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Parallel and Series circuits w/ multiple batteries

  1. Feb 26, 2010 #1
    Hello, this is my first time posting on this forum and I have seeked help on this forum before on previous assignments for my physics class.

    There are four problems that I do not understand clearly or do not understand how the image is portrayed in a blueprint format

    Problem 1
    1. The problem statement, all variables and given/known data
    http://img291.imageshack.us/img291/4382/problem39.jpg [Broken]
    Find i

    2. Relevant equations
    Kirchhoff's rules


    3. The attempt at a solution
    I can find 2 loops with Kirchhoff's rules, but it is not enough to find i.
    6 - 3.5i2 - 2.2i = 0; 6 - 3.5i1 - 2.2i = 0; I understand I need to find a loop without i so I can eliminate i2 or i1 from one of the equations, but I do not understand the concept of two batteries facing each other.
    I tried using
    i2=i3+i or i = i2 - i3 and i just comes out to 0
    Problem 2
    1. The problem statement, all variables and given/known data
    http://img175.imageshack.us/i/problem30.jpg/" [Broken]

    If the emf of the battery is 15 V, and each resistance is 2, what is the power consumed by bulb B?

    Bulb D is then removed from its socket.How does the brightness of bulb A change?

    How does the brightness of bulb B change when bulb D is removed from its socket?
    2. Relevant equations
    not sure?

    3. The attempt at a solution
    The picture itself is confusing to me. I do not get how to put the light bulb picture into a blueprint format. Any information on simplifying this image would be amazing.

    Problem 3
    1. The problem statement, all variables and given/known data
    http://img687.imageshack.us/i/problem25.jpg/" [Broken]
    Find the equivalent resistance R between points A and B of the resistor network.
    2. Relevant equations
    series circuit resistance = R1+R2+R3....
    Parallel circuit resistance= [(1/R1) + (1/R2) + (1/R3)]-1
    3. The attempt at a solution
    (1/(41+21+26)+1/33)-1= 264/11
    [(1/(21+ 264/11)) + 1/36]-1 = 1980/59 = 33.55932203

    Solved Final answer = 20


    Problem 4
    1. The problem statement, all variables and given/known data
    http://img85.imageshack.us/i/problem22.jpg/" [Broken]

    R = [(675-486)/5.8] - .36 - .16 - 1.15= 30.9162069

    Find the potential difference Vxy = Vx − Vy between points X and Y

    How much energy UE is dissipated by the 1.15 in 55 s ?
    2. Relevant equations
    Vxy = Vx − Vy
    P=IV
    V=IR

    3. The attempt at a solution
    Once again the negative sides facing each other completely throw me off and I have no Idea what to do.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 26, 2010 #2

    vela

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    There's nothing particularly special about having the two batteries in the circuit. Your equations are fine.

    You can get the third equation using Kirchoff's current law to relate i, i1, and i2. In this particular case, because of the symmetry of the circuit, you might see that i1 and i2 have to be equal.
     
    Last edited by a moderator: May 4, 2017
  4. Feb 26, 2010 #3

    vela

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    Your method is fine. You just made a mistake somewhere churning out the final answer.
     
    Last edited by a moderator: May 4, 2017
  5. Feb 28, 2010 #4
    Problem 1 and 3 are done, can anyone help with the last two?
     
  6. Feb 28, 2010 #5

    vela

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    There are only two loops in the circuit. Go around each one, placing and connecting the resistors as appropriate.
     
    Last edited by a moderator: May 4, 2017
  7. Feb 28, 2010 #6

    vela

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    Like I said for problem 1, you don't have to do anything differently when analyzing a circuit with two voltage sources. You seem to have a decent grasp of the material. Just get past that mental block and just do what you've been doing.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 28, 2010 #7
    I have tried drawing it and I only need to figure out how much Power is consumed by the B bulb. My design is an attachment below

    (each resistor A,B,C, and D are 2 Ohms)
    And I keep getting PB = 18 which is not correct
    Did I set up my circuit right?
     

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    Last edited: Feb 28, 2010
  9. Feb 28, 2010 #8
    I found the Vx= 189 and I subtracted [5.8(1.15+.36)] because of voltage drops and I got 180.242.
     
  10. Feb 28, 2010 #9

    vela

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    Yes, you drew the circuit correctly. Show us how you calculated PB.
     
  11. Feb 28, 2010 #10

    vela

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    This doesn't make sense. It's 189 volts relative to what?

    I assume you got the 189 V by subtracting 486 V from 675 V. That 189 V would be the voltage difference between the point to the left of the 675-V battery and the point to the right of the 486-V battery.
    Try applying Kirchoff's voltage law going around the loop starting from point Y, jumping straight up to point X, and then going to the left through the 675-V battery and the two resistors.
     
  12. Feb 28, 2010 #11
    I first calculated the equivelent resistance:
    (1/(2+2) + 1/2)-1 = 4/3
    4/3 + 2 = 10/3
    Req= 10/3

    With the total resistance I found the whole circuit's current:
    V=IR
    I=V/R
    I= 15 * 3/10= 4.5

    next, I found the allocated towards the parallel circuit so I subtracted the voltage dropped from A immediately.
    VA=IR
    VA= 4.5*2 = 9

    So, 6 volts allocated towards B,C, and D.
    PB=IV
    PB=V2/RB
    PB= 36/2
    PB=18
     
  13. Feb 28, 2010 #12
    So what you're saying is to do 675 - 5.8*.36 - 5.8*1.15 = the potential difference? How is there a loop for the electricity to flow?
    It is right but I do not understand how it works.
     
    Last edited: Feb 28, 2010
  14. Feb 28, 2010 #13

    vela

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    Perfect up to here.
    6 volts is the drop the series combination of B and C, not across B and C individually. Your answer would be correct for the power dissipated by D since the entire 6-volt drop appears across that resistor, but with B and C, the voltage drops across the individual resistors add up to 6 volts.
     
  15. Feb 28, 2010 #14
    This makes PERFECT sense! Thank you so much! I was thinking the same thing but I got confused and I forgot the square the V so instead of 4.5 I ended up with 9.
     
  16. Feb 28, 2010 #15

    vela

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    Yes, that's the correct equation.

    You get used to going around the actual loops in a circuit because that's how you usually get your equations, but KVL actually applies to any loop. As long as you end up in the same place as you started, the voltage differences around the loop must sum to zero.
     
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