calculating current with parallel batteries

Discussion in 'Homework Help' started by msedtal, Nov 11, 2009.

  1. msedtal

    Thread Starter New Member

    Nov 11, 2009
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    0
    i am in a physics 2 course, and i am having trouble calculating current with two batteries connected in parrallel. I am at a complete loss of where to start. The problem is probably simple to most of you.

    From the problem i am thinking that i should use I=V/R ; but am unsure if V=9+9; and am also unsure about R, i am thinking it should be R=1+ (1/(1/2)+(1/3))

    Please help me get on the right track

    [​IMG][​IMG][​IMG]
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
    658
    85
    Start by noticing that the circuit can be re-drawn. See Diagram.

    From circuit B, work out the current through R3. Then work back.

    There is a more mathematical way to do it but because both battery voltages are the same this is the easy way. Only Ohms law needed.
     
  3. msedtal

    Thread Starter New Member

    Nov 11, 2009
    2
    0
    thanks for the help, it put me on the right track and i got the questions correct.
     
  4. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    It appears that the top 1Ω Resistor is the load, the other two resistors represent the internal resistance of the batteries (which goes up as a battery goes dead).

    Help for the bonus point: P_{ower}=I^2\cdot R
     
  5. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,040
    287
    Number one rule of Electronics.....never make anything more complicated than it really is!

    In a PARALLEL circuit, the voltage is the same across all components. The battry will appear as a single voltage source. Ohm's law does the rest.


    eric
     
  6. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
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    I see it differently;
     
  7. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    718
    Are you sure about the polarity of the batteries in the OP, to be parallel?

    They appear to be connected in series, as the post above shows.
     
  8. Quintilis_Telescope

    New Member

    Oct 13, 2009
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    0
    Loop 1 = 3Ω*I1 + 1Ω(I1-I2) = 9V
    Loop 2 = 2Ω*I2 + 1Ω(I2-I1) = 9V

    I1 = 3.2727A
    I2 = 4.0909A
    I3 = 3.2727A-4.0909A = -818mA (flowing up)
     
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