super position worksheet question

Discussion in 'General Electronics Chat' started by Liav, Jun 24, 2009.

  1. Liav

    Thread Starter New Member

    Jun 10, 2009
    2
    0
    Hi all. I feel like I have a basic understanding of superposition, but keep crashing and burning on some of the worksheet questions. Question five from http://www.allaboutcircuits.com/worksheets/super.html ask us to find the current through the load.

    Removing the current source. we get:

    Total circuit resistance = R1 || RL + R2 = ~350.4 ohms
    R1 || RL = 130.4 ohms.

    Voltage across R1 || RL = (130.4/350.4) * 7.2 = 2.68 V
    Current across RL = 2.68V / 1000 ohms = .00268 A

    ----
    Removing the voltage source we get:

    Total resistance = R1 + R2 || RL = ~330.0 ohms
    R2 || RL = 180.3 ohms
    Total circuit volage = .004A * 330.3 ohms = ~1.3213 V
    Voltage across R2 || RL = (180.3/330.3) * 1.3213 V = .721 V
    Current across RL = .721V / 1000 ohms = .000721 A

    Total current = .0034 A

    The listed solution, however, is 6.623 mA.

    Any insight into what I'm doing wrong here?

    Regards,
    Liav.
     
  2. bill2009

    Active Member

    Apr 17, 2009
    31
    0
    isn't the resistance R2 + R1||Rload?
     
  3. AdrianN

    Active Member

    Apr 27, 2009
    97
    1
    The first part is wrong. When you remove the current source the only circuit remaining is the voltage source and R2 in series with R3. R1 circuit is open, so there is no current going through R1. Therefore the current through Rload is

    7.2V / (R2 +Rload) = 5.902 mA

    Add to this 0.721 mA you found out in the second part and the result is 6.623mA.
     
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