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Homework Help: Internal resistance circuit problem

  1. Apr 1, 2007 #1
    1. The problem statement, all variables and given/known data

    In a circuit, two identical batteries with EMF of 22 V and internal resistance of 7 Ohm are placed in parallel to provide power to a load resistance of RL = 35 Ohms. What is the power transfered from the current to RL to heat the resistor?

    2. Relevant equations
    Is R(L) (Resistance load) mean the same thing as R(total)? I know EMF = Isys * Rsys as well.

    3. The attempt at a solution

    Only drew the diagram, labeled areas and put the RL = [1/R(1) + 1/R(2)] ^ -1I also know that Itot = I1 + I2. I'm assuming here that EMF of 22V is for each battery???
  2. jcsd
  3. Apr 1, 2007 #2
    The load resistance is the resistance of the resistor that is added to the circuit. In this case, the total resistance is not the same as the load resistance.
  4. Apr 1, 2007 #3
    So RL = ((1/R1) + (1/R2))^-1 is an incorrect statement?
  5. Apr 2, 2007 #4


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    Your equation RL = ((1/R1) + (1/R2))^-1 only equates the total (internal) resistance of the batteries. The load resistance is a separate load (resistor) in the circuit which is in series with the two batteries. R(total)=(total internal resistance)+(load resistance RL).
  6. Apr 9, 2007 #5

    If i've got the right circuit in mind, you could use Thevenin's Theorem to find the thevenin equivalent circuit then determine I and P.

    Peace out.

  7. Apr 9, 2007 #6

    Me again,
    Forget my last comment. I was thinking of the wrong circuit. Use Superposition.
  8. Apr 10, 2007 #7
    Don't recognize either of those, Fez07
  9. Apr 10, 2007 #8

    Firstly, is this the circuit you have?

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