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Homework Help: Voltage B to D in a simple Circuit

  1. Jan 6, 2013 #1
    1. The problem statement, all variables and given/known data

    For the network, calculate the voltage B to D


    2. Relevant equations
    KVL, and Ohms law.

    3. The attempt at a solution
    This is what i have:

    Rt=4k+3K+2K=9k ohms
    From here on i am not sure what i am supposed to do.
  2. jcsd
  3. Jan 6, 2013 #2


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    Gold Member

    Well, you have the total voltage around the loop and the total resistance around the loop. Think you could get the current flowing? If you had that, what might you do next?
  4. Jan 6, 2013 #3
    I = V/R It = 20/9k=2.2mA
    Now that I have that would i just use current divider formula?:

    I= Rt/Rx*It Iab = 9k/2k*2.2mA = 9.9mA? Then would i just continue to find the currents?
  5. Jan 6, 2013 #4


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    Uh ... WHAT currents? You've already found the current. Do you have some reason to think there is more than one?
  6. Jan 6, 2013 #5
    This is frustrating and no help what so ever, obviously i am struggling with the problem and haven't been able to figure out what i need to do. Thanks for your time, I know mine has been wasted.
  7. Jan 7, 2013 #6
    The current in the circuit is (45v - 25v)/(2k+3k+4k) = 2.2mA.

    The voltage across r2 is (3k)(2.2mA) = 6.666v, with r2's positive at C.

    So voltage across BD is 45v - 6.666v = 38.333v with positive at B.
  8. Jan 7, 2013 #7


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    It's good that you want to be helpful but please re-read the forum rules. It is not the purpose of this forum to spoon-feed answers but rather to help folks learn how to solve problems.

    I can see that the OP's frustration might have led you to charitably just give him the answer rather than continue to try to lead him to it, but again ... that's not what we do here.
  9. Jan 7, 2013 #8
    Oh OK, no problem, makes sense; I shall be more Socratic in future :)
  10. Jan 7, 2013 #9
    Thank you very much Minki, this has help a great amount. I now know how to solve this as well as similar problems
  11. Jan 8, 2013 #10
    Glad to be of help :)
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