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Find current

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data
    What is ix (in μA)? Given all R and V.

    1.jpg

    2. Relevant equations
    V=IR


    3. The attempt at a solution
    I reduced the circuit into just one Req = 20.778 kohms and find the current through the reduced circuit to be 240.6375 uA (5V/20778 ohms). Since current flowing in series is the same, Ix through the 10 kohms is also 240.6375 uA.

    Would anyone please tell me where I am doing wrong here?

    Many thanks in advance!
     
  2. jcsd
  3. May 22, 2013 #2

    NascentOxygen

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    Staff: Mentor

    The element carrying Ix has a path in parallel to it: the current through the 15k resistor is shared between two parallel paths.

    What did your circuit look like just before you collapsed it all into one resistance? :smile:
     
  4. May 22, 2013 #3
    what u r finding is the total current. the current however splits into 3 branches at the junctions. i wouldn't recommend this problem until u have covered kirchoff's law
     
  5. May 22, 2013 #4

    It was 15 kohms + the reduced resistance (including 10, 10, 4, and 47 kohms resistors). Do you mean that the current I found has to be divided between the 10 and the rest of the reduced resistors (10, 4, 47)?
     
  6. May 22, 2013 #5
    I did. I just need to practice more. So I thru 15 = I thru 10 + I thru combined (10, 4, 47)?
     
  7. May 22, 2013 #6

    NascentOxygen

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    Staff: Mentor

    Yes.
     
  8. May 22, 2013 #7
    I got I = 139.0438 μA.
     
  9. May 22, 2013 #8

    haruspex

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    Looks right.
     
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