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Internal resitance and electromotive force

  1. Aug 8, 2009 #1
    1. The problem statement, all variables and given/known data

    1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

    a. across the terminal of the battery when the switch is open
    b. across the resistor when the switch is open
    c. across the terminal of the battery when the switch is closed
    d. across the resistor when the switch is closed

    2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.


    2. Relevant equations
    Terminal Potential Difference= -Ir+ electromotive force
    Terminal Potential Diff= IR
    emf=W/q


    3. The attempt at a solution
    @1 would it be Terminal Voltage= IR? but i kinda don;t know how to apply it with the voltmeter placed in different places.
     
  2. jcsd
  3. Aug 9, 2009 #2
    Just do it the same way as the first one. You can treat the internal resistance as a separate resistor.
     
  4. Aug 9, 2009 #3

    vk6kro

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    Science Advisor

    In each case, you have the following arrangement (all in series).....

    Battery positive voltage.......internal resistance.......switch.........load resistor.......battery negative.

    Bearing in mind that there is no voltage across a resistor if there is no current flowing through it, what would be the open circuit voltage of the battery in each case?

    Now, close the switch and consider there are just two resistors (internal resistance and load resistance) in series across the EMF of the battery. The total current would just depend on the total resistance, so what would be the voltage across each resistor?
     
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