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Circuit problem with two voltage source

  1. May 28, 2016 #1
    1. The problem statement, all variables and given/known data

    inu1s2.jpg

    R1 = 2 ohms
    R2 = 4 ohms
    R3 = 6 ohms

    E1 = 10 V
    E2 = 14 V

    What is the voltage between point A and B ?
    A. 0.18 V
    B. 0.98 V
    C. 1.34 V
    D. 1.64 V
    E. 3.28 V

    2. Relevant equations
    KVL

    3. The attempt at a solution

    I tried using KVL
    I assume that both mesh currents will go in clock-wise direction.
    x = current goes in lower mesh
    y = current goes in upper mesh
    So, the current that goes through AB or I3 is (x-y)

    For the lower mesh,

    E2 - I3R3 - x R1
    14 - 6(x-y)-2x = 0

    For the upper mesh,

    4y - 10 - 6 (y-x) - 14 = 0

    Solve for x and y..

    x = 5.8 A
    y = 5.4 A

    So, the current goes through AB is 0.4 A

    So,
    Vb = 14 - 6 (0.4) = 11.6 V
    Va = 11.6 - 2(5.8) = 0 V

    So, I think Vab = Va - Vb = -11.6 V
    But, it's not in the option..

    However, this is the book answer

    mr7l36.jpg

    I have no idea what methods are used to solve this very quickly.
    Please help
     
    Last edited: May 28, 2016
  2. jcsd
  3. May 28, 2016 #2

    Orodruin

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    You need to reconsider this equation.

    Generally, it is easier if you wait with inserting values until you have the final expression. It makes it much easier to follow what you are doing.
     
  4. May 28, 2016 #3
    What's wrong with that equation??
     
    Last edited: May 28, 2016
  5. May 28, 2016 #4

    Orodruin

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    Try doing it again. Mark the direction of the assumed currents at each wire and follow the loops.
     
  6. May 28, 2016 #5
    Alright. I try again.

    For the lower mesh,

    I start at point A and going clockwise

    E2 - I3R3 - I1R1 = 0

    E2 - (I1 - I2) R3 - I1R1 = 0

    14 - 6 (I1 - I2) - 2 I1 = 0


    For the upper mesh,

    I also start at point A and going clockwise

    - I2R2 - E1 - (I2 - I1)R3 - E2 = 0
    - 4 I2 - 10 - 6 (I2 - I1) - 14 = 0

    (sorry, I made the mistake at this upper mesh equation)

    Solving both equation :

    I1 = -0.0909 A
    I2 = -2.4545 A

    which means that both current goes in counter-clockwise direction, right ? (since I assumed the currents are in clockwise)

    I3 = I1-I2 = 2.3636 A

    Vab = Vb- Va = I1R11 = 0.0909 A * 2 ohms = 0.1818 V
    Thanks for help!

    Anyway, what's the name of method my book using? It's so simple and I'd like to learn about it.
    Indeed, using calculator is not allowed in the test so I need to know better methods to solve this.
     
    Last edited: May 28, 2016
  7. May 28, 2016 #6

    ehild

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    It is Thevenin's theorem. The upper part of the circuit between A and B (with R1 removed) is replaced with a voltage source with emf equal to the open-circuit voltage between A and B and the equivalent resistance RAB, when the sources are replaced with shorts.
    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html
     
  8. May 28, 2016 #7
    Alright, I'm interested in this approach. I've just watched basics of Thevenin's theorem video.

    It says that I should turn the 'mess' into a Thevenin voltage and a Thevenin resistor
    In the video, it's asked to determine current, but in my question, it is voltage between A and B, so I'm a bit confused.

    So, what is the 'mess' here? I can't decide if it's the upper part or the lower part.
    Please help me.. The example of the video I watched is far easier than this problem
     
  9. May 28, 2016 #8

    ehild

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    It is the upper part. Ignore R1.

    upload_2016-5-28_17-27-36.png


    Determine the voltage Uo between A and B. That will be the emf of the Thevenin-equivalent voltage source. The internal resistance of the source is obtained if you connect A and B with a zero ohm wire and determine the current Is flowing through that "short". The internal resistance is Ri=Uo/Is. But it is the same as the equivalent resistance between A and B if you replace the batteries by shorts, (simple wires). When you know the emf and internal resistance of the Thevenin-equivalent source, you need connect the resistor R1 to its terminals, and find the voltage across it.
     
