text:
Current through each resistor

I'm solving this high-school level problem, where I have to find the current through each resistor.

simulate this circuit – Schematic created using CircuitLab

There aren't any resistors in parallel or series, which could be combined, so I tried using Kirchoff's law for junctions and loops. The resulting system of linear equations doesn't have any solution according to Mathematica. $I_0$ is the current through the source, $I_{15}$ is an auxiliary current through the empty wire in the middle of the circuit. My resulting system of linear equations:

My questions are:

• Is my approach to this problem right? Maybe I chose wrong loops. Is there any rule for which loops I should choose?
• There are 16 unknown currents, I have 10 junction equations. Does it mean I have to create at least 6 loops, to find all the unknowns?
• If my approach is wrong, what is the first step or direction I should use to solve this problem?

EDIT: There are a few mistakes in the equations:

• equation 3 is $I_3=I_5+I_7$
• equation 15 and 16 don't include $I_{15}$ (there is no voltage change over the wire without any resistor)
• With these changes, the system can be solved by satisfying $\mathbf{A}^\intercal \mathbf{A}\vec{x}=\mathbf{A}^\intercal\vec{b}$. However, the accepted answer shows a much better approach to this problem (the results are the same).
• Current isn't "on" a resistor. Dec 29 '17 at 14:35
• @OlinLathrop sorry, I'm not native Dec 29 '17 at 14:36
• For starters I3=I5+I7, not I5+I9. Dec 29 '17 at 14:43
• @Finbarr thank you so much, this solves it. Do you want to post it as an answer? Dec 29 '17 at 14:46
• Now that is some highschool. Dec 29 '17 at 19:48

Extending Olin's answer for those that want to see how much of a difference the art makes.

simulate this circuit – Schematic created using CircuitLab

Once you see it that way it's pretty easy to tell, or should be, what voltage those nodes in the middle are at. After that it's simple Ohm's Law stuff.

You should also be able to see that the nodes I marked as $VT$ have the same voltage, as do those I marked with $VB$.

Since there is no voltage across $R6$ and $R9$,

$I6 = I9 = 0A$.

Because of that, $R6$ and $R9$ can be considered as shorts, so $R4$ is effectively in parallel with $R2$ ($R24 = 50R$). Which means $VT$ is $50/150$ or $1/3$ of $12.5V$ above the mid-point, that is,

$VT = 12.5 + 12.5/3 = 16.667V$

Similarly $R8$ is effectively in parallel with $R10$, making

$VB = 2/3 * 12.5V = 8.333V$.

So,

$I1 = I3 = I12 = I14 = 8.333/100 = 83.33mA$ $I2 = I4 = I5 = I7 = I8 = I13 = I11 = I10 = I1/2 = 41.667mA$

$Io = I1 + I2 = 166.67mA$

The puzzle is actually delightful in it's symmetry. Of course this only falls out like this because all the resistor values are identical. Had they not been, your original method, or some other analysis method, would be required.

• I think you've made one slip. You say "Since there is no voltage across R6 or R9, I6=I9=0A. Because of that, R6 and R9 can be considered as shorts." No, they can be considered as open (and so removed from the circuit entirely). I haven't worked this through, but it probably makes little difference to the rest of your answer. Dec 30 '17 at 1:04
• What does node voltage mean? Is it the voltage between the node and the grounded node? Dec 30 '17 at 11:13
• @Dreamer actually, no. Although they can be considered both shorts or open - because there's no voltage difference and no current, it really doesn't matter if you envision them as one or the other - it won't change neither the equations or the working logic a bit. Still, what Trevor says is true - if two points have the very same voltage regardless of operating conditions, considering them shorted is IMVHO much more sensible than considering them open - since for two open points, there no guarantee of 0A and deltaV = 0, while a short essentially guarantees it.
– user20088
Dec 30 '17 at 12:29
• @Dreamer as vaxquis mentioned then can be considered as either open or short, but I did not mention that in it's entirety. Since the voltage is the same at the ends of R4 and R2, it changes nothing if you short out or remove R6. Dec 30 '17 at 12:47

The first step in solving these kinds of problems is to redraw the schematic in a more logical layout. A main component to these problems is the deliberately obfuscated circuit. Often the problem becomes easier to solve, and certainly easier to see, when its schematic is drawn logically.

Put the power supply at left. Draw the + power as a line across the top, and the - power as a line across the bottom of the schematic. Now draw any obvious + to - resistors or strings of resistors vertically. Higher voltages go higher on the page than lower voltages. Show any resistor connected directly to + power vertically, going down from the + horizontal line. Similarly any resistor connected directly to - goes up vertically from the - line. Try to visually simplify the remaining resistors as much as possible.

