Superposition Thereom

Discussion in 'Homework Help' started by dav_mt, Nov 14, 2004.

  1. dav_mt

    Thread Starter New Member

    Nov 14, 2004
    4
    0
    Can somebody please guide me of how I can obtain the voltage across the 25Ohm Resistor using the Superposition Thereom. I rewrote the circuit leaving one source every time but my answer does not seem to match with the lecturer's answer(-9.139V). I can't understand what I'm doing wrong, any help will be appreciated
    Thanks
     
  2. vineethbs

    Well-Known Member

    Nov 14, 2004
    56
    0
    yes , the answer -9.13 is correct .

    using superposition , the 3V source can be replaced by a source as the ckt is linear and the source has no internal resistance .

    then we have a 20ohm // 25ohm combo .
    // parallel combo :)
    20//25 = 500/45 =11.11111111
    so voltage across 20/25=V due to 14 V = 14*11.11111/(11.111111+5)
    which is abt 9.65V(note the polarity of this voltage , if the nodes on the bottom are connected to the grnd assumption then this is -9.65 V)

    now take the 3V source and replace 14 V by short

    again a 5//25 ohm combo = 125 /30 = 4.16666

    so voltage across that is 4.1666/(20+4.16666) * 3V which is abt .5 V

    now this is +ive when the nodes at the bottom are at 0V (assumption)
    so voltage across the 25 ohm is -9.65+.52 = -9.13

    well , the values are approximated here ,but u will get the given ans
     
  3. dav_mt

    Thread Starter New Member

    Nov 14, 2004
    4
    0
    Thanks a lot, I solved it :D
    my problem was that i didnt see 5 and 25 resistors in parallel when removing 14V source
    once again many thanks
     
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