A simple diversionary puzzle

[IanB]

I sometimes like to think about things like this, to help keep my mind sharp. It is of no particular relevance here, other than amusement.

Solving such a problem is not itself the challenge, but rather it is seeing how to break the problem apart and render it easier to solve. Can you look at a problem, and instead of working it through laboriously, can you see how to decompose it, to break it apart, and turn it into a much simpler problem?

Yes, this example may not challenge the EEs on this forum, but I see lots of more complex problems in my work that would similarly benefit from being deconstructed into simpler problems with smaller, more efficient solutions. It is a skill that should fill a valuable space in any engineer's armory.

So here is a simple puzzle problem, maybe a diversion over a glass of beer. Can you solve it in your head, without pencil and paper? Can you do it in a minute or two?

Obviously you could solve it on a computer or calculator, but that's not the point. Many engineering design problems are not simple numerical problems and don't have a single well defined answer. Computers can't solve those for you.

Don't give away working or clues in replies below, just mention whether you did in fact see a rapid way to work it out in your head, and approximately how long it took you to arrive at the solution.

[c4757p]

2min. I did jot a couple numbers on paper, just because I have terrible memory for numbers, but no diagramming or anything.

And I had mine over coffee, beer is gross.

[Rerouter]

as you ruled out calculators and pen and paper, it ended up taking me closer to 7 minutes, sadly i think I've missed the some obvious solution that you have nutted out

[IanB]

2min. I did jot a couple numbers on paper, just because I have terrible memory for numbers, but no diagramming or anything.

I thought you might get it pretty quickly   

I also think you should have no problem landing a job when you graduate. If I were hiring in the EE space I would certainly invite you for interview 

[Phaedrus]

Yeah, I got it. Cute. Not too difficult though. Give us a hard one.

[mariush]

I cheated... took me a minute to draw the circuit in http://www.falstad.com/circuit/ and simulate it.

[c4757p]

I cheated... took me a minute to draw the circuit in http://www.falstad.com/circuit/ and simulate it.

Hey, I think I know you from school!

[IanB]

as you ruled out calculators and pen and paper, it ended up taking me closer to 7 minutes, sadly i think I've missed the some obvious solution that you have nutted out

The existence of a simple and direct solution method to an apparently complex problem that gives the same answer as more elaborate methods is the kind of thing I find fascinating. In fact I tried to check my simple answer using node equations and I kept making mistakes in the arithmetic and getting it wrong. So finding simple shortcuts can not only save you from effort,  they can also save you from error.

[c4757p]

using node equations

*shudder*

[Phaedrus]

I sent you a message with my guess, but I'm not sure if it got sent or not (this forum has weird software). Do I win a cookie?

[IanB]

I sent you a message with my guess, but I'm not sure if it got sent or not (this forum has weird software). Do I win a cookie?

Sorry, no cookie 

The current is a simple round fraction.

[Phaedrus]

I know what I did. 

[nuhamind2]

I did this.Tie 12V together,tie GND together . Then it looked like a Wheatstone bridge with balanced resistor, but unfortunately not (hoping it was a trap question).I'm thinking of delta-Y transformation, find the current,find voltage across 13Ohm,find the current on it but I don't think it's the most direct solution.

Perhaps a clue 

[c4757p]

Perhaps a clue 

How's "delta-Y is about as far from what you want as possible" for a clue?

[IanB]

Perhaps a clue 

Well a transformation is along the right lines. There is a way of looking at the circuit that makes it appear like a different, simpler circuit that is easier to solve.

[nuhamind2]

Thevenin ?
0.2A ?

[c4757p]

Shh....

[AndyC_772]

Can you look at a problem, and instead of working it through laboriously, can you see how to decompose it, to break it apart, and turn it into a much simpler problem?

Yes - it took me a few minutes, mental arithmetic not being my strong point first thing in the morning when my coffee hasn't really soaked in. The basic approach was immediately obvious, and you've been very kind with the numbers.

Anyone who solves it the same way I did will spot a certain symmetry in the puzzle which isn't immediately apparent.

[johnwa]

Fairly simple using equivalent circuits for the voltage dividers. (I did use the back of an old post-it note though, didn't want to risk a stack overflow!)

I always enjoy little puzzles like this though. The ones they ran each month in 'Electronics Australia' were often quite challenging. Anyone know any other good ones?

[c4757p]

I always enjoy little puzzles like this though. The ones they ran each month in 'Electronics Australia' were often quite challenging. Anyone know any other good ones?

Yeah, I like them too. I'm always careful not to specialize too much - I don't want to work myself into a corner where I'm very familiar with one small area and suck at everything else. So even though I think they're easy, they keep me thinking about things I don't do very often. I'd love to have a little book of these, like a Sudoku book for engineers...

