Another black box problem

[kl7aj]

This one showed up in Popular Electronics ages ago, but it's still pretty clever.


You have a black box. Four leads are poking out of the box. You discover that if you apply a certain voltage to any pair of adjacent leads, you will get precisely one quarter of that voltage on the remaining two leads. Exactly what is inside the box, and how is it wired?

Eric

[WA9SVD]

Is that AC or DC?

[kl7aj]

Is that AC or DC?


either one

[ve2nsm]

either one

So it's resistive. Clue # 1
It has to be linear, so I rule out any silicon junction. Clue # 2
If it's from any leads it had to be symetrical. Clue # 3

I checked with a bridge of 4 equal resistors but I end up with 1/3 of the voltage.
I keep looking :)

[K8ERV]

Simple, just get out a razor knife and look.

TOM K8ERV Montrose Colo

[ve2nsm]

Don't tell me I'm gonna have to dig out the star/delta resistor network calculations?
I thought I could forget about that after getting out of school :p

[ab1ga]

Wild guess, still doing the math:

Six resistors, connected as a tetrahedron.


(Later) Nope, scratch that, my stupid.

[ve2nsm]

Wild guess, still doing the math:

Six resistors, connected as a tetrahedron.


(Later) Nope, scratch that, my stupid.

That's was thinking about but my math book is too far :D

[kl7aj]

Wild guess, still doing the math:

Six resistors, connected as a tetrahedron.


(Later) Nope, scratch that, my stupid.


This might be ONE solution, but not the simplest....I'll have to work on that......stay tuned.

[kl7aj]

Wild guess, still doing the math:

Six resistors, connected as a tetrahedron.


(Later) Nope, scratch that, my stupid.


Alas, your solution won't work...the voltage ratios come out wrong....however, this is clever enough to be a different problem! You get a bronze star for creativity.


Eric

[VE7DCW]

Ok Eric...whats in the "the little black box" can be calculated by Thevinen's thereom.I did'nt do too well at in school.My guess is if the box was a bit bigger,it would probably have a cb linear in it:D no??

73

[KD6NIG]

If AC, it could be a transformer.

Won't work on DC though.

[ve2nsm]

This might be ONE solution, but not the simplest....I'll have to work on that......stay tuned.

I was thinking on something like this, is this it?

[kl7aj]

I was thinking on something like this, is this it?


Actually there is an even simpler solution..but yours is definitely okay.

[n4cd]

Well, one solution is using six resistors.

The four resistors conntecting the terminal are all of the same value

Two others are connected catty corner.

I think the two catty corner are twice the resistance of the resistors on the outside edges.

That should work, assuming you pull no current while you measure it.

The voltage across any two terminals is 1/4 the total, but if the insides are to be symmetical, that means that if we take one edge, say A, and B, and meaure the voltage at c and D with respect to B, where we put 1 volt across AB, you will measurre .25v across CD, but C will be at 0625v wiht respect to B and D will be at 0.375v with respect to D. Then you can figure out the divider ratio from the current flow.


Since we don't know what is in the box, any solution that solves the problem should be acceptable. The original quesiton did not specify anything about what could be in the box. You could have 1000 resistors in there.

[ab1ga]

'NSMs latest solution _is_ a tetrahedron of resistors.

If you take a close look, you'll see that the circuit is symmetrical about a vertical line.

If you apply the test voltage to the upper and lower connections, then the voltages at the remaining two terminals must be equal, by symmetry.

Therefore, the output voltage across those terminals must always be zero.

[ab1ga]

Eric, why do I always see these problems toward the end of lunch time? I won't be able to concentrate on anything else this afternoon!

;)

[N2RJ]

Bridge rectifier with four equal value resistors in series with the leads.

[kl7aj]

However, the SIMPLEST is a pentagon of resistors. ONE internal node connection actuall goes to two opposite external terminals. It's unsymmetrical and goofy looking but it works. :)

eric

[kf6rdn]

Since we don't know what is in the box, any solution that solves the problem should be acceptable. The original quesiton did not specify anything about what could be in the box. You could have 1000 resistors in there.


OR a small dwarf with a battery and a potentiometer.
:p

[n4cd]

"OR a small dwarf with a battery and a potentiometer"

Dang, that small dwarf would be challenged to come up with a 4 MHz AC signal, wouldn't he?

That dwarf would also need to have one heck of a battery if someone put 100v or 200v across the terminals. The dwarf would also need a voltmeter to know where to set the pot - since you assume that since he has a pot, that it must be adjusted.

You fell into your own trap.