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Emf and internal resistance of a battery

  1. Oct 25, 2008 #1
    When switch S in the figure is open, the voltmeter V of the battery reads 3.07 V. When the switch is closed, the voltmeter reading drops to 2.95 V, and the ammeter A reads 1.70 A. Assume that the two meters are ideal, so they do not affect the circuit.
    Find the value of the emf. (see picture posted below)

    the equation would be

    E-Ir-IR
    or E= I(r+R)
    E=Ir
    however I am totally confused on how to find the internal resistance value,r, when the switch is open and the value of R when the switch is close. Any help would be appreciated. Thanks. I apprecaite the help.

    Stephen
     

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    Last edited: Oct 25, 2008