How to solve this series-parallel connection?

Discussion in 'Homework Help' started by soumy, Dec 4, 2009.

  1. soumy

    soumy Thread Starter New Member

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    Dec 4, 2009
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    7
    please explain how to solve this kind of problem. where to start and what are the rules to be considered.

    [​IMG]
  2. Firestorm

    Firestorm Senior Member

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    Jan 24, 2005
    Messages:
    351
    I would probably use mesh. Try to write down all of the mesh equations and I can check what you have. If you choose a different way (there are other ways), post that up here too so we can see.
    The idea is to get as many things related to each other as you can. Then you can simplify your equations and solve for the variables.
  3. Dave

    Dave Retired Moderator

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    Nov 17, 2003
    Messages:
    6,961
    Can I suggest you start by having a look at our DC Network Analysis section. You will then have a better understanding of the different techniques available and your questions will be more focused and relevant to the problem in the OP.

    Dave
  4. soumy

    soumy Thread Starter New Member

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    Dec 4, 2009
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    thank you! i'll post my solution when i'm done. :D thanks a lot!
  5. soumy

    soumy Thread Starter New Member

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    Dec 4, 2009
    Messages:
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    Here's what I did, not sure about it though. I used mesh current method.

    I1 -> CURRENT THROUGH TOP LOOP (COUNTER-CLOCKWISE)
    I2 -> CURRENT THROUGH BOTTOM-LEFT LOOP (CLOCKWISE)
    I3 -> CURRENT THROUGH BOTTOM-RIGHT LOOP (CLOCKWISE)

    The equations I got:
    -15 + 42I1 - 24I3 = 0
    27I
    2 + 18I3 = 0
    12I
    3 + 24I1 = 0

    Solving, I got the following:
    I
    1 = 1/6
    R
    1 = 9
    V
    1 = 3/2

    12
    = 1/6
    R
    2 = 9
    V
    2 = 3/2

    I3
    = 1/2
    R
    3 = 24
    V
    3 = 12

    I4
    = 2/9
    R
    4 = 9
    V
    4 = 2

    I
    5 = 1/9
    R
    5 = 18
    V
    5 = 2

    I7
    = 1/3
    R
    7 = 18
    V
    7 = 6

    I need help, please check it.
  6. GetDeviceInfo

    GetDeviceInfo Senior Member

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    Jun 7, 2009
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    1,503
    Location:
    Canada
    because it's single source, I'd just reduce the circuit.

    - parallel R4 and R5
    - add in series to R7
    - parallel this with R3
    - add your series
    - calc your current flow
    - work back with voltage drops/currents
  7. soumy

    soumy Thread Starter New Member

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    Dec 4, 2009
    Messages:
    7
    thanks! our prof. taught us that yesterday, too.
  8. zgozvrm

    zgozvrm Member

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    Oct 24, 2009
    Messages:
    115
    Or, just stick with good old Ohm's Law:

    1) R4 is in parallel with R5. Calculate and replace them with an equivalent resistance R8.

    2) R8 is in series with R7. Find the equivalent resistance R9 to replace them.

    3) R9 is in parallel with R3. Replace with equiv. resist. R10.

    4) R1, R10, and R2 are all in series.

    5) From here, you can calculate R(total) and I(total)

    6) Working backwards, you can find all the missing values:

    7) I1 = I10 = I2 = I(total), so now find V1, V10, & V2

    8) R10 represents the parallel connection of R3 and R9, therefore, V3 = V9 = V10. Now find their currents (I3 and I9)

    Continue until you're back to the original 7 resistors.
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