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Guru

Join Date: Oct 2006
Location: MSP, MN
Posts: 718

# A Christmas Present

12/24/2017 4:19 PM

I found this little puzzle on the wall of the men's room in the EE building at Purdue 50 years ago. As I recollect it took me most of the day to solve it.

A ladder of 1 ohm resistors extending to infinity. What is the resistance R looking into the ladder? Hint: while you may be rational, the answer is not.

Merry Christmas!

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Guru

Join Date: Apr 2010
Location: About 4000 miles from the center of the earth (+/-100 mi)
Posts: 9193
#1

### Re: A Christmas Present

12/24/2017 5:31 PM

The resistance R is (1 + 1) + (1 ohm in parallel with R) .

R=2 + R / (1+R)

R-2 = R/(1+R)

R2-R+2 = R

R2-2R+2=0

Guru

Join Date: Apr 2010
Location: About 4000 miles from the center of the earth (+/-100 mi)
Posts: 9193
#2

### Re: A Christmas Present

12/24/2017 5:47 PM

R=1+sqrt(3).

Guru

Join Date: Oct 2006
Location: MSP, MN
Posts: 718
#3

### Re: A Christmas Present

12/24/2017 5:58 PM

Spot On. Reduces to R²-2R-2=0, and we all know how to solve that. Looking forward to many more discussions in the New Year!

Guru

Join Date: Jun 2006
Location: Toronto
Posts: 3968
#4

### Re: A Christmas Present

12/25/2017 10:06 AM

not correct, the two input resistors do not reduce.

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Guru

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#10

### Re: A Christmas Present

12/30/2017 11:10 AM

So that would make the answer 1+ SQRT(2)

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Guru

Join Date: Oct 2006
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Posts: 718
#11

### Re: A Christmas Present

12/30/2017 12:17 PM

Randall, the +2 should be -2. And the answer is 1+sqrt (3).

Guru

Join Date: Oct 2014
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#5

### Re: A Christmas Present

12/25/2017 1:14 PM

I would be looking on the wall for 867-5309, for a good time call.

Guru

Join Date: Dec 2010
Posts: 1623
#6

### Re: A Christmas Present

12/26/2017 11:48 AM

It is a symmetrical T attenuator chain with each section made up of 5 resistors of 1 ohm and a source/load resistance of 2√2 ohms = 2.82843... ohms. An infinite chain will be 2.8243 ohms input

As a check, if you terminate one section of 5 resistors of one ohm with 2.83 ohm, you have (1 + 1 + 2.83) // 1 ohm = (4.83*1)/(4.83 + 1) = 0.8285 ohm: then +1 + 1 = 2.83 ohms input resistance.

Guru

Join Date: Dec 2010
Posts: 1623
#7

### Re: A Christmas Present

12/26/2017 12:58 PM

Urgh. Complete goof by me.

The repeating "symmetric T" is 4 of 0.5 ohm series with a shunt of 1 ohm, having terminal resistance of √3 = 1.732 ohms, add 1 ohm at the input end gives 2.732 ohms.

But Richter also suffered from the Christmas surfeit, his formula should end

R2 -2R -2 =0, so 7.464 - 5.464-2 =0 when R = 2.732..

Guru

Join Date: Jan 2006
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Posts: 1781
#8

### Re: A Christmas Present

12/26/2017 1:18 PM

My shaky memory of maths means I can't put it in a neat formula - but by inspection the first parallel resistor is in parallel with an infinite series of 2 ohm resistors all in parallel - which if taken to the limit will get to as near zero as you like. This leaves 2 x 1 ohm resistors in series.

I plonk for 2 ohms.

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Guru

Join Date: Jan 2007
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#9

### Re: A Christmas Present

12/27/2017 4:32 PM

<...on the wall of the men's room in the EE building at...>

On the wall of the men's room in the Mechanical Engineering building at Exeter adjacent to the "confectionery dispenser" was the expression, "Not to be used above 400rpm"!

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