circuit

Discussion in 'Homework Help' started by Nina Ricci, Sep 20, 2004.

  1. Nina Ricci

    Thread Starter New Member

    Sep 20, 2004
    2
    0
    [attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125][attachmentid=125]I would be so pleased if somebody can help me with this problem presented in the attachment file
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    HI,

    Is there some figure for the resistance of the ammeter? If it's a classroom perfect ammeter, then it has no resistance, and so Vo is zero.
     
  3. Mjollnir

    Member

    Apr 22, 2004
    27
    0
    just use node voltage analysis

    node with 2 resistors = V1

    node at current source = Vo

    KCL at Vo

    (Vo - V1) / 10000 + (Vo / 10000) - 0.004 = 0

    2Vo - V1 = 40

    Current Ix can be expressed in terms of Vo and V1

    Ix = (Vo - V1) / 10000

    Now u can express V1 in terms of Ix

    V1 = -4000Ix (note the reference of the dependant voltage source)

    V1 = -4000 * (Vo - V1) / 10000

    2 unknowns, 2 equations..

    I think this is right.. V1 = -10V, Vo = 15V
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    The ammeter is in parallel with the output resistor. The combination of the 10K resistor and the internal resistance of the ammeter has to be much less than 10K. Ammeters don't measure accurately if they have more than negligable resistance. Let's say 10 ohms, for an example. In parallel with the 10K resistor, the value of Rt will be something over 9 ohms. With 4 mills current, the figure for Vo will be on the order of 36 millivolts.
     
  5. Mjollnir

    Member

    Apr 22, 2004
    27
    0
    umm that looks like an independant current source?
     
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