Simple Series/Parallel Equivalent Resistor Question

Discussion in 'Homework Help' started by werd2501, Aug 28, 2014.

  1. werd2501

    Thread Starter New Member

    Aug 28, 2014
    4
    0
    This one-port network is given and I have to find the equivalent resistance in it.

    [​IMG]

    I already have the answer for it but I'm still not quite understanding something about it.

    My first attempt at this was:

    300//200 + 400//100 = 120 + 80 = 200Ω

    It seemed to me that the 300 was in parallel with the 200 resistor, and that was in series with a 400//100 resistor. But I realise why this is not the case.

    I don't know anymore. Someone enlightened me today it was 100//400 in series with 200, in parallel with 300: (100//400 + 200)//300, which got the right answer.

    I guess, can someone just explain to me how I can interpret stuff like this and what I should know/understand to avoid getting it wrong in the future?

    Thanks.


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    Offtopic, but is it me, or is the forums loading really slowly?
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
    2,499
    380
    hi,
    Why not calc' the 100R || 400R first.?

    then that || combination value in series with the 200R

    then the final || parallel value with the 300R.?

    What do you make the final Requivalent value.?
    E

    EDIT:
    I always look for the obvious 'parallel' resistor pairs in the circuit, then if necessary re sketch the circuit using a single resistor for the || pairs.

    So its a simple method of reducing each parallel or series resistor path to its equivalent value.
    If you find it difficult, do a simple circuit redraw for each reduction.

    BTW: the site access appears normal to me.
     
    Last edited: Aug 28, 2014
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
    5,450
    1,066
    Stand on the top right node in the circuit. Look left. Where does the current flow.

    There are only two parallel branches (upper and lower) starting at that node. The "upper" branch is 100//400 + 200. The "lower" branch is just 300Ω.

    This is just a consequence of KCL
     
  4. Lestraveled

    Well-Known Member

    May 19, 2014
    1,957
    1,215
    When I solve networks like this my first step is to simplify the circuit and components.

    1. The 100 and the 400 ohm resistors are in parallel. They resolve to a single 80 ohm resistor.
    2. The 200 ohm resistor is in series with the 80 ohms which equals 280 ohms.
    3. This leaves you with 300 ohm resistor in parallel with 280 ohms which equals 144.827 ohms.

    Turn many components into fewer components.
     
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