What is the net resistance of this circuit if each resistance has $R=2.8 k\Omega$?

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It looks a lot easier if you show it like this. Sorry for the poor graphics - the PC I'm at is starved of good s/w.

     |
     R
     |
  -------
  |      |
  |      R
  |      |
  R    -----
  |    |    |
  |    R    R
  |    |    |
  |    |    R
  |    |    |
  -----------
       |
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akot717
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akot717

Updated on March 28, 2020

Comments

  • akot717
    akot717 over 3 years

    enter image description here

    "What is the net resistance of this circuit connected to the battery in the figure above? Each resistance has $R=2.8k\Omega$?"

    We've recently started circuits in my senior year (U.S. High School) A.P. Physics 1 class, and this question came up. While I understand how to calculate resistance for series circuits and parallel circuits separately, this is a sort of "compound" circuit (if that makes any sense), and seems to involve both.

    Here's what I know:

    In a series circuit, $R_{eq}=R_1+R_2+R_3+...$

    In a parallel circuit, $\frac1{R_{eq}}=\frac1{R_1}+\frac1{R_2}+\frac1{R_3}+...$

    Should I try rearranging the circuit into a more palatable shape? What is the most efficient way to approach this problem?

    Thanks!

    This is from Giancoli's "Physics" 6th Edition, pg. 548 #20 (in case anyone is curious).

  • akot717
    akot717 over 6 years
    So, just to be clear, every time you saw a path diverge into two, you drew two branches (regardless of the original "direction" or "angle") to make the above image? And yeah, this shape is quite a bit easier to deal with. The first resistor will be in series with the rest of the circuit. It will split off into two branches, and these two branches have two resisters parallel with each other. One of those two branches has another two branches of resisters, with one resistor on one branch and two together on the other. Phew. That was wordy, and probably confusing.
  • ZeroTheHero
    ZeroTheHero over 6 years
    the "angles" or "directions" do not matter at all. That's just a device to confuse students.
  • akot717
    akot717 over 6 years
    ZeroTheHero - I thought as much...damn teachers. So, (numbering the R's by closeness to the beginning of the circuit), the equivalent resistance will be: $R_1 + (1/R_3 + 1/(R_2 + (1/R_4 + 1/(R_5 + R_6)))$ No, that was probably wrong. Is there away to do this without drowning in parentheses? Should I work backwards (from the last resistor to the beginning, rather than the other way around)?
  • rob
    rob over 6 years
    @akot717 Units! You can't add $R_1 + (1/R_3 + \cdots)$; you can, however, add $R_1 + (1/R_3 + \cdots)^{-1}$, assuming the units are consistent in all the other terms you add.
  • user1583209
    user1583209 over 6 years
    @akot717 Yes working backwards (from inside to outside) is a good idea. You could even draw intermediate diagrams, where you combined some resistances already. Here I'd start with adding $R_5+R_6$ which are in series and just call this new resistance something like $R_{56}$, then combine it with $R_4$ which is in parallel, i.e. you replace this whole parallel circuit of $R_4,R_5,R_6$ with a single resistance $R_{456}=(1/R_4+1/R_{56})^{-1}$ and so on....