# Given five 10-Ohm resistors, design a network with an equivalent resistance of 10-Ohm

Trying to help a friend with their circuits homework, but it's been a good 6 years since I took this stuff at Uni.

The problem is: given five 10-Ohm resistors, design a network with an equivalent resistance of 10-Ohm. You are not allowed to short any of the resistors (wouldn't that make life easy!) and both ends of every resistors must be connected to at least one other resistor.

I've come up with over 20 different configurations (including deltas, and figuring the equivalency with delta-Y conversion formulas), but none of them are equivalent to 10-Ohm. Clearly it's the one I haven't thought of ;)

I feel like I must not be approaching this in the correct manner (drawing out random unique configurations as I think of them and figuring the equivalent resistance). Any ideas?

Edit: left out a piece of the problem >_<

• What's the name of this game: resistor puzzle or R Mikado. Commented Feb 9, 2016 at 22:32
• You've tried a Y, can you think of any other capital letters which might work? Commented Feb 9, 2016 at 22:37
• Finding the right wording for such challenges is a challenge! I think the wording doesn't rule out that you connect one resistor to the two connections, and the other 4 in parallel, with one end of that set to one of the poles. Commented Feb 10, 2016 at 8:26

Doing it with 4 resistors is pretty trivial. But what about having to include a fifth...

I don't want to give the answer, but I will pose this question which should be enough information to figure it out.

What is the current through a resistor which has $0\mathrm{V}$ across it?

You updated your question, but it doesn't change this answer at all.

You have to figure out how you get $0\mathrm{V}$ across one resistor. There is one topology of the 5 that will achieve this.

As @LP has now worked out the solution, I might as well include it here:

simulate this circuit – Schematic created using CircuitLab

I've actually drawn the same circuit in two different ways for reasons I will explain in a moment.

As to how it works, well, basically you have two identical sets of series resistors which will be $20\Omega$ each. These in parallel gives $10\Omega$ as required. But we need to add a fifth resistor in a way which will not affect the overall resistance, but in which the resistor is neither shorted nor has any open end.

The trick here is to realise that the centre point of the two series resistor chains will be at the same voltage because the two chains contain the same values of resistance. Because these points are the same, we can add anything (short circuit, resistor, open circuit) between them, and it won't have any affect - there is no voltage drop, so no current will flow through the central link. So we simply stick the fifth resistor in there and the problem is solved.

As to why I drew it in two ways. This bit goes a bit off topic from the question, just some bonus info that the question reminded me of.

The second drawing (on the right) is shown in a way that anyone in EE should instantly recognise. The question is actually touching on something known as a Wheatstone Bridge, whether intentionally or not. If you replace R5 with a Galvanometer, and R3 with a variable resistor, you could find the value of R4 if it was unknown by adjusting R3 until no current flows through the middle branch. This uses the same principal as used in the question - knowing that if you get to a point where no current is flowing, and if the top two resistors are the same as each other, then the bottom two must now also be the same as each other.

• The 5th is soldered like a spike in between. Commented Feb 9, 2016 at 22:33
• That was a joke, but now I realy figured out where it should be. Commented Feb 9, 2016 at 22:44
• Heh, fun quiz. All puns intended. Commented Feb 9, 2016 at 22:52
• Oh duh. Alright, I had tried that one earlier but I did the math wrong and so it was thrown out. Got it. Thanks for the hint!
– L P
Commented Feb 10, 2016 at 3:37
• Yeah I remembered the diamond shape, and I'm decent at redrawing circuits so the layout is different but the connections are still the same. I had just done the math wrong in my head when figuring the equivalent resistance of the two parallel 40/3-ohm branches (forgot to flip the fraction, so I was ending up with 3/20-ohms + 10/3-ohms! Derp!) Your discussion definitely reminded me of an important conceptual point that I'd like to make sure my pal understands (and that I had forgotten), the 0 voltage across the middle of a wheatstone bridge composed of equivalent resistors!
– L P
Commented Feb 10, 2016 at 15:08