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Circuit analysis questions

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey, attached is a word document with the three problems. The third problem gave me no grief but the first two i'm unsure about.


    2. Relevant equations
    V=IR


    3. The attempt at a solution

    Question 1:

    let I1 = current flowing through 20kohm resistor
    let I2 = current flowing through 6kohm resistor

    by KCL:

    I1 + I2 = 40,

    From here i'm not sure what to do, can I simplify the circuit perhaps? Also the current source confuses me, doesn't KVL say that the voltage generated will equal the voltage dissipated? if there is no voltage generated how can there be a voltage drop?

    Question 2:

    I honestly have no idea what to do here it all looks like a big jumble.
     

    Attached Files:

    • Q.doc
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  2. jcsd
  3. Feb 3, 2009 #2
    There is a voltage generated, but it doesn't state it since it's not at all necessary to solve the problem.

    To go about Q1, you've stated the correct equation (V=IR), now use that + some knowledge about resistors in series/parallel and voltage drop in series/parallel
     
  4. Feb 3, 2009 #3
    If V is the voltage at the junction of the 5 K, 20 K and 6 K resistors, can you write a formula to find I1 and I2 in terms of V?
     
  5. Feb 3, 2009 #4
    Q1, 6k and 4k resister are in series = 10k resister
    the current through the 5k is 40amps then it splits between the new 10k and 20k resister (current divider)

    I(through the 20K) = 40 (10/10+20) = 40(1/3) = 13.33 amps
     
  6. Feb 4, 2009 #5
    To person above; are you sure about that? 'Cause that's not the current I get.. and I can't quite understand what you'e done either.
     
  7. Feb 4, 2009 #6
    for Q1:
    the right side of the circuit had a 6k and 4k resister in series, these 2 resistors can be replaced with a 10k resister (eqiv resistance between 2 resisters in series is the sum)

    so the circuit is a 40 a current source then a 5k resister then has 10k and 20k resistors in parallel

    the current through resisters in series is the same so think there is 40amps going into the node after the 5k resister (doesn't affect the current through the 20k b/c is in series with it)

    that means 40amps is going to go either through the 10k resister or 20k resister = current divider

    wikipedia does a decent explanation
    http://en.wikipedia.org/wiki/Current_divider

    so I(through 20k) = 40 amps ( 10k/ (10K + 20k)) = 13.33 amps

    hope this does a better job, if you still don't understand tell me exactly what part and I'll try to do a better job. What were you doing to get a different answer, and if you did get a different answer what do you see wrong with what I do.
     
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