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EMF and internal resistance!

  1. May 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A 5.60ohm resistor is connected to a battery that has 0.200ohm internal resistance, and a terminal voltage of the battery is 10.0V. What is the current in the resistor? What is the emf of the battery?

    2. Relevant equations

    EMF=terminal voltage+(I)(Rint.)
    V=RI

    3. The attempt at a solution
    E-I1Ri-I1R1=0
    10=5.8I
    I=1.72.
    EMF=10+0.344=10.34

    Help guys? Exam tomorrow at nine and this is the last problem I don't understand! Very much appreciated :)
     
  2. jcsd
  3. May 14, 2009 #2

    Doc Al

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    Staff: Mentor

    Looks good. (But is the resistor 5.6 or 5.8 ohms?)
     
  4. May 14, 2009 #3
    5.6+0.2 for R1+ri... no?
     
  5. May 14, 2009 #4

    rl.bhat

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    Homework Helper

    E-I1Ri-I1R1=0
    10=5.8I
    I=1.72.
    EMF=10+0.344=10.34

    The terminal voltage is the voltage across the resistance. So V = IR1.= ER1/(R1 + Ri ).
     
    Last edited: May 14, 2009
  6. May 14, 2009 #5

    Doc Al

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    Staff: Mentor

    No, 10V is the terminal voltage, which equals EMF - Iri. Thus 10 = I*5.6. (From your equation: E-I1Ri-I1R1=0.)
     
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