  10. May 28, 2016 #9
    2m6u881.jpg

    Alright, This is what I've done.

    I just know how to find the current that loops through the upper part.

    But, I don't know how to find the emf (Thevenin-voltage). Please help

    EDIT : I made a mistake

    The current should be 10+14/(4+6) = 2.4 A
     
    Last edited: May 28, 2016
  11. May 28, 2016 #10

    ehild

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    Yes, it is the current that flows through the resistors, anti-clockwise. Consider the potential zero at A. What is the potential at B?
     
  12. May 28, 2016 #11
    Since the current goes anti-clockwise, I think there will be potential rise across the 14V E2 and potential drop across 6 ohms resistor.

    So, the potential will be = 14 - 2.4 * 6 = -0.4 V, right?

    And, for the resistor, I changed all the voltage sources into a short circuit/wire.
    Then, I get the resistance is equal to the parallel combination of 6 ohms and 4 ohms, which is 2.4 ohms.
    Then, I reconnect the resistor R1 to the Thevenin circuit. So, the voltage across the R1 is 0.4 * 2 / (2 + 2.4) = 0.18 V
    Thanks for helping me through this

    But, I'm still confused about one thing.
    The problem asks about the potential between point A and B.
    Why the answer is not 0.4 V which is the potential difference between point A and B ?
    Why we instead find the potential across R1 resistor ??

    And, one more question ._.
    Why the book approach in finding the Thevenin emf is just (E1 R3 - E2R2)/(E3-R2)? It seems like a mean(average-value), but why it substracts?
    Both E1 and E2 are running currents in the same direction (anti-clockwise), but why it substracts in the book approach? Or, is there a theorem about this?
    Please help
     
    Last edited: May 28, 2016
  13. May 28, 2016 #12

    SammyS

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    That looks good!

    Don't forget about the Thevenin resistor in series with the v0.4V source.
    Slight typo (corrected in red.)

    In relation to points A and B, E1 and E2 are in opposite directions.
     
  14. May 28, 2016 #13
    Can the potential difference between point A and point B be I2R2 + E1 ?? (if we refer to the upper part)
     
    Last edited: May 28, 2016
  15. May 28, 2016 #14

    SammyS

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    Try it & see.
     
  16. May 28, 2016 #15
    Yes, it can.. Thanks
     
  17. May 28, 2016 #16

    ehild

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    It is the potential difference between B and A.
    0.4 V was the voltage without load, the emf of the Thevenin source. If you connect the resistor R1, the current flowing will cause potential drop across the internal resistor and the voltage on R1 becomes less than the emf. It happens every time you load a voltage source. The terminal voltage is always less than the emf.
    You would get the same formula if you did the derivation symbolically. I=(E1+E2)/(R3+R2) , UAB=E1-R2I or UAB=R3I-E2. Bring to common denominator and simplify.
     
  18. May 29, 2016 #17

    gneill

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    Have you considered applying nodal analysis (KCL) rather than loop analysis?

    A very quick solution can be obtained if you apply nodal analysis. If you choose the reference node to be B then A is the only essential node. That means you can solve for the potential at A with respect to B using a single node equation.
     
  19. May 29, 2016 #18
    ##\frac{V_b-E_1}{R_2} = \frac{V_b-E_2}{R_3+R_1} \\
    \frac{(V_b-10)}{4} = \frac{(V_b-14)}{8}##

    But, I get ##V_b## equals to 6 V.
    Sorry, I'm bad at nodal analysis. Please help
     
  20. May 29, 2016 #19

    gneill

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    There are three branches between nodes A and B (Don't forget the R1 branch!), so your equation will contain three terms, one for each branch leaving node A. Re-drawing your circuit to show the reference node as B:
    upload_2016-5-29_11-44-29.png

    You want to find the potential at node A. Write your equation in the form ##\sum ~ I = 0##.
     
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