We can get into your actual problem after you've posted the de-obfuscated schematic as described above.

• thanks for these tips for drawing circuits, we don't learn that at school Dec 29 '17 at 14:48

You have shown

$I_3=I_5+I_9$

$I_3=I_5+I_7$

• This makes my system of linear equations solvable: oi68.tinypic.com/f3tez4.jpg Dec 29 '17 at 14:57
• I suggest you try following the instructions from @OlinLathrop anyway. You will see that this diagram has a very neat symmetry to it when drawn better, and having all the resistors identical makes it easy to solve without relying on clever tools. Dec 29 '17 at 15:06

The key principle of this problem is symmetry.

[The other key principle I have demonstrated is not being lazy about a problem that looks easy! I previously got sloppy and tried to make it TOO simple. I believe I've corrected the errors.]

I would further say that the purpose of this problem is precisely to make you think clearly about the circuit, rather than try to solve it mathematically. I'm sad to hear that you didn't learn techniques for drawing clear circuit diagrams in school, because that's maybe the most useful practical skill in solving this problem. (When I saw your question, and decided to try to solve it for myself before reading the answers, redrawing it clearly was the first thing I did.)

Others have already given some useful information about the solution, but I'd like to try to give you my intuition for how to approach it.

When I first saw the diagram, I was immediately suspicious that it was probably more symmetrical than it looked. You described it as a 'high school' problem; sometimes school exercises will involve doing gruntwork, but more often they will involve clever tricks you have to spot. And though the drawing obscures this, it's pretty easy to see that each terminal of the power supply is connected to exactly two resistors, each of which is connected to exactly two more resistors; and all the resistors in the circuit have the same value.

Symmetry is a critical principle to keep in mind, across all kinds of problems in all kinds of classes, and beyond that into the working world. You may never have to solve the current flow through a resistor network once you graduate, but you can amaze your coworkers all your life by making complex problems trivial using hidden symmetry. It often lets you prove that dramatic simplifications of a problem are possible with very little work.

In this case, as you can see in Trevor's beautiful drawing, the resistor network in this problem is extremely symmetrical. Trevor uses this to prove that the voltages in the middle nodes must be 12.5 V -- can you see why? Because, whatever the total resistance comes out to, between those nodes and V+/V-, it is obvious that it's the SAME above and below. It's then easy to see that no current can flow in resistors r6 and r9, since as Trevor has shown, the voltages across them are 0 by symmetry.

Now we can continue using the left-right and top-bottom symmetry of the whole diagram to see that the voltages must also be the same at all the mirror points on the left and the right, and the currents must be the same through the paired resistors r1-r3-r12-r14, r2-r7-r8-r13, and r4-r5-r10-r11. That alone reduces this to a fairly simple set of equations to solve.

But we can go a little further. Given two points with the same voltage, it won't change the current flows to connect them; and given a wire through which no current flows, it won't change anything to disconnect it. So we can make the diagram even more symmetric in a couple of ways, the easiest of which is this: Since we know no current flows left-right or right-left across the center (by symmetry), we can disconnect r4-r6-r10 from r5-r9-r11. This reduces the problem to a pair of parallel circuits, so you can solve one and then apply the result to both (and within each one, there are parallel strings of identical resistors where you can do the same thing again.)

• I don't think this is correct, There is no potential across R6 and R9, so no current flow, but there is potential across R4, R5, R10, and R11 Dec 29 '17 at 19:13
• Hmmm, argh, I think you're right. I will think about whether my answer can be salvaged or not. Thanks. Dec 29 '17 at 19:16


text:
# apa6 package "man" document mode produces unwanted page 3

I'm trying to write my thesis as closely conforming to APA6 style as reasonable. I've scraped together a preamble from answers on this page that is messy, but works pretty well.

The only problem is that this code produces an unwated page 3 between the abstract page 2 and the table of contents page 4 (against APA guidelines). This page 3 only has the title of the thesis on it and I can't seem to get rid of it. It also appears in the table of contents.

This page disappears if I switch document mode to doc, so man mode seems to be the problem. However man mode most closely resembles what the APA6 guidelines dictate (running head and page number on page 1, abstract on page 2, amongst other things), so I don't want to switch to doc.

So how to I get rid of this pesky page 3 in man mode?

