Series-parallel help

Discussion in 'Homework Help' started by monkeystevie, Jun 5, 2012.

  1. monkeystevie

    Thread Starter New Member

    Jun 5, 2012
    HI folks,

    I've stumbled across my first series-parallel circuit on my distance learning course. I have very limited tutor support and this forum has been a gold mine so far.

    My current stumbling block is the attached schematic where I have to find "Ix".

    Do I have to simplify the circuit or can I work out Ix directly?

    Many thanks
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  2. StayatHomeElectronics

    AAC Fanatic!

    Sep 25, 2008
    You definitely need to simplify the circuit first. The first current that you will solve for is I, followed by I1 and only then Ix.

    Give it a try and let us know if you run into any problems.
  3. WBahn


    Mar 31, 2012
    You seldom, if ever, HAVE to simplify the circuit. It is sometimes easier to use something like Node Voltage or Mesh Current analysis directly (which I'm guessing you haven't gotten to, yet).

    Given what you probably know at this point, simplifying the circuit, at least partly, is probably your best approach.

    However, instead of simplifying the circuit and then trying to walk yourself back to the quantity you are trying to find, try a more focused and directed approach. This is the approach that I usually take:

    I want to know Ix. What one piece of information that I don't presently know would let me immediately calculate Ix when combined with what I do already know. In this case, it is the voltage on the top-right node since it would let me directly calculate the current in the far right resistor, which is Ix. I then ask myself if, given that peice of information, are there any KNOWN quantities that I already have that I could calculate if they weren't a known quantity. If so, I am done because I have identified an equation in which I can set a known quantity equal to an expression involving my one unknown quantity (and other known quantities). If not, then I proceed by asking myself what other of my unknown quantities could I directly calculate if I were to have that key same additional piece of information. In this case, you could compute the current in the middle vertical resistor. I repeat this process until I get to a point where I can create that equation using a known and an unknown. So, in this case, once I had the current in both of the right-most vertical resistors, I could directly calculate I1. Once I had I1, I could directly calculate the voltage between the two horizontal resistors. Once I had that I could directly calculate the current in the left-most vertical resistor. Once I had that I could calculate the current in the left-most horizontal resistor. Once I have that, I can calculate the voltage on the left side of the left-most horizontal resistor and, finally, I have reached a point where I can calculate a known quantity (namely 17V).

    It sounds a lot more cumbersome than it frequently is. As you go you are constructing the equation needed for each step so, in the next step, you are just expanding it a little bit. And always remember that you do not have to do it this way. If it starts turning into a complicated mess using this approach, try a different approach. In that regard, engineering is very much like being a cabinet maker or other skilled craftsman: There are multiple ways to get any given task accomplished and you generally have many tools and techniques available to you. The more tools you have and the more skilled you are at using each of those tools, the better your options are in picking the best tool and technique that will produce a quality result in a reasonable amount of time. But if the only tool you have (or know how to use) is a hammer, then everything looks suspiciously like a nail.