Отговори Решение

USAPhO 2004, Полуфинал — Задача 1. A1

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Всички материали от USAPhO 2004

AAPT UNITED STATES PHYSICS TEAM AIP 2004 — 2004 Semi-Final Exam · 25 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. 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

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
Coulomb’s constant	$k = 1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\ \mathrm{N\cdot m^2/C^2}$
Biot-Savart constant	$k_\mathrm{m} = \mu_0/4\pi = 10^{-7}\ \mathrm{T\cdot m/A}$
Speed of light in a vacuum	$c = 3.00 \times 10^{8}\ \mathrm{m/s}$
Boltzmann’s constant	$k_\mathrm{B} = 1.38 \times 10^{-23}\ \mathrm{J/K}$
Avogadro’s number	$N_\mathrm{A} = 6.02 \times 10^{23}\ \mathrm{(mol)^{-1}}$
Ideal gas constant	$R = N_\mathrm{A}k_\mathrm{B} = 8.31\ \mathrm{J/(mol\cdot K)}$
Stefan-Boltzmann constant	$\sigma = 5.67 \times 10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
Elementary charge	$e = 1.602 \times 10^{-19}\ \mathrm{C}$
1 electron volt	$1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}$
Planck’s constant	$h = 6.63 \times 10^{-34}\ \mathrm{J\cdot s} = 4.14 \times 10^{-15}\ \mathrm{eV\cdot s}$
Electron mass	$m = 9.109 \times 10^{-31}\ \mathrm{kg} = 0.511\ \mathrm{MeV/c^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$
Small angle approximations	$\sin\theta \approx \theta$ , $\cos\theta \approx 1 - \frac{1}{2}\theta^2$

Условие

The capacitors in the circuit diagrammed to the right are initially uncharged.

a. What is the current in each resistor immediately after the switch labeled S is closed? [10 т.]

After a very long time, the capacitors are fully charged.

b. What is the current in each resistor at this time? [5 т.]

c. What is the charge on each capacitor? [10 т.]

Circuit: a 6.00V battery with switch S and a 10.0 Ω resistor in the left branch; a 20.0 Ω resistor in the middle branch; in the right branch a 4.00 μF capacitor, a 30.0 Ω resistor and a 2.00 μF capacitor in series.

Отговори

Решение

a. At the instant the switch is closed, there is no charge on the capacitor and no voltage across it. Applying Kirchhoff’s junction rule $I_{10} = I_{20} + I_{30}$ .

Applying Kirchhoff’s loop rule to the right loop $0 = (20.0\ \Omega)I_{20} - (30.0\ \Omega)I_{30}$ .

Solving for $I_{20}$ and then $I_{10}$ , $I_{20} = 1.5I_{30}$ $\quad I_{10} = 1.5I_{30} + I_{30} = 2.5I_{30}$

Applying Kirchhoff’s loop rule to the left loop

0 = 6.00\ \mathrm{V} - (20.0\ \Omega)I_{20} - (10.0\ \Omega)I_{10} = 6.00\ \mathrm{V} - (20.0\ \Omega)(1.5I_{30}) - (10.0\ \Omega)(2.5I_{30})

0 = 6.00\ \mathrm{V} - (30.0\ \Omega)I_{30} - (25.0\ \Omega)I_{30} = 6.00\ \mathrm{V} - (55.0\ \Omega)I_{30}

Solving for the currents $I_{30} = \dfrac{6.00\ \mathrm{V}}{55.0\ \Omega} = 0.109\ \mathrm{A}$ . $\quad I_{20} = 1.5(0.109\ \mathrm{A}) = 0.164\ \mathrm{A}$

and $I_{10} = 2.5(0.109\ \mathrm{A}) = 0.273\ \mathrm{A}$

$\{$ A solution using parallel and series resistor combinations is equally valid. $\}$

b. When the switch has been closed for a very long time, the capacitors are fully charged. Current no longer flows in the capacitor branch and $I_{30} = 0$

Applying Kirchhoff’s loop rule to the left loop

0 = 6.00\ \mathrm{V} - (20.0\ \Omega)I_{10} - (10.0\ \Omega)I_{10} = 6.00\ \mathrm{V} - (30.0\ \Omega)I_{10}

I_{20} = I_{10} = \frac{6.00\ \mathrm{V}}{30.0\ \Omega} = 0.200\ \mathrm{A}

c. The two capacitors are in series. Both have the same charge. The equivalent capacitance is

\frac{1}{C_{eq}} = \frac{1}{C_2} + \frac{1}{C_4} = \frac{1}{2.00\ \mu\mathrm{F}} + \frac{1}{4.00\ \mu\mathrm{F}} = \frac{3}{4.00\ \mu\mathrm{F}}

C_{eq} = 1.33\ \mu\mathrm{F}.

The voltage across the equivalent capacitance is the same as that across the 20.0 Ω resistor.

V_c = V_{20} = (20.0\ \Omega)I_{20} = (20.0\ \Omega)(0.200\ \mathrm{A}) = 4.00\ \mathrm{V}.

So the charge

Q_2 = Q_4 = Q_{eq} = C_{eq}V_c = (1.33\ \mu\mathrm{F})(4.00\ \mathrm{V}) = 5.33\ \mu\mathrm{C}

Оригинал в Архива: Semifinal 2004.pdf · официални решения: Semifinal 2004 Solutions.pdf