[Bored@Work]

I always enjoy little puzzles like this though. The ones they ran each month in 'Electronics Australia' were often quite challenging. Anyone know any other good ones?

http://www.eeweb.com/electronics-quiz/
http://www.eeweb.com/electronics-quiz/browse/all/

[nessatse]

http://www.eeweb.com/electronics-quiz/
http://www.eeweb.com/electronics-quiz/browse/all/

Like this novel approach to current limiting: http://www.eeweb.com/electronics-quiz/regulated-power-supply/ 

[MacAttak]

Mind posting the actual answer for those of us who think they might know, but aren't confident?

I suppose I could always just construct and measure to confirm - but thats cheating.

[IanB]

The correct answer is 0.2 A.

If you are on the right path you will find all the numbers work out very neatly, as Andy mentioned above.

[AlfBaz]

Mind posting the actual answer for those of us who think they might know, but aren't confident?

I suppose I could always just construct and measure to confirm - but thats cheating.

The way I worked it out can be done mentally in under a minute or two but the way in which the question was posed and talked about makes me think there is some other short cut I just cant see

[IanB]

The way I worked it out can be done mentally in under a minute or two but the way in which the question was posed and talked about makes me think there is some other short cut I just cant see

No, I had no special trick in mind other than standard EE techniques. Part of the amusement is to solve it in your head without writing down any intermediate results.

[AlfBaz]

without writing down any intermediate results.

My mind has either gotten leaky or weak from the lack of mental arithmetic. If its the latter then these sorts of puzzles are indeed a good exercise to keep mentally sharp

[IanB]

Here's the solution I was thinking of: (in invisible ink to avoid spoilers)

Find the Thevenin equivalent of the circuit seen by the 13 ohm resistor.

The open circuit voltage V_Th is 8 V - 3 V = 5 V.

The Thevenin resistance R_Th is 24 ohms in parallel with 8 ohms, in series with 9 ohms in parallel with 18 ohms.

To work out 24||8, compute (24 x 8 ) / (24 + 8 ) = 24/4 = 6 ohms. Similarly 9||18 = (9 x 18 ) / (9 + 18 ) = 18 / 3 = 6 ohms.

Then the total resistance in the circuit is 6 + 6 + 13 = 25 ohms. The current is 5 V / 25 ohms = 0.2 A.

[AlfBaz]

I worked that part out in perhaps a lazier way, like this...

[AlfBaz]

Damn it, I'm no good at this internet stuff

[mechelec]

A nice trick with working out parallel resistors (which I presume you intended) is:

x||x = 1/2 x
x||2x = 2/3 x
x||3x = 3/4 x

I confess I had to work it out before I saw the simple answer.

[IanB]

A nice trick with working out parallel resistors (which I presume you intended) is:

x||x = 1/2 x
x||2x = 2/3 x
x||3x = 3/4 x

I confess I had to work it out before I saw the simple answer.

Good catch. I am familiar with the x||x case of course, but I had never quite picked up on the pattern in the x||2x and x||3x cases. Sometimes you can miss what is right in front of you...

[IanB]

I worked that part out in perhaps a lazier way, like this...

Damn! Now I want to know what the lazier way is? 

[AlfBaz]

I worked that part out in perhaps a lazier way, like this...

Damn! Now I want to know what the lazier way is? 

I tried your invisible ink trick but it didn't work so I went back and deleted it. It was what mechelec posted, a little different though

x||yx = yx/y+1

In English, what ever the ratio is add 1 and divide the larger resistor by it

[IanB]

In English, what ever the ratio is add 1 and divide the larger resistor by it

Cool. That definitely short circuits a bunch of mental arithmetic 

[AlfBaz]

path of least resistance... bada tsh

[c4757p]

Mental arithmetic tricks have never really worked for me, I just can't memorize them. I much prefer a familiar calculator that I can work very quickly. x||y = (x, inv, y, inv, +, inv) on an RPN calculator. Extends very easily to large resistor networks and works as easily with E24 values as with "nice" values. The trick is to tell nobody and keep your embarrassment to yourself when you realize you just used a calculator to find 15k || 10k...

[MacAttak]

Ah, so I had half of it right then. V_Th was easily done in my head, but I couldn't figure out R_Th. I knew it was more complicated than V_Th/13 and the other resistors had to be accounted for. There had to be something simple that I was forgetting.

This circuit is a Wheatstone Bridge, yes?

I was going to look up how to do network analysis on bridge circuits (since I don't have it memorized), but that felt too much like cheating.

[AlfBaz]

Only because algebra is the only math I feel competent in and I like latex (not in a kinky way) 

Given 2 resistors in parallel where one is x and the other a y multiple of x