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Why formula for internal resistance for charging battery v = e + Ir

  1. Aug 14, 2014 #1
    1. The problem statement, all variables and given/known data

    Why is the formula for internal resistance for a charging battery v = e + Ir?

    When the battery is supplying I to the circuit I understand why the terminal voltage, V, equals emf - IR (voltage lost by internal resistance), but if a battery is charging, why is the voltage lost by internal resistance being added to the ideal voltage, e? The resistance is still there right, even if it is charging, so shouldn't the sign in front of Ir still be negative?
     
  2. jcsd
  3. Aug 14, 2014 #2

    gneill

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    Staff: Mentor

    Consider the direction of the current flowing in each case. What's the direction of the resulting voltage drop across the internal resistor in each case?
     
  4. Aug 14, 2014 #3
    In the first case if we use conventional current, the positive charges will be flowing toward the negative terminal. In the second case the positive charges will be flowing toward the positive terminal. (positive charges spontaneously move towards the terminal of lower electric potential). In the first case the positive charges are moving spontaneously toward the negative terminal (delta electric potential or voltage = - and delta PE = - ), but in the second case, energy is required to move the positive charges toward the positive terminal, resulting in delta electric potential or voltage to be positive and delta PE to be positive. That makes sense because you are storing electric energy into potential energy to be used at a later time which is what charging a battery is. So voltage drop is -ve in the first case and +ve in the second case. I'm still having some trouble seeing how this connects with the change in sign in the formula for terminal voltage?
     
  5. Aug 14, 2014 #4

    gneill

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    Staff: Mentor

    It's really very simple. Draw the two cases and note the directions of the potential drop on the internal resistance.
     
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