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Circuit with resistors in series and two power supplies

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data

    There is a current of 0.25 A in the circuit of the figure

    [​IMG]

    What is the value of the resistance R?


    2. Relevant equations


    V = IR


    3. The attempt at a solution

    since the circuit is in series:

    V = 6+ 12 = 18V
    Req = R + 6 + 12
    V = IReq
    Req = 18/0.25 = 72

    72 = 18+ R
    R = 54

    but the answer is wrong !
    help
     
  2. jcsd
  3. Mar 8, 2008 #2
    is the answer 30?
     
  4. Mar 9, 2008 #3
    current of the whole circuit =0.25A

    I=V/R = (12-6)/(18+R) = 6/(18+R)
    4.5+0.25R=6
    0.25R=1.5
    R=6 ?
    I am not sure ....
     
  5. Mar 9, 2008 #4
    I'm sure you know Kirchoff's voltage law - "The sum of all voltage drops around any closed loop must be equal to zero.

    Applying Kirchoff's voltage law gives ->
    0 = Power supply driving current (largest voltage) - Power supply opposing flow of current (smallest voltage) - Voltage drop across resistance in circuit (6Ohm, 12Ohm and ROhm).

    0 = 12 - 6 -(6*0.25) - (12*0.25) - 0.25R
    Thus..
    0 = 6 - 1.5 - 3 - 0.25R
    So..
    0= 1.5 - 0.25R
    0.25R = 1.5
    R = 6Ohms
     
  6. Mar 9, 2008 #5
    Ohm's law is something in this world i guess....
     
  7. Mar 9, 2008 #6
    Ohm's law is something i guess.
     
  8. Mar 9, 2008 #7
    Ohms law is something - but in this case Kirchoff's voltage law is something also.
     
  9. Mar 10, 2008 #8
    Utility of the either one counts.
     
  10. Mar 10, 2008 #9
    My take on this is that because there is a charge flow from the 12V battery and the potential of this battery is higher than the 6V, the 6V does not release any electrons at all and acts as a conductor.
     
  11. Dec 24, 2009 #10
    the circuit is an illusion.

    the effective voltage is not 18 volts, it is 6 volts.

    redrawing circuit with resistors in one leg and batteries
    in another gives a better understanding of actual circuit.

    6 ohm 12 ohm Rx
    +----VVVV----VVVV----VVVV---+
    | |
    __|__ + | E/I = Rt
    ___ 12 V | 6V/.25A = 24 OHM total
    _____ | therefore;
    ___ | 6 + 12 + Rx = 24 OHM
    | | Rx = 6 OHM
    | 6 volts effective |
    _|_ |
    _____ |
    ___ |
    _____ 6 V |
    | + |
    | |
    +-------------------------------------+
     
  12. Dec 24, 2009 #11
    please excuse previous post. seems that character graphics can not be used on this forum.

    repeating text of previous post and adding a graphic to show redrawn circuit.

    ++++
    the circuit is an illusion.

    when 2 batteries connected with like polarities connected cause lesser voltage to be subtracted
    from higher voltage. therefore, the effective voltage is not 6 volts + 12 volts = 18 volts.

    it is 12 volts - 6 volts = 6 volts.


    attached redrawing of circuit with resistors in one leg and batteries in another leg
    gives a better of what is actual circuit.

    using formula,
    E/I = Rt
    6V/.25A = 24 OHM total
    therefore;
    6 + 12 + Rx = 24 OHM
    Rx = 6 OHM
     
  13. Dec 24, 2009 #12
    You are right. A person should go around the loop in one direction, say, arbitrarily choosing clockwise. Then we would encounter one battery in the polarity plus-minus and we would encounter the other battery in the polarity minus-plus. Therefore the batteries oppose one another. We have to subtract the batteries, not add them.
     
  14. Dec 24, 2009 #13
    this is true. considering as a loop is one way to look at it.

    when i first looked at circuit, i saw reason "743344"'s formula did not work just from fact being that
    'load' side, [normal of convention of supply to left, load to right of a circuit] had an opposing voltage.

    "mysqlpress" replied with correct answer in formula with out a corrected drawing.

    "Mike Cookson" replied with an implication of "Kirchoff's voltage law", no drawing.

    24 Dec 2009, 06:48 PM UTC, i attempted an 'ascii graphic' of circuit, which failed due to
    compressing of spaces used.

    after seeing this, i submitted again, 24 Dec 2009, 07:46 PM UTC, with an 'attached' drawing,
    which for some reason or other, did not get shown. [i just now tried an edit and was able to get
    attachment to save]

    all in all, another way to spend December 24th UTC.

    happy Christmas to all.
     

    Attached Files:

    Last edited: Dec 24, 2009
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