Resolving those simple circuits

Discussion in 'Homework Help' started by activee, Jun 11, 2014.

  1. activee

    Thread Starter Member

    Jan 16, 2014
    39
    0
    Hello,
    I did this exercice now 3 times and for I2 i got I2 = -270µA but my book says it's 250µA. Is there an error ? what I wrote on the page I put in attached file is not correct. there is a sign error somewhere.

    It would be nice if someone could point out my mistake(s). Thanks in advance.
     
    Last edited: Jun 11, 2014
  2. pwdixon

    Member

    Oct 11, 2012
    488
    56
    Without doing any calculations at all it is easy to see that the current I2 will be negative as defined in your schematic as the 10V will dominate and push current upwards through R2. After that it's maths.
     
  3. pwdixon

    Member

    Oct 11, 2012
    488
    56
    Pretty sure the answer should be -0.5mA for I2. Are you sure you drew the circuit correctly?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    That's incorrect. I2 is -250uA as stated.
     
  5. pwdixon

    Member

    Oct 11, 2012
    488
    56
    I'll do the maths again.
     
  6. crutschow

    Expert

    Mar 14, 2008
    13,009
    3,233
    It is simpler to do the Thevenin equivalent of the two voltage source branches in parallel. Then use that to calculate I2.
     
  7. syed_husain

    Active Member

    Aug 24, 2009
    61
    5
    I got I2 = -250 uA. Try calculate the circuit using Superposition method.
     
  8. activee

    Thread Starter Member

    Jan 16, 2014
    39
    0
    thanks I found where I was doing wrong.
     
  9. Fibonacci

    New Member

    May 23, 2014
    25
    5
    The answer is I2= -250μA and Uab=-1.25 V, as has been established. The simplest way is by using mesh method and the solution of simultaneous equations resulting. In this simple case is an array of 2 x 2, which casts I1=625μA and I3=875 μA, from the circuit I2=I1-I3.
     
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