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Internal Resistance problem when batteries are connected in opposition

  1. Mar 12, 2012 #1
    A simple circuit with 2 batteries and a resistor connected in series. The positive terminals are connected to each other via a 4.5 resistor. Their emfs and internal resistances are below. What is the resultant emf?
    battery 1: ε=6V r=1Ω
    battery 2: ε=24V r=0.5Ω
    resistor connecting the two batteries is 4.5Ω


    (I've attached a picture of the circuit diagram)

    The answer is 18V. I know that resultant emf =ε - Ir. But I have no idea how to achieve an answer when their are two batteries present, let alone connected in opposition. I've spent the past two hours research books and the internet with nothing to show of it. The only way I can see to get 18V would be to simply minus emf of battery 1 from battery2, if this is correct rather than a coincidence, what is the rule behind it?
     

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    Last edited: Mar 12, 2012
  2. jcsd
  3. Mar 12, 2012 #2

    tiny-tim

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    welcome to pf!

    hi andipandi! welcome to pf! :smile:
    i've no idea why they tell you all about the resistances :confused:

    the emf in a circuit (a loop) is simply the sum of the individual emfs :wink:

    (and you can see from the long and short lines that the batteries are "facing" opposite ways, so one of them will be minus the other)
     
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