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Basic Circuit Diagram - Im stuck

  1. Sep 30, 2007 #1
    This isnt a hard question im sure, but im new to circuit diagrams and need them as part of my product design course, so any help would be much appreciated.

    1. The problem statement, all variables and given/known data

    Find the currents which are flowing in the 14Ohm and 10Ohm resistor.

    [​IMG]

    2. Relevant equations

    V = I x R, but im sure you knew that already ;)

    3. The attempt at a solution

    Im not really sure. I guess i would combine the 15Ohm and 10Ohm resistor to give 6Ohm overall. But then i dont know where to go from there. If you could please explain the steps you took to come up with the answer so i know how to do it in the future, it would be much appreciated.

    James
     
  2. jcsd
  3. Sep 30, 2007 #2

    berkeman

    User Avatar

    Staff: Mentor

    If you have learned about how to use Kirchoff's Current Law (KCL), then writing those node current equations is the simplest way to solve this. If you haven't learned to use the KCL yet, then you can use the "fold up, and then re-expand" method.

    Use parallel and series combinations of the resistors to simplify the network to just the voltage source and the equivalent overall resistance. This gives you the current flowing through the 8 Ohm load. Then start incrementally re-expanding the resistor network, solving for voltages and currents as you go. That method will eventually get you to the point where you know all of the node voltages and branch currents in the resistor network. Does that make sense?
     
  4. Sep 30, 2007 #3
    It does make sense, and i used that to find out that the value for the total resistance in the circuit is 20 Ohms (i hope!). However, now that i know the current I is 1amp, when i go back to my original model, i dont know how to divide that 1 amp into the various resistors to find out what resistor has what current. I have tried reconstructing the circuit back to its original form from the simplified form, but i keep hitting dead ends where x + y should (but dont) equal z.

    Can you please confirm the below is correct, where i have simplified it, and tried to find how much of the current is going in each direction:

    [​IMG]
     
    Last edited: Sep 30, 2007
  5. Sep 30, 2007 #4

    dynamicsolo

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    Homework Helper

    Here's a quick way you can check it yourself. The 30 Ohm and (14+6) Ohm branches are in parallel, so they should have the same "voltage drop" across them. We find

    0.4 A x 30 Ohm = 12 V ; 0.6 A x (14+6) Ohm = 12 V.

    Upstream of the parallel branches is an 8 Ohm resistor with 1 Amp through it, so its voltage drop is 8 V. The total voltage drop is 20 V, which is equal in magnitude to the voltage rise through the battery.

    The sum of the currents in the parallel branches is (0.4 + 0.6) = 1 Amp, which is what the current is before it branches.

    Everything checks out! (so far)

    NOW, what is the current through the 10 Ohm and 15 Ohm resistors? (That's another set of parallel branches.)
     
  6. Oct 1, 2007 #5
    luks fine to me...
     
  7. Oct 1, 2007 #6
    After looking at this again this morning, i got the values to be:

    I(14 Ohm) = 0.6 Amps
    I(10 Ohm) = 0.36 Amps

    This seems to check out as i got the 15 Ohm resistor as 0.24 Amps, which equals 0.6 Amps when added with the 10 Ohms resistors current.

    Thanks you again for the help, and sorry i put this in the wrong catagory. The questions get a lot harder, but since i never did physics at AS / A level, it was a little overwhelling. Also, i found this, which i can check my answers with:

    http://www.walter-fendt.de/ph14e/combres.htm
     
    Last edited: Oct 1, 2007
  8. Oct 1, 2007 #7

    dynamicsolo

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    Homework Helper

    You can also check it this way:

    The voltage drop through the branch with the 14 and "6" Ohm resistors is 12 V. Of this, the drop across the 14 Ohm resistor is 0.6 Amp x 14 Ohm = 8.4 V, leaving 3.6 V across the "6" Ohms, which is your parallel branching with the 15 and 10 Ohm resistors.

    The current through each of those is thus:

    3.6 V / 10 Ohm = 0.36 Amp and 3.6 V / 15 Ohm = 0.24 Amp (or 0.6 - 0.36 = 0.24 A).

    Great!


    It's nice to have an online tool for checking homework problems, but it's also useful to know other quick ways to check your work (or at least judge its credibility) when you don't have access to them, such as while sitting for exams...
     
  9. Oct 1, 2007 #8
    :D Thank you. Ill try not to rely too much on the online checker, and use the ol' noggin.
     
  10. Oct 1, 2007 #9

    dynamicsolo

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    Homework Helper

    Fair enough! [sermon] We live in a wonderful time where computer software can solve circuit diagrams for us (we wouldn't be able to have much of the complex electronics that exist now without it). But if we don't work to develop our own skills and experience to have a sense of what the answers should be, we eventually become no smarter than our tools... [/sermon] ;)
     
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