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Parallel voltage sources circuit

  1. Femme_physics

    Femme_physics 2,555
    Gold Member

    [​IMG]

    given this circuit find the current on R1,R2 and R3.

    My attempt


    First of, I wanna know-- did I build the equations correctly?

    [​IMG]

    [​IMG]
     
    Last edited: Jul 7, 2011
  2. jcsd
  3. gneill

    Staff: Mentor

    Your equations look fine, although you've provided more equations than are strictly necessary to solve the circuit.

    EDIT: Actually, check the sign of the I1 term in your second equation. In your first equation you assumed that I1 was flowing from A to B, so you must maintain that convention throughout. \EDIT

    What exactly are you trying to solve for in this circuit, and are you expected to use any particular method of solution?
     
    Last edited: Jul 7, 2011
  4. Femme_physics

    Femme_physics 2,555
    Gold Member

    I see, so the second equation really should be:

    Sigma = 0; 20 - I2x2 - 14 +I1x4 = 0

    Yes?
     
  5. I like Serena

    I like Serena 6,193
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    Yes.
     
  6. Femme_physics

    Femme_physics 2,555
    Gold Member

    Then I think I got I1, yes? :smile:

    [​IMG]

    [​IMG]

    [​IMG]

    [​IMG]
     
  7. I like Serena

    I like Serena 6,193
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  8. Femme_physics

    Femme_physics 2,555
    Gold Member

    Thanks! :smile:

    w00t! If you say so, then I must really am!

    Easy peasy!

    [​IMG]

    [​IMG]
     
  9. I like Serena

    I like Serena 6,193
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    Good! :smile:

    I'll post another problem in your thread, since the second problem there probably won't hold you up for long! :biggrin:
     
  10. Femme_physics

    Femme_physics 2,555
    Gold Member

    Hehe. I was actually hoping you'd let me find the voltage difference between two points! Can you come up with two points for me?
     
  11. I like Serena

    I like Serena 6,193
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    Sure! VAB?

    And can you also give me PAB?
     
  12. Femme_physics

    Femme_physics 2,555
    Gold Member

    Let me start with Va,

    Am I correct with this result for Va?

    Va = 22.84 [V] based on the calculations below

    [​IMG]

    [​IMG]
     
  13. I like Serena

    I like Serena 6,193
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    Sorry, but I can't follow your reasoning here.

    Anyway, you can't calculate the voltage at a point without defining a reference voltage first (usually earth).
    But then, you usually have no need to know the voltage at a point.

    You're only interested in the voltage difference between 2 points.
    That is what you use and get from KVL.
    And also what you use and get from Ohm's law (the voltage "across" a resistor based on current).
     
  14. Femme_physics

    Femme_physics 2,555
    Gold Member

    What I did is calculate the voltage drop from each voltage loop UNTIL point A, then I added all the remiain voltages together (since I've discounted all the voltage drops).

    So when calclating voltage difference I have to set a zero point otherwise I can't solve the thing? And just what exactly is "earth"? Is it A? Is it B?
     
  15. I like Serena

    I like Serena 6,193
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    I still don't understand. :(


    :uhh: Noooo! You don't set a zero point at all. You don't need it to calculate a voltage difference.
    (You only need it to calculate the voltage at a point, and you don't need that.)

    Could you perhaps look at the resistor between A and B, look up the current "through" it, and apply Ohm's law to find the voltage "across" it? :shy:


    Usually a circuit in reality is grounded.
    That is, some point of the circuit is literally attached to earth.
    We have a nice symbol for it (actually 2 symbols that mean the same thing).

    But without an actual earth, you are free to choose any point to be the zero point.
    You can choose A and you can choose B (but only 1 of the two! :wink:).
     
  16. Femme_physics

    Femme_physics 2,555
    Gold Member

    If we look at each loop individually, the same loops that I composed my equations with to solve for the current, we can see that that there "voltage drops throughout the loops, yes?

    For instance, if we take the first loop of the 36V, the voltage drop of that 5 ohm resistor will be Io x 5 = 31.58

    When it reaches point A the 36 voltage LOST an amont of 31.58 voltage. Therefor, the remainder is 4.42.

    Same with loop 2, until I reach point A, and same with loop 3, until I reach point A. Then I add the remaining voltages together.

    Before that let me ask you a question, please.

    I present to you case (1) and case (2)

    [​IMG]

    In case 1, the same voltage that drops through R1 also drops through R2. In case 2, it's not the case, right? Or, otherwise, not "necessarily" the case, because there are other voltage sources complicating things up, aye?


    You mean as long as the circuit is not on a airplane/spaceship or anything that flies?
     
  17. I like Serena

    I like Serena 6,193
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    Aye! :smile:





    Now let me see if I can follow your reasoning.
    Yes. :)

    Wait! Stop!
    Yes, so the voltage difference "across" the 5 ohm resistor is 31.58 V.

    Wait! Stop!
    There! You have your voltage difference between points A and B! :smile:

    Now why would you want to do that?
    Did you perhaps look at another solution where they did something like that?
    Actually, I remember Gneill did something like that, but he also defined his currents differently.
    His method was quite nice, eliminating an equation and an unknown before he even started.
    But you have to choose, you shouldn't mix them.






    *grins*
    Did you never see the long copper wire connecting a plane to the ground? :confused:

    Actually on a plane, the circuits would be connected to the frame of the plane.
    That is big enough to function as an earth.
     
  18. Femme_physics

    Femme_physics 2,555
    Gold Member

    Is it not also the "voltage drop"?

    What?!? How?!? I didn't even get to point B!!!

    No, I don't have any solutions, just trying to reason out things on my own...apparently not working out so well :tongue:

    Oh, so by "earth" you don't really mean "the ground of planet earth"? It's also the "air of planet earth"?
     
  19. I like Serena

    I like Serena 6,193
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    Yes. It's the same thing.
    "voltage drop" does suggest a bit that the voltage is going down.


    Whaaaat? Didn't you?
    I think you did! :smile:


    Keep reasoning! At some point it will say *click*, and then you'll breathe hotly *Aaaaaaaaaaaah*. :blushing:


    "Earth" or "ground" is a sufficiently large metal (conducting) body.
    Not air, but a plane will have metal stuff attached, that can be used.

    It means electrons can flow more or less freely in or out as needed.
     
  20. Femme_physics

    Femme_physics 2,555
    Gold Member


    Did I? Does that mean point B is my zero point, or my "ground point", if you will?

    hotly eh? :wink:

    I will keep reasoning, then :approve:

    Really? That's how "earth" is defined? I didn't know that...I wish we were told that.

    So a circuit in space can't have a "grounding point"?
     
  21. I like Serena

    I like Serena 6,193
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    Hmm, B is still not your zero point.
    But since you have a voltage "difference", it will look a bit as if B is zero I guess.
    That is, you also have the voltage that A would have if B were the zero point.




    There! :smile:
    It should be somewhere in your textbook if you have one.


    Space...... *looking dreamily up*

    Well, we'd need to build a spaceship then, wouldn't we?
    Let's put some conducting metal in!
    If only to ground our artificially intelligent computer with the hot voice! :approve:
    Would you be willing to lend your voice to make it happen? :blushing:
     
    Last edited: Jul 8, 2011
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