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Homework Help: Voltage in a 2 battery series circuit

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known dataWhat is the magnitude and direction of the current in the following circuit?
    [URL=http://imageshack.us][PLAIN]http://img234.imageshack.us/img234/7587/circuituk2.gif[/URL][/PLAIN]
    resistance for series: Req=R1+R2+R3+R4....etc
    Kirchhoff's rules --> Loop Rule: sum of potential drops must equal sum of potential rises
    V=IR



    3. The attempt at a solution
    1st added all the voltage drops: I(5+27+12+8) = I52
    2nd added all the Voltage rises: 30 + 10 = 40V

    3rd solved for V when V=IR
    40 volts/52 ohms= I
    *since resistors are in series they experience the same current right?

    my answer .77 Amps clockwise [which is wrong] the right answer is .38 Amps clockwise

    My question is this: Why am i not supposed to add the voltages (30+10) together when solving this problem?? If i solve for this problem in the following way, 30V-10V = I(52ohms) then i get .38 Amps clockwise (which is right) but i don't understand why i have to subtract the voltages!:confused:
     

    Attached Files:

  2. jcsd
  3. Feb 13, 2007 #2
    Mark the poisitive side of the cells in your image.
    Does that give you a clue?
     
  4. Feb 13, 2007 #3
    Yes. It seems the only logical explanation i can think of is that when 2 batteries "oppose" each others flow; then one battery will reduce the voltage of the other.
     
  5. Feb 13, 2007 #4
    Yes, the equivalent voltage will either be a sum or (in this case) difference of the 2 voltages. Going in a clockwise loop from A, a particle will go from low potential to high potential (+30V) and then go from high potential to low potential (-10V). If the batteries were arranged + to - in series, you could add the voltages because there would be a rise followed by another rise
     
  6. Feb 13, 2007 #5
    i see; i mistakenly counted the 2nd battery as a rise instead of a drop. Thanks again!
     
  7. May 9, 2010 #6
    i'm starring at the same problem and yet don't get how to solve it ? can anyone offer me any help?! I might have this problem on my final exam this coming monday may 10th , 2010. help please
     
  8. May 9, 2010 #7
    Which part do you find confusing?
     
  9. May 9, 2010 #8
    the kirchoff's rule as a whole. I tried solving this problem in the following way :
    R= 1/32 + 1/20 = 0.083 ohms
    I= V/R = 30/0.083= 360 amps
    which I know is wrong so what can I do, I spent the last hour watching a professor from MIt working a similar problem yet he did not show it with numbers or real answers.
     
  10. May 9, 2010 #9
    Resistance in series adds like this:

    R1=32
    R2=20

    Rsum=32+20
     
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