Monday, April 11, 2011

Equivalent Resistor Twister

Recently, I've been having a lot of discussions with colleagues about equivalence resistance, and the difference between teaching the concept as a mathematical tool for thinking (and problem-solving) vs. teaching the parallel and series equations in order to calculate resistances. So far, not a single colleague has demonstrated much facility with mathematical thinking about equivalent resistance situations without resorting to the equations. Once I show them how, however, they are off to the races, and often pretty excited about their new mathematical tool.

Here are two problems that are fun if you think about the concept and simply miserable if you approach it using guess-and-check-with-equation.

Problem 1
Using only 10-Ohm resistors, come up with 4 different ways of hooking them up to a single (ideal) battery such that the equivalent resistance across the battery is 10 Ohms.

Problem 2
Using only 3-Ohm and 2-Ohm resistors, come up with 4 ways of making a 1-Ohm equivalent resistor circuit element.


  1. I love this concepttest from Mazur about this concept.

  2. Yeah, that's a great problem--noticing the 24-8 ratio makes it easy.

    A comment I've gotten from more than a few folks is concern that this way of thinking isn't helpful when the numbers don't have an easy least common multiple. I'm not sure how best to respond to that. I've tried telling them I'd rather use simple numbers that emphasize the concept (and its applicability for mathematical thinking), than more realistic numbers that mask the fundamental idea.

  3. actually I think you don't need to do the ratio. You can just say that the parallel part on top has to be less than 8 and then you can say that the whole thing has to be greater than 16 and less than 24 (without the top part in series and with). Yeah, you can't get closer than that but it's a multiple choice and you can see that the others can't be true.

  4. Nice way of thinking about it, especially with multiple choice you can reason through it without any calculation. I think what's nice about your thinking is it simply reflects the idea that the equivalent resistance for the branch must be less than the each branch.

    For me, noticing the common multiple lets me think of the 8-Ohm resistor as three 24-Ohm resistors in parallel, making it a total of 4-24 Ohm resistors in parallel. Now that the branches all have the same resistance, we just divide by numbers of branches to get 6 Ohms.