\documentclass[12pt,a4paper,man,notimes,draftfirst,noextraspace]{apa6}

\usepackage[american]{babel}
\usepackage[style=apa,sortcites=true,sorting=nyt,backend=biber]{biblatex}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{upgreek}
\usepackage{paralist}
\usepackage{csquotes}
\usepackage{lscape}
\usepackage[doublespacing]{setspace}

\usepackage{authblk}
\usepackage{nextpage}

\usepackage{calc}

\usepackage{epstopdf}

\usepackage{rotating}

\usepackage{hyperref}
\hypersetup{
citecolor=black,
filecolor=black,
urlcolor=black
}

\setlength{\oddsidemargin}{0cm}   % A4-paper format
\setlength{\evensidemargin}{0cm}  % A4-paper format
\setlength{\voffset}{0cm}       % A4-paper format
\setlength{\hoffset}{0cm}       % A4-paper format
\setlength{\textwidth}{15.92cm}        % A4-paper format
\setlength{\textheight}{24.62cm}       % A4-paper format

\clubpenalty = 10000
\widowpenalty = 10000
\displaywidowpenalty = 10000
\hyphenpenalty = 10000
\exhyphenpenalty = 10000

\raggedright
\setlength\parindent{.5in}
\bibindent=.5in
\setlength{\itemindent}{.5in}
\parskip 0pt

\captionsetup{justification=justified,font=singlespacing}

\DeclareLanguageMapping{american}{american-apa}

\DefineBibliographyStrings{american}{
}

\title{The effects of intervention on dependent variable}
\shorttitle{Effect of intervention}
\author{Yours Truly}
\affiliation{LaTeX University}
\date{}

\abstract{placeholder abstract}

\begin{document}

\maketitle

\clearpage

\tableofcontents

\clearpage

Theory

\clearpage

\section{Method}
Method

\clearpage

\section{Results}
Results

\clearpage

\section{Discussion}
Discussion text

\clearpage
\section{References}

\end{document}

• Welcome to TeX.SX! My doubt is: why using the man option that produces the worst typesetting on the face of earth? Jul 6 '15 at 13:37

Add donotrepeattitle to the \documentclass options:

\documentclass[
12pt,
a4paper,
man,
notimes,
draftfirst,
noextraspace,
donotrepeattitle
]{apa6}

• Thank you very much, that solved it. I wonder why this is not in the official documentation. Jul 6 '15 at 13:49

format: plain


text:
### Double Digit

Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?

### Repeaters

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

### Big Powers

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

# More Mathematical Mysteries

##### Age 11 to 14Challenge Level

Write down a three-digit number

Change the order of the digits to get a different number

Find the difference between the two three digit numbers

Add together the digits of your answer. If you arrive at a two-digit answer, add the two digits to reduce your answer to a single digit answer

Subtract $5$ from the number you have

Find that equivalent number letter in the alphabet (1=A, 2=B etc.)

Name a country that begins with letter

Using the second letter of the name of the country that you chose, name a jungle animal that starts with that letter.

Name the colour of that animal.

Using the final letter of the name of the country that you chose, name a different animal that starts with that letter.

Name a fruit that starts with the last letter of that animal's name.

Did you end up with orange or apple?

How did I know?


text:
# A clock is a one-dimensional subgroup of the torus

Recently, my partner and I installed a clock in our home. The clock previously belonged to my grandparents and we have owned it for a while. We hadn’t put it up earlier because the original clock movement ticked and the sound would disrupt our small studio apartment. After much procrastinating, I bought a new clock movement, replaced the old one and proudly hung up our clock.

When I first put on the clock hands I made the mistake of not putting them both on at exactly 12 o’clock. This meant that the minute and hour hands were not synchronised. The hands were in an impossible position. At times, the minute hand was at 12 and the hour hand was between 3 and 4. It took some time for me to register my mistake as at some times of the day it can be hard to tell that the hands are out of sync (how often do you look at a clock at 12:00 exactly?). Fortunately, I did notice the mistake and we have a correct clock. Now I can’t help noticing when others make the same mistake such as in this piece of clip art.

After fixing the clock, I was still thinking about how only some clock hand positions correspond to actual times. This led me to think “a clock is a one-dimensional subgroup of the torus”. Let me explain why.

### The torus

The minute and hour hands on a clock can be thought of as two points on two different circles. For instance, if the time is 9:30, then the minute hand corresponds to a point at the very bottom of the circle and the hour hand corresponds to a point 15 degrees clockwise of the leftmost point of the circle. As a clock goes through a 12 hour cycle the minute-hand-point goes around the circle 12 times and the hour-hand-point goes around the circle once. This is shown below.

