How can you tell whether sealed black boxes contain three identical resistors connected in Delta or in Wye?
physics/delta.wye New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
How can you tell whether sealed black boxes contain three identical resistors connected in Delta or in Wye? A Black Box Problem - For those knowing Delta from Wye A man has a warehouse full of identical black boxes. They can not be opened or penetrated and are not susceptible to x-ray or other imaging techniques. Each contains three identical resistors in either Delta or Wye configuration connected to three outside terminals. All boxes have the same configuration, but without identification. A buyer offers to purchase half of the boxes, if the owner can demonstrate beyond a doubt which configuration they have. He does it. How did he do it? -- Ernest R. Steele <ernsteele@aol.com>
Show solution
OK... So let's see if I understand the problem.
There are 2 possible types of black box. Each contains 3 identical
resistors (same resistance, same maximum power rating).
Type 1 contains 3 resistors of value R1 connected as
A o---/\/\/--+
|
B o---/\/\/--+
|
C o---/\/\/--+
Type 2 contains 3 resistors of value R2 connected as
A o-+--/\/\/---+
| |
\ |
/ |
\ +----o C
/ |
| |
B o-+--/\/\/---+
Where R2//(2*R2) = 2*R1 (so the terminal-terminal resistances of the 2
types of boxes is the same).
I _think_ any non-destructive electrical measurement can't tell them
apart - that, after all, is the point of star-delta (the UK term for
Wye-delta) equivalent circuits. So I propose the following.
Connect a DC PSU and an ammeter to terminals A,B. Slowly increase the
voltage of the PSU. The current will rise linearly with the voltage,
but at some point, one (and, I think only 1) resistor will burn out. In
the case of the delta configuration, the resistor between terminals A,B
will have twice as much voltage across it as the other two, and thus
will dissipate 4 times as much power. It will burn out first, leaving a
resistor from A to C and a resistor from B to C.
In the case of the star/wye network, either _both_ the resistors (from
A to the Wye point and B to the Wye point) burn out at once, or more
likely one of them goes first and opens the circuit. You thus either
have a total open circuit between all 3 terminals or an equivalent of
2*R1 from terminal C to the one of A,B (and the other of A,B totally
disconnected)
Distinguishing between the 2 damaged boxes is, of course, trivial.
Since you mentioned there was a warehouse full of these things, and
that only half of them were to be sold, it seems reasonable you can
damage one of them to figure out what it is.
Now, of course, if they contain 1M Ohm 2W resistors, you have a
_problem_... :-)
-- A.R. Duell <ard12@eng.cam.ac.uk>
This has gone on so long with so many conflicting answers, and I can't
understand why it's so difficult.
The puzzle states that ALL the boxes have the same configuration
(either Delta or Wye), so if you can determine the configuration of
just one box, you know the rest are the same. It's not quite clear if
all the boxes have the same resistance value, but we aren't asked to
determine that, and it doesn't matter anyway, excluding the obvious 0
(short) and infinity (open).
It takes two simple resistance measurements, and for sake of example
I'm going to assume a resistance value of 10 ohms. This doesn't
invalidate the method, just makes it easier to follow along. In fact,
after determining the configuration, you will also have determined the
resistance value used.
MEASUREMENT 1: Measure the resistance between any two terminals. If
it's a Wye, this will be 20 ohms, if a Delta, 10 ohms.
MEASUREMENT 2: Short any two terminals together, and measure the
resistance to the third terminal. If it's a Wye, this will be 15 ohms,
if a Delta, 5 ohms.
Examine the ratio of measurement 1 to 2. It must be either 4/3, or
2/1. If 4/3, then it's a Wye; if 2/1, a Delta.
So, not knowing the resistance beforehand is no problem. The ratio of
the two measurements tells you which configuration. Once that is
known, then you can say that the three resistors values are either
measurement 1 divided by 2 (if Wye), or measurement 1 (if Delta).
The only problem left is: How do I get rid of the other half of the
boxes?
