Съдържание

Instructions Table of Constants Условие Решение

INPhO 2018, Теория — Задача 3. An infinite ladder of resistors

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Всички материали от INPhO 2018

Indian National Physics Olympiad – 2018 · 28 януари 2018 г. · 12 т.

Внимание: Бележка към темата

All seven matching archived keys are retained, including alternative ladder methods, unequal-resistor generalization, accepted ranges and the full handwritten graph. Printed refraction, hourglass, Fibonacci and notation inconsistencies remain literal with separate source notes. The full original includes nineteen blank detailed-answer sheets and two graph papers.

Native cover and instructions

Indian National Physics Olympiad – 2018

Date: 28 January 2018. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,8\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ .

I permit/do not permit (strike out one) HBCSE to reveal my academic performance and personal details to a third party.

Besides the International Physics Olympiad (IPhO) 2018, do you also want to be considered for the Asian Physics Olympiad (APhO) 2018? For APhO 2018 and its pre-departure training, your presence will be required in Delhi and Vietnam from April 28 to May 15, 2018. In principle, you can participate in both Olympiads. Yes/No.

Full Name (BLOCK letters) Ms./Mr.: [blank].

Extra sheets attached: [blank box]. Date: [blank]. Centre (e.g. Kochi): [blank]. Signature: [blank].

(Do not write below this line)

Instructions

This booklet consists of 27 pages (excluding this sheet) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Rest mass of proton	$m_p$	$1.67\times10^{-27}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
Molar mass of water		$18.02\ \mathrm{g\cdot mol^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	4
2	8
3	12
4	9
5	10
6	17
7	15
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover and marking policy

Solutions of Indian National Physics Olympiad – 2018

The key cover repeats the date, roll-number form, APhO participation question, nine instructions, thirteen constant rows, seven-question score table and HBCSE address of the question cover, but prints “16 pages (excluding this sheet)”. The retained key has eighteen PDF pages: the cover, sixteen numbered pages and one appended handwritten graph sheet.

On its next page the key states: ANY ALTERNATIVE METHOD OF SOLUTION TO ANY QUESTION THAT IS SCIENTIFICALLY AND MATHEMATICALLY CORRECT, AND LEADS TO THE SAME ANSWER WILL BE ACCEPTED WITH FULL CREDIT. PARTIALLY CORRECT ANSWERS WILL GAIN PARTIAL CREDIT.

Условие

Consider an infinite ladder of resistors. The input current $I_0$ is indicated in the figure.

Complete original infinite resistor ladder, input I0, horizontal r, vertical R, In−1/In and primed branch currents, all ellipses.

a) Find the equivalent resistance of the ladder.

Summary answer field: Equivalent resistance = [blank]. [2 т.]

b) Find the recursion relation obeyed by the currents through the horizontal resistors $r$ . You will get a relationship where $I_n$ will be related to (may be several) $I_i$ s, $i<n$ , $n>0$ .

Summary answer field: Relation: [blank]. [2 т.]

c) Solve this for the special case $R=r$ to obtain $I_n$ and $I'_n$ as explicit functions of $n$ . You may have to make a reasonable assumption about the behaviour of $I_n$ as $n$ becomes large.

Summary answer fields: $I_n=$ [blank]; $I'_n=$ [blank]. [4 т.]

d) If the ladder is chopped off after the $N$ -th node (so that $I_{N+1}=0$ ) what will the form of $I_n/I_N$ be for $n\le N$ ?

Summary answer field: $I_n/I_N=$ [blank].

Detailed answers can be found on page numbers: [blank]. [4 т.]

Решение

(a)

Whole native equivalent ladder circuit, Req=X, r/R/X, open terminal circles and connecting wires.

Let equivalent resistance be $X$

r+\frac{RX}{R+X}=X

X=\frac{r\pm\sqrt{r^2+4rR}}2

Taking positive sign

X=\frac{r+\sqrt{r^2+4rR}}2

(b)

From Kirchhoff’s circuit law

I_{n-1}=I_n+I'_n\qquad (1)

Using Kirchhoff’s voltage law

rI_n+RI'_{n+1}-RI'_n=0\qquad (2)

From Eq. (1) we see that

I'_{n+1}-I'_n=(I_n-I_{n+1})-(I_{n-1}-I_n)=2I_n-I_{n+1}-I_{n-1}

So that Eq. (2) becomes

I_{n+1}-\left(2+\frac rR\right)I_n+I_{n-1}=0\qquad (3)

which is the required recursion relation.

(c)

First method:

One can see that $I_{n-1}$ , $I_n$ , $I_{n+1}$ form geometric progression. Also

Both whole native I1/I2 geometric-progressions circuit panels and all currents/resistances.

