Series DC problem

Discussion in 'Homework Help' started by TwoPlusTwo, Mar 2, 2012.

  1. TwoPlusTwo

    Thread Starter Member

    Oct 14, 2010
    51
    0
    I got stuck on this problem on last year's exam, and apparently I still can't solve it :mad:

    The text reads:

    a. How big is Rx if Va = - 40V
    b. Find I.

    The only way I can make it work is if Rx is not a resistor but a 5V battery.

    I get I = 30/4 = 7.5A because the "drop" across the top resistor is 30. But then the next drop has to be 45, and the two add up to more than 70. And I can't have a drop in the opposite direction across Rx, because there is no such thing as a negative resistance.

    What am I missing here?
     
    • exam.jpg
      exam.jpg
      File size:
      150.3 KB
      Views:
      27
  2. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    The analysis you gave is exactly what I would have done. Did you get an answer from the instructor?
     
  3. TwoPlusTwo

    Thread Starter Member

    Oct 14, 2010
    51
    0
    That's a relief. I'll ask my teacher on Monday and see what he has to say for himself..
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    Let us know. Never too old to learn something new!
     
  5. Paul Kerry

    Member

    Jan 9, 2012
    36
    4
    Just had a play with this and the voltage at Va is -42V with Rx at 0Ω and
    the current flowing is 7A. I think Mr. Instructor may have made a mistake with the question!
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    In your circuit we can get -40V at point Va if we replace Rx with negative resistance.

    Rx = - 2/3Ω ≈ -0.666Ω

    In theory and in real live also we can build a device that has a negative resistance.
    Here you have a diagram. I use a VCVS to convert positive resistance Rx to the negative one.

    [​IMG]
     
    • 27.png
      27.png
      File size:
      7.6 KB
      Views:
      30
    TwoPlusTwo likes this.
Loading...