-- Fritz Whittington <fritz@ti.com>
[ Editor: in a Delta of 10 ohm resistors, the resistance between two
terminals is 10 ohms in parallel with 20 ohms, i.e. 6.67 ohms, not 10.
Measurement 2 (5 ohms) is right, so the Delta ratio is 6.67/5 = 4/3,
the same as the Wye's 20/15, and the method cannot tell them apart. ]
Doug McKean wrote:
> Start by assuming you have a WYE circuit in the box.
> Arbitrarily label the terminals A, B, and C.
> Measure the resistance between A and C.
> Call this measurement MRW1.
> This will be MRW1 = 2R, (for a WYE circuit).
> Locate an external resistor RXW = (MRW1)/2 = R.
> Connect RXW between terminals A and B.
> Repeat the resistance measurement between A and C.
> Call this measurement MRW2.
> This will be MRW2 = (5/3)R. (I'll leave the algebra to you.)
> The difference between the first measurement MRW1 and MRW2 is
> MRW1 - MRW2 = 2R - (5/3)R = (1/3)R
> In other words, the second measurement will DROP by 1/3 from
> the first measurement.
It will 1/3 R which is 1/6 of the first measurement MRW1.
> Now, say you have a DELTA circuit instead of a WYE circuit.
> Arbitrarily label the terminals A, B, and C.
> Measure the resistance between A and C.
> Call this measurement MRD1.
> This will be MRD1 = (2/3)R, (for a DELTA circuit).
> Locate an external resistor RXD = (MRD1)/2 = (1/3)R.
> Connect RXD between terminals A and B.
> Repeat the resistance measurement between A and C.
> Call this measurement MRD2.
> This will be MRD2 = (5/9)R. (I'll leave the algebra to you.)
> The difference between the first measurement MRD1 and MRD2 is
> MRD1 - MRD2 = (2/3)R - (5/9)R = (1/9)R
> In other words, the second measurement will DROP by 1/9 from
> the first measurement.
This is 1/9R drop, which is 1/6 of MRD1. No difference.
You may try another approach. Let U1, U2, and U3 are voltages on the
terminals, and I1, I2, I3 are currents. You cannot read more
information, can you? Since all elements in circuit are linear, the
relation between currents and voltages is also linear for any possible
circuit inside black box. Here it is
I1 S11 S12 S13 U1
I2 = S21 S22 S23 x U2
I3 S31 S32 S33 U3
which is conventional tensor form for
I1 = S11*U1 + S12*U2 + S13*U3
I2 = S21*U1 + S22*U2 + S23*U3
I3 = S31*U1 + S32*U2 + S33*U3.
The Sij are "mutual conductances" between terminals, they can all be
measured by grounding two terminals, applying 1V to third, and
measuring all three currents. These currents are numerically equal to
S13,S23,S33. The rest six Sij could be measured in the same manner.
This tensor Sij contains *all* information about the black box and it
is impossible to learn more. On the other hand this information is
enough for predicting result of *any possible* measurement.
Now for Y the tensor is
2/3R -1/3R -1/3R 2 -1 -1
-1/3R 2/3R -1/3R = 1/3R x -1 2 -1
-1/3R -1/3R 2/3R -1 -1 2
[Editor: in the left-hand matrix the second and third diagonal entries
were 3/2R and the entry in row 3, column 2 was -3/4R]
and for Delta
2/R -1/R -1/R 2 -1 -1
-1/R 2/R -1/R = 1/R x -1 2 -1
-1/R -1/R 2/R -1 -1 2
The two tensors are identical. No solution is possible without tricks.
-- Sasha Semenov <sasha@qv3pluto.LeidenUniv.nl>
[ Editor: with the same R the two tensors differ by the factor 3 (a
Wye of R behaves exactly like a Delta of 3R); since R is unknown, no
linear measurement distinguishes them, which is Semenov's point. In
the thread Ken Shirriff asked for an intuitive explanation of why the
two configurations are equivalent, and Fritz Whittington pointed to
physics/resistors in the archive. ]