I_1=I_0\frac R{R+X}\qquad\text{(see the figure to the left)}

I_2=I_1\frac R{R+X}=I_0\left(\frac R{R+X}\right)^2\qquad\text{(see the figure to the right)}

Hence

I_n=I_0\left(\frac R{R+X}\right)^n

For $r=R$ ,

I_n=I_0\left(\frac{3-\sqrt5}2\right)^n=I_0k^n=I_0\left(\frac{\sqrt5-1}2\right)^{2n}

Using Eq. (2)

I'_n=I_0\left(\frac{\sqrt5-1}2\right)\left(\frac{3-\sqrt5}2\right)^{n-1}=I_0\left(\frac{\sqrt5-1}2\right)^{2n-1}

Second method:

For $R=r$ , Eq. (3) becomes

I_{n+1}-3I_n+I_{n-1}=0\qquad (4)

We can solve this linear recursion relation by assuming the ansatz $I_n\sim\rho^n$ . This leads to

\rho^2-3\rho+1=0

so that we have

\rho=\frac{3\pm\sqrt5}2

This means that the general solution to (4) is

I_n=A\left(\frac{3-\sqrt5}2\right)^n+B\left(\frac{3+\sqrt5}2\right)^n\qquad (5)

The second term grows without bound with increasing $n$ - so that for an infinite ladder we must have $B=0$ . Also when $n=0$ , $I_0=A$ . Thus

I_n=I_0\left(\frac{3-\sqrt5}2\right)^n

I'_n=I_{n-1}-I_n=I_0\left(\frac{\sqrt5-1}2\right)\left(\frac{3-\sqrt5}2\right)^{n-1}

(d)

First method: Consider the end part of the ladder as shown below.

Whole native finite ladder end geometry, ABCDEF, primed/unprimed currents, all arrows/resistors/ellipses.

Voltage drop across AB $=2RI_N=I'_NR\Rightarrow I'_N=2I_N$ and $I_{N-1}=3I_N$ .

Voltage drop across CD $=\frac{5R}3\,3I_N=I'_{N-1}R\Rightarrow I'_{N-1}=5I_N$ and $I_{N-2}=8I_N$ .

Voltage drop across EF $=\frac{13R}8\,8I_N=I'_{N-2}R\Rightarrow I'_{N-2}=13I_N$ and $I_{N-3}=21I_N..$

This is depicted in figure below.

Whole native finite ladder with Fibonacci current multipliers21/13/8/5/3/2/1 and ABCDEF nodes.

This looks like the terms of Fibonacci sequence. Fibonacci sequence is 1,1,2,3,5,8,13...etc in which $n$ th term of Fibonacci sequence is the sum of previous two terms i.e. $F_n=F_{n-1}+F_{n-2}$ , $F_1=F_2=1$ . In other words:

I_{N-1}=F_4,\ I_{N-2}=F_6,\ I_{N-3}=F_8,\ I_{N-4}=F_{10}\ldots

In general

\frac{I_{N-r}}{I_N}=F_{2r+2}\qquad (6)

\frac{I_n}{I_N}=F_{2(N-n)+2}\qquad (7)

Second method: See Eq. 5. If $I_{N+1}=0$ , we must have

A\rho_1^{N+1}+B\rho_2^{N+1}=0

where $\rho_{2,1}=(3\pm\sqrt5)/2$ . Thus $B=-A(\rho_1/\rho_2)^{N+1}$ and thus

I_n=A\left(\rho_1^n-\rho_2^n\left(\frac{\rho_1}{\rho_2}\right)^{N+1}\right)

and thus

\frac{I_n}{I_N}=\frac{\rho_1^n-\rho_2^n(\rho_1/\rho_2)^{N+1}}{\rho_1^N-\rho_2^N(\rho_1/\rho_2)^{N+1}}=\frac{\rho_1\rho_2}{\rho_2-\rho_1}\left[\rho_1^{n-N-1}-\rho_2^{n-N-1}\right]

Substituting the values, we get

\frac{I_n}{I_N}=\frac1{\sqrt5}\left[\left(\frac{3+\sqrt5}2\right)^{N+1-n}-\left(\frac{3-\sqrt5}2\right)^{N+1-n}\right]\qquad (8)

Equations (7) and (8) are equivalent.

Generalization

if $r\ne R$ :

Solution to Eq. (3) is

\rho_{2,1}=\frac{b\pm\sqrt{b^2-4}}2\quad\text{where}\quad b=2+\frac rR

Thus

\frac{I_n}{I_N}=\frac{\rho_1\rho_2}{\rho_2-\rho_1}\left[\rho_1^{n-N-1}-\rho_2^{n-N-1}\right]\qquad (9)

=\frac1{\rho_2-\rho_1}\left[\rho_2^{N+1-n}-\rho_1^{N+1-n}\right]\qquad (10)

Alternate form of Eqs. (7), (8) and (10) are also accepted.

Source note: Before Eq. (6), the source equates the currents $I_{N-1},I_{N-2},\ldots$ directly to dimensionless Fibonacci terms without an $I_N$ factor. The subsequent equations explicitly use current ratios. Both printed forms are retained.

Оригинал в Архива: INPhO2018-Question.pdf · решение от архива: INPhO2018-Solution-20180213.pdf