A Christmas Present

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

R=1+sqrt(3).

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!

not correct, the two input resistors do not reduce.

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

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

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

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.

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..

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.

<...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"!