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Find voltages across resistors using voltage division

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    7a from picture

    2. Relevant equations
    V=IR
    Voltage division

    3. The attempt at a solution
    I am not sure how to apply voltage division here. It is a new concept for me and I'm not really sure how to apply it here. voltage division can only be used in series, but these resistors are not in series. I also am not sure how to deal with the 2 voltage sources. Is there a way to add them to make this simpler? What do I do with the 10V source exactly? I could find a current, but wouldn't it be only through the one loop on the left? I=V/R=.08A.
    Screen Shot 2016-01-30 at 1.14.06 PM.png
     
  2. jcsd
  3. Jan 30, 2016 #2

    cnh1995

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    Assume the negative terminal of the battery at 0V. Can you find the current through the two branches?
     
  4. Jan 30, 2016 #3

    cnh1995

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    There is only one source, 10V. V is the voltage between the two points shown in figure, which they're asking. Assume it's a voltmeter.
     
  5. Jan 30, 2016 #4
    What do you mean by that? "Assume the negative terminal of the battery at 0V"? Should I ignore the spot on the right where I am supposed to find the voltage for now? And just try finding the equivalent resistance and current?
     
  6. Jan 30, 2016 #5

    cnh1995

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    Let's keep ground aside for a minute. You've found the current in one branch correctly. What will be the current through the other branch?
     
  7. Jan 30, 2016 #6

    Merlin3189

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    You can simply use Ohms law to find the current through each resistor.

    You can then use Ohms law to find the voltage across each resistor.

    (BTW. I would stick to the original concept of voltage between two points and not assign any absolute potentials.)
     
  8. Jan 30, 2016 #7
    Ok, so I just ignore the V on the right for now? I thought that was like a wire and it would mess with the circuit. The current for the first loops is .08A and for the second loop it is .05 A.
    image (1).png
     
  9. Jan 30, 2016 #8

    cnh1995

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    Correct. Now as Merlin said earlier, find the voltage drops across 50Ω and 100Ω using Ohm's law.
     
  10. Jan 30, 2016 #9

    Merlin3189

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    Yes. Now you know the current through each resistor, you can find the voltage across them.
     
  11. Jan 30, 2016 #10
    Ok. I got 8V for the 100ohm and 2.5V for the 50ohm using ohms law. So now how can I find the voltage V across those 2 wires?
    Edit: Voltage in parallel is the same. So what would happen if you connect those to wires and try to find the voltage across them?
     
  12. Jan 30, 2016 #11

    cnh1995

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    Yes. If you assume 0V at the negative terminal of the battery, what will be the potentials of the two points( upper ends of the resistors)?
     
  13. Jan 30, 2016 #12
    I am not sure. What do you mean by upper ends of the resistors?
     
  14. Jan 30, 2016 #13

    cnh1995

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    Label the points as A and B between which the voltage is asked(upper ends of 50ohm and 100 ohm). Label the negative terminal as G. You have VA-VG and VB-VG. Can you find VA-VB from this?
     
  15. Jan 30, 2016 #14

    Merlin3189

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    V is not the voltage at a point. It is the voltage difference between the points labelled + and -
     
  16. Jan 30, 2016 #15

    cnh1995

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    If you connect those points by a wire, it will be a short and voltage between + and - will be 0. It will change everything.
     
  17. Jan 30, 2016 #16
    Ok so you just treat this like you are using a voltmeter and measuring the voltage across those 2 wires right?

    I am not really following what you are doing here. Do I find the voltage at point by using equivalent resistances and then the same at point B and then subtract? So at point A the voltage is V=IReq= (.08A)(125ohms). and then do the same at point B?
     
  18. Jan 30, 2016 #17

    cnh1995

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    Right.
    No. If you labelled correctly, + is A, - is B and negative terminal of battery(where lower ends of 100Ω and 50Ω meet)is G.
    You got voltage across 100Ω as 8V. Isn't it VA-VG?
     
  19. Jan 30, 2016 #18
    I'm still not following where you are getting these equations or how they are related. Is there a different approach? Or can you explain you approach more? image (2).png
     
  20. Jan 30, 2016 #19

    cnh1995

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    Ok. I got your confusion. G is the negative terminal of the battery i.e. the 10V source.
     
  21. Jan 30, 2016 #20

    cnh1995

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    It is the only approach. I am sorry I am unable to draw anything but it is not really difficult. 100 ohm and 50 ohm have a common terminal, don't they? It is connected to the -ve tetminal of 10V source, isn't it? That common terminal is G. Now, voltage "across" 100Ω means VA-VG i.e. potential difference between A and G, right? Same goes with VB-VG. It is the voltage across 50 ohm resistor.
     
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