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Circuit help (parallel and series together)

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A 50V independent voltage source supplies power to three resistors in the circuit shown below. For each resistor find the voltage drop and current.

    Circuit Diagram:
    https://webwork2.uncc.edu/webwork2_files/tmp/Fall2014-Engr1201-Common/img/f65baf2e-12f5-3b7e-898f-18c0cd5a2145___71184e03-0c3d-3804-97e6-b93ed961be0a.gif [Broken]

    RA = 55 Ω, RB = 310 Ω, and RC = 24 Ω.

    Vdrop across RA = ____V
    Vdrop across RB = ____V
    Vdrop across RC = ____V

    Current through RA = ____A
    Current through RB = ____A
    Current through RC = ____A


    2. Relevant equations
    V = RI

    V = Voltage
    R = Resistance
    I = Current

    ____________________________

    R = r1 + r2 + r3.... (series)
    1/R = 1/r1 + 1/r2 + 1/r3... (parallel)

    ____________________________


    3. The attempt at a solution
    I have tried solving this by solving the voltage and current at each individual resistor and munipulating formulas. I haven't really gotten anywhere. We haven't been over this stuff in class and the handout doesn't cover this, but they refused to change the due date.

    I am so lost.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 2, 2014 #2

    berkeman

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    Staff: Mentor

    Please show us your work so far. It's pretty hard to help you without seeing how you are approaching the problem.
     
    Last edited by a moderator: May 6, 2017
  4. Sep 2, 2014 #3
    Well, I have tried applying Kirchoff's Laws and tried to find the currents of the first two resistors, but I don't think I can without knowing all but one of them.

    50V - IR1 - IR2 - IR3 = 0
    50V - 55I1 - 310I2 - (24)(.707090275) = 0

    I found the third current by taking the voltage/overall resistance. (The third current and voltage drop and they were correct)

    I don't think I can calculate the voltage drop without knowing the current or vice versa.
     
  5. Sep 2, 2014 #4
    Oh and I also tried getting the percentage of resistance of both sides of the parallel resistors and tried applying it to the current of both respectively.
     
  6. Sep 2, 2014 #5

    FOIWATER

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    Gold Member

    Why not find the total current coming from the voltage supply? (simplify the resistor network) then you know the total current through Rc.

    With the current through Rc, you can find the voltage drop across Rc. Then 50-Rc voltage drop is the voltage across both Ra and Rc. So you can find the current through each of those as well.

    Does anything here not make sense?
     
  7. Sep 2, 2014 #6
    Well I got Rc, but basically you are saying that I can find Ra voltage drop by:

    50V = Ra - (Rc voltage drop)?
     
  8. Sep 2, 2014 #7

    FOIWATER

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    Gold Member

    no Ra = 50- Rc voltage drop

    50V would actually be the sum, that is: 50=(Ravoltage drop)+(Rcvoltage drop)
     
  9. Sep 2, 2014 #8

    FOIWATER

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    Gold Member

    you know all the resistances so that's not an issue, the first trick is to find the total current from the source.
     
  10. Sep 2, 2014 #9

    FOIWATER

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    Gold Member

    find an equivalent circuit resistance (the resistance seen by the source)

    That is, Rc in series with (Ra and Rb in parallel)

    then use ohms law to find total current. total current = 50v/total resistance
     
  11. Sep 2, 2014 #10
    I got it! Thanks guys!
     
  12. Sep 2, 2014 #11
    The issue for me was that I thought that 50V was the sum of the voltage of the entire circuit, as in Ra V + Rb V + Rc V. Again, thank you!
     
  13. Sep 2, 2014 #12

    FOIWATER

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    Gold Member

    ahh, I see.

    Most welcome
     
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