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Homework Help: Simple circuit question

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    If a 12 V supply is placed across AD, find effective resistance of whole circuit, current through resistor 3, and voltage reading on resistor 5 with polarity

    2. Relevant equations
    Rn= Resistor number n
    I= current
    V= volts

    3. The attempt at a solution

    I could not figure if the current would flow from c to b, or b to c, because there is no potential difference across it, and therefore could not solve the problem

    help please!!!

    Attached Files:

  2. jcsd
  3. May 23, 2010 #2
    help? anyoen?
  4. May 23, 2010 #3
    How about drawing it like this...

  5. May 23, 2010 #4
    yeah, after drawing like that I have no clue how to simplify the diagram....my teacher never even went over it.
    I also don't know which direction the current flows through in the middle.
    if you could show me how, I would appreciate it so much!
  6. May 23, 2010 #5

    Now when you look at this, realize that if Vg exists, then you will have current flowing through that middle resistor in your problem. The nice thing about your problem is you know all of the resistors R1-Rx... They are the ones on the "outside".
  7. May 23, 2010 #6
    hey I plugged this equation in: 69bb1ade5a91b6a6f807585132362585.png
    but it doesnt work...
    you get 4 Volts, when the answer is 6.86 V...
  8. May 23, 2010 #7
    The equation tells you which way the current will flow through the middle resistor. Once you know the direction of the through the various parts of the circuit, you need to use kirchoffs rules. The 5th resistor also comes into play. There is a 5th resistor in the diagram. The way I have it drawn the current should flow down throught the middle resistor.
    Last edited: May 23, 2010
  9. May 23, 2010 #8
    The equation implies that Vg is always determined by subtracting R2/(R1+R2) from Rx/ (R3+Rx) before multiplying Vs. Why can the equation not be (R2/(R1+R2)- Rx/ (R3+Rx) )Vs instead?
    I also dont understand how there is a 5th resister...isn't the 5th resister simply R3??
    I'm really confused
    could you please show me your work??? Thanks
  10. May 23, 2010 #9
    And I do get a 6.86V drop across the 12 ohm resistor drawn in your diagram.

    This problem requires some time and some calculations. It is difficult for a basic circuit question, but not so if you guys have done Kirchoffs rules. I will put arrows on my diagram that show the flow of current that I figured had to be true due to that link I put up.
  11. May 23, 2010 #10
    ok I think i understand kirschoffs rule better now...just basically plug in all the equations

    now how do i determine how much current will enter each branch??Is there a rule for determining this?
  12. May 23, 2010 #11
  13. May 23, 2010 #12
    Yes. Kirchoffs loop rule helps you determine the voltage drops. But in order to know these you must know the current. This is found using some algebra and variables for the current in various wires. I will do one example of an equation. But I have to label the different parts of the wires. I will do so using the resistor numbers in your problem.
  14. May 23, 2010 #13
    no Im asking about the exact quantity of Amperes that go into each branch...

    for example, how do I determine how much amps will go into the initially divided branch that has a 6 ohm resistor on the top, and 12 ohm resistor on the bottom...
  15. May 23, 2010 #14
    I understand this.


    So I = I1 + I2
    I2 = I3 + I5
    I4 = I1 + I3
    I = I5 + I4

    oops relabel I6 as I4; I will fix in the next picture.

    And doing one Kirchoff drop from one end of the battery back to that same point the total voltage drop equals zero: Remember V = R*I

    0= 12V - 6 ohms(I2) - 12ohm(I5) this is only one equation. I will now post the path I took in Green.

    I get a Voltage drop each time I cross a resistor that has current running through it.
    Last edited: May 23, 2010
  16. May 23, 2010 #15

    From this we can also conclude that I2 = I4 and I1 = I5

    Like I said. Its not hard but it takes some algebra work.
    Now if you do another path like I did in the post above, you can figure out exact current numbers and thus by I*R get voltage drops.
    Last edited: May 23, 2010
  17. May 23, 2010 #16
    thx for keeping up with my slow learning

    do you get I2 = I4 and I1 = I5, from this system of equations?:
    I = I1 + I2
    I2 = I3 + I5
    I4 = I1 + I3
    I = I5 + I4
  18. May 23, 2010 #17

    This somewhat of a pain in the rear. When I first looked at it and no one had posted, I thought it might be because its a bit of an algebra/system of equations pain. Some people love these things though. Maybe your teacher does.
  19. May 23, 2010 #18
    !!! I speant my past 30 min trying to solve the system of equations!!!!!!
    can u show me how to do this one?
    Maybe one example could help me solve all the other questions
    ]thanks so much
  20. May 23, 2010 #19
    Start with I1 + I2 = I5 + I4 because both sides equal I...
  21. May 23, 2010 #20
    I tried.. still cant solve it.....
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