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Circuit Problem -- Two voltage sources and some resistors

  1. Oct 15, 2015 #1
    1c6275e4ee362437ec20e92c3d7ee54d_zpssy2lfb7h.jpg

    Let's say I start at the high voltage part of the Emf source on the right and go counter-clockwise; the first thing I hit is the 10 Ohm resistor, and the current would not split up the junction because the wire is open-ended, correct? So the next thing I would hit is the 8 Ohm resistor, then the 15 V Emf source, before getting back to the same voltage source.

    So if we call that current I1, then we get 15-V - I1 x 10.0 Ohms - I1 x 8.00 Ohms - 15.0-V = 0, correct?

    The difficulty I run into is that I do not know which direction is in the "same" direction or "against" the direction of the resistor. How would it be possible to hit a resistor going in the opposite direction of the current?
     
  2. jcsd
  3. Oct 15, 2015 #2

    berkeman

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    Welcome to the PF.

    I have moved your thread to the Homework Help section of the PF. All schoolwork-type questions (even for self study) need to go in the HH forums.

    Now on your questions -- the two voltage sources are pointing in the same direction, so you add their two voltages to make an overall 30V source. You are then correct that the 5 Ohm resistor carries no current because it is not connected to anything on the left.

    Where are the points A and B supposed to be in the figure?
     
  4. Oct 15, 2015 #3
    My professor e-mailed us yesterday and said that that the "top left" is A and the "top right" is B. So I'm guessing Vb - Va is supposed to be the voltage drop across the 10 Ohm resistor?

    I know how to add resistors and capacitors, but I do not know how to add voltage sources. Can you please elaborate as to what you mean by the voltage sources "point in the same direction"? (E.g., do you mean that the high voltage source of one is oriented to the low voltage source of the other, or what?).
     
    Last edited: Oct 15, 2015
  5. Oct 15, 2015 #4

    berkeman

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    Just think of them as batteries. When you put two AA batteries in series with the polarities pointing the same way (1.5V each), what do you get total?

    And to finish your problem, since there is no current through the 5 Ohm resistor, what is the voltage drop across it. Can you solve the problem now? :smile:
     
  6. Oct 15, 2015 #5
    It would be 3.0 V total. But what if the batteries were oriented such that the + end of one faced the + end of the other? Also, does it make a difference if they're parallel rather than in a series?

    Yes; Vtotal = 30V, and Rtotal = 18 Ohms, so I = 1.67 Amps. The voltage drop across the 10 Ohm resistor would therefore be (10 Ohms) (1.67 Amps) = 16.7 Volts.
     
  7. Oct 15, 2015 #6

    JBA

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    Your two voltage sources are in series so what does that tell you about the their total voltage?
    There is no connection to the end of the wire at the left end of the 5.00 ohm resistor, what does that tell you about the current though that resistor and therefore the voltage on each side of that resistor regardless of the amount of voltage?

    For my information, where are the missing marked A and B points on your circuit

    (You and Ellatha solved the problem while I was writing my post)​
     
  8. Oct 15, 2015 #7

    berkeman

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    Connecting power supplies (or batteries) in opposition is not a good idea -- it's likely they will be damaged.

    Same thing with trying to parallel power supplies or batteries. It can be done in special cases, but in general is not done.
     
  9. Oct 15, 2015 #8
    Okay, so like in this circuit:

    _zps9acnx1k7.jpg

    The voltage sources are parallel and there are three possible circuit paths so we would have to use Kirchhoff's Rules to determine the the value of the currents and it would be a much more difficult problem, correct?
     
  10. Oct 15, 2015 #9

    berkeman

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    Not *much* more difficult, but you would end up with two KCL equations that you would solve simultaneously. For that circuit, I would flip it vertically and put the "ground" node at the - terminal of the 12V supply. Then write the two KCL equations for the other two nodes in the circuit (+ side of the 12V supply and the - side of the 10V supply). Makes sense?
     
  11. Oct 15, 2015 #10
    I don't think this kind of problem will be on my exam tomorrow (there will only be 6 problems; 4 from 4 chapters and 2 from 1 chapter), but we did go over the material.

    The only thing that confused me about these KCL problems is when to add or subtract when tracing the paths of the current. I was told to "subtract the potential difference when the current hits the resistor in the same direction as it was defined." But it seems to me that the current would always hit the resistor in the same path that it was defined. Can you give me an example of a current hitting a resistor in the path that it wasn't defined (so that we would add I x R instead of subtracting)?

    Like let us say that we wanted to go counter-clockwise all the way around the circuit from the 10 V voltage source; the first thing we would pass is the 8 Ohm resistor, then we would hit the 12 V Emf source, then the 4 Ohm resistor, before making our way back to the original voltage source.

    So that would give the following: 10 V - (8 Ohms)I2 - 12 V - (4 Ohms)I1. My confusion arises because I do not know the difference between when I need to add or subtract voltage sources or resistors (I just guessed).
     
    Last edited: Oct 15, 2015
  12. Oct 15, 2015 #11

    berkeman

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  13. Oct 16, 2015 #12

    CWatters

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    You are correct. You always do that for resistors. However for voltage sources you add or subtract according to the polarity.... eg Go around the circuit in the direction that you defined I to be +ve. When you encounter a voltage source if you hit the +ve terminal first then you subtract the voltage. If you hit the -ve first you add it.
     
  14. Oct 16, 2015 #13

    CWatters

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    First think to do is mark up the circuit with current arrows showing I1 and I2. I assume I1 is the current in the left hand loop and I2 in the right hand loop and that anti clockwise is +ve current.

    Then starting from the top left corner you do indeed get..

    +10 - (8*I2) - 12 - (4*I1) = 0
     
  15. Oct 22, 2015 #14
    85% on the first exam; thanks for the help!
     
  16. Oct 22, 2015 #15

    CWatters

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    Well done.
     
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