If you take the collection of all pairs of points on a circle you get what mathematicians call a torus. The torus is a geometric shape that looks like the surface of a donut. The torus is defined as the Cartesian product of two circles. That is, a single point on the torus corresponds to two points on two different circles. A torus is plotted below.

To understand the torus, it’s helpful to consider a more familiar example, the 2-dimensional plane. If we have points $x$ and $y$ on two different lines, then we can produce the point $(x,y)$ in the two dimensional plane. Likewise, if we have a point $p$ and a point $q$ on two different circles, then we can produce a point $(p,q)$ on the torus. Both of these concepts are illustrated below. I have added two circles to the torus which are analogous to the x and y axes of the plane. The blue and red points on the blue and red circle produce the black point on the torus.

### Mapping the clock to the torus

The points on the torus are in one-to-one correspondence with possible arrangements of the two clock hands. However, as I learnt putting up our clock, not all arrangements of clock hands correspond to an actual time. This means that only some points on the torus correspond to an actual time but how can we identify these points?

Keeping with our previous convention, let’s use the blue circle to represent the position of the minute hand and the red circle to represent the position of the hour hand. This means that the point where the two circles meet corresponds to 12 o’clock.

There are eleven other points on the red line that correspond to the other times when the minute hand is at 12. That is, there’s a point for 1 o’clock, 2 o’clock, 3 o’clock and so on. Once we add in those points, our torus looks like this:

Finally, we have to join these points together. We know that when the hour hand moves from 12 to 1, the minute hand does one full rotation. This means that we have to join the black points by making one full rotation in the direction of the blue circle. The result is the black curve below that snakes around the torus.

The picture above should explain most of this blog’s title – “a clock is a one-dimensional subgroup of the torus”. We now know what the torus is and why certain points on the torus correspond to positions of the hands on a clock. We can see that these “clock points” correspond to a line that snakes around the torus. While the torus is a surface and hence two dimensional, the line is one-dimensional. The last missing part is the word “subgroup”. I won’t go into the details here but the torus has some extra structure that makes it something called a group. Our map from the clock to the torus interacts nicely with this structure and this makes the black line a “subgroup”.

### Another perspective

While the above pictures of the torus are pretty, they can be a bit hard to understand and hard to draw. Mathematicians have another perspective of the torus that is often easier to work with. Imagine that you have a square sheet of rubber. If you rolled up the rubber and joined a pair of opposite sides, you would get a rubber tube. If you then bent the tube to join the opposite sides again, you would get a torus! The gif bellow illustrates this idea

This means that we can simply view the torus as a square. We just have to remember that the opposite sides of the squares have been glued together. So like a game of snake on a phone, if you leave the top of the square, you come out at the same place on the bottom of the square. If we use this idea to redraw our torus it now looks like this:

As before we can draw in the other points when the minute hand is at 12. These points correspond to 1 o’clock, 2 o’clock, 3 o’clock…

Finally we can draw in all the other times on the clock. This is the result:

One nice thing about this picture is that it can help us answer a classic riddle. In a 12-hour cycle, how many times are the minute and hour hands on top of each other? We can answer this riddle by adding a second line to the above square. The bottom-left to top-right diagonal is the collection of all hand positions where the two hands are on top of each other. Let’s add that line in green and add the points where this new line intersects the black line.

The points where the green and black lines intersect are hand positions where the clock hands are directly on top of each other and which correspond to actual times. Thus we can count that there are exactly 11 times when the hands are on top of each other in a 12-hour cycle. It might look like there are 12 such times but we have to remember that the corners of the square are all the same point on the torus.

So far I have ignored the second hand on the clock. If we included the second hand, we would have three points on three different circles. The corresponding geometric object is a 3-dimensional torus. The 3-dimensional torus is what you get when you take a cube and glue together the three pairs of opposite faces (don’t worry if you have trouble visualising such a shape!).

The points on the 3-dimensional torus which correspond to actual times will again be a line that wraps around the 3-dimensional torus. You could use this line to find out how many times the three hands are all on top of each other! Let me know if you work it out.

I hope that if you’re ever asked to define a clock, you’d at least consider saying “a clock is a one-dimensional subgroup of the torus” and you could even tell them which subgroup!

