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Circuit Question - Driving me crazy

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    The problem is "What is the current through resistor g when switch B has been closed and switch A has been open for a long time?" (see diagram)

    2. Relevant equations

    V = IR
    I/Reff = 1/R1 + 1/R2
    Reff = R1 + R2


    3. The attempt at a solution

    Because f+g are in series, the Reff = 6 ohms. Since f/g is in parallel with c, the new Reff is 4 ohms. Then because f/g/c are in series with a/b, the new total resistance is 10 ohms. As a result, we can find through V = IR that the current is 9A (90 V = I x 10 ohms).

    So then, if V = IR, to find the current through each respective resistor (in parallel), I did 90 V = I x 12 ohms, and 90 = I 6 ohms, however, this gives me currents that are incorrect. Instead, the answers are 6 A, and 3 A through the 6 ohm resistor and 12 ohm resistor. I understand this conceptually, i.e. yes, the resistor 12 ohm is 2X the resistor 6 ohm, therefore current would be half as large - however, I can't figure this out mathematically. Literally, I have spent an hour on this problem, I've read various tutorials, watched khanacademy, read the solution in the back of the book (which gives a conceptual answer) but I can't figure this out.

    Please help, because I really want to figure this out and I can't move on until I do.
     

    Attached Files:

  2. jcsd
  3. Jul 14, 2014 #2
    Correct till here .

    The P.D across 12 and 6 Ω is not 90 V . Look carefully in the picture .
     
  4. Jul 14, 2014 #3
    How do you find the potential difference? I thought it was the same between any set of resistors in series (i.e. only the current changes with differing resistances). Please help, at this point I am totally lost.

    Thanks
     
  5. Jul 14, 2014 #4
    You have correctly calculated the current 9A flowing through the battery.The current 9A from positive terminal of battery passes through the series combination of 'a' and 'b' and then divides in the parallel combination of 12 Ω and 6Ω (f+g) .

    How does current divide in the parallel resistors ?
     
  6. Jul 14, 2014 #5
    I = V / R ? Then the currents add up to the total current, but I still can't get the right answers.

    I know total current in is equal to the total current out..?
     
  7. Jul 14, 2014 #6
    I thought maybe, okay find a new voltage? But V= IR --> (9A) (4 ohms) where R = Total effective resistance. Then it's 36 V, so V = IR --> 36/12 = 3 A for c? Right?

    But why do you find a new V and do the problem this way? But 4 ohms is for f, and not for C. I'm so confused :( :(
     
  8. Jul 14, 2014 #7
    The P.D across parallel resistors is same.

    Let the potential difference across the parallel resistors 12 Ω and 6Ω be V .

    Then V=I1(12) and V=I2(6) , where I1+I2 =9A .

    Now solve for I2 .

    Does this make sense ?
     
    Last edited: Jul 14, 2014
  9. Jul 14, 2014 #8
    Yes, got that. Then what?
     
  10. Jul 14, 2014 #9
    Solve and obtain value of I2 .This is what you want.
     
  11. Jul 14, 2014 #10
    Why are we solving for I2 though?
     
  12. Jul 14, 2014 #11
    Because this is what the question is asking .I2 is flowing through the resistor g .
     
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