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Current Electricity

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Three resistances of 4,6,12 ohms are connected in parallel and the combination is connected in series with a 4V battery with internal resistance 2 ohms The battery current is what


    2. Relevant equations
    No relevant question.


    [b]3. The attempt at a solution[/b]
    I dont know
     
  2. jcsd
  3. Oct 23, 2009 #2
    replace parallel or series combinations of resistances by their equivalent until you have only 1 resistance left.
     
  4. Oct 23, 2009 #3
    I = [tex]\frac{E}{R}[/tex] = [tex]\frac{4V}{Ri+(R1||R2||R3)}[/tex] = [tex]\frac{4V}{2 ohm + (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})}[/tex] = [tex]\frac{4V}{2 ohm}[/tex] = 2A
     
  5. Oct 23, 2009 #4
    What happened to this? [tex] (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm}) [/tex]
     
  6. Oct 27, 2009 #5
    thank u
     
  7. Oct 27, 2009 #6
    I didn't notice it the first time but (R1||R2||R3) =

    [tex]
    1 / (\frac{1}{4 ohm}+\frac{1}{6 ohm}+\frac{1}{12 ohm})
    [/tex]

    The answer is still right.
     
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