## 2 thoughts on “A clock is a one-dimensional subgroup of the torus”

1. graham says:

awesome! pls make a toroidal clock! i think it would look awesome on a coffee table

Like


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This tool is very convenient to help you preview the symbol, including viewing the details of the symbol display and the effect displayed on the web page. Using Square root symbol shortcuts. Quadratic Equation. Use sqrt() Example: sqrt(9) link. Square Root Info. The square root of a number answers the question "what number can I multiply by itself to get this number?". If you enter a negative number, Minitab returns a missing value. The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. X 2 – Square-The x2 button computes the square of the number currently displayed. This section will focus on that. sqrt (Square Root) (Example: sqrt(9)) < (less than) > (greater than) <= (less than or equal to) >= (greater than or equal to) Need Help? Please feel free to Ask MathPapa if you run into problems. Update: ok thanks! And 4 divided by 2 is 2, and so on. The square root function is used by clicking on the "x" button or using the keyboard shortcut "r". So, a square looks like this: 4 2. The square root symbol is: $$\sqrt{16}$$ When we multiply 4 by itself, we get 16. Plus, if you ever want to square a number, your iPhone has a hidden scientific calculator … To find the square root of a number you would use the square root symbol on the calculator, √, then enter the number. A list of symbols will appear, one of which is the square root symbol. square root key missing on scientific calculator Problem using Windows Calculator I'm trying to find square root on scientific calculator; "help" sez use the "sqrt" key. With this tool, you can adjust the size, color, italic, and bold of Square Root(symbol). Can I use A Calculator with my … SQRT(number) For number, specify the number or column of numbers. Square root calculator online. 4 years ago. So √16 is 4 (and not -4). √ Note : This method works on stock Android keyboard or G-Board. 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text:
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2021-09-17, 04:09   #753
petrw1
1976 Toyota Corona years forever!

"Wayne"
Nov 2006

3×1,613 Posts
I guess as the "inventor" of this project...

Quote:
 Originally Posted by chalsall Would anyone like to coordinate the last bit of P-1'ing needed in 22.1, 22.7, and 22.8M? Oliver "did his thing".
Most importantly speak for your ranges to avoid toe stepping

There are actually 21 ranges that Oliver has completed to 74 bits (22.8 is special ... he is finished).
P.S. within a few weeks he will probably complete the rest (fingers crossed)

I have ordered them by "ToGo".
Then I have suggested P1 effort.
Always work on the exponents with B1=B2 first.
Do not fear that some ranges have NO B1=B2 remaining.
Most of these exponents have had very meagre P1 based on today's hardware.
In these cases look for low B1/B2 here .
Sort by B2 (click on the column) and start at the top.
The top exponents will likely have a current P1 success rate of under 2%; most will be under 3%
Choosing a B1/B2 that give 4% (+2%) for example will require on average 50 exponents per factor.
Each assignment will take about 2 GhzDays ... not a whole lot, right?

Code:
Range	To Go	B1=B2	Low P-1
This first bunch have so few remaining; don't pick B1/B2 with over 3-4% success rate
22.8M	1		302
29.6M	1	579	122
25.1M	2	497	162
24.6M	3		189
22.1M	4		265
23.6M	7	602	258
27.4M	9	761	163
22.7M	10	130	322
26.3M	10	722	145
28.1M	17	679	161
27.3M	17	712	153
27.7M	23	770	159
28.2M	24	610	157
28.5M	24	330	153
28.9M	25	643	154
27.5M	26	382	175

For these we'll need about a 5-5.5% success rate
27.8M	41	791	179
29.0M	47	715	176
28.0M	50	531	163

These 7-8% but probably better another bit of TF and then something more reasonable
27.2M	89	821	150
29.8M	113	789	100
Does anyone need actual recommended B1/B2 for PMinus1 or B1 for PFactor?


text:
## Isotropic Random Walks on Affine Buildings

### J. Parkinson

#### Abstract

Recently, Cartwright and Woess provided a detailed analysis of isotropic random walks on the vertices of thick affine buildings of type \tilde{A}n. Their results generalise results of Sawyer, where homogeneous trees are studied (these are \tilde{A}1 buildings), and Lindlbauer and Voit, where \tilde{A}2 buildings are studied. In this paper we apply techniques of spherical harmonic analysis to prove a local limit theorem, a rate of escape theorem, and a central limit theorem for isotropic random walks on arbitrary thick regular affine buildings of irreducible type, thus providing a broad generalisation of the \tilde{A}n case.

Keywords: Affine buildings, random walks, Macdonald spherical functions.

: Primary 20E42; secondary (60G50 33D52).

This paper is available as a gzipped postscript (332kB) file and a PDF (396kB) file.
 Tuesday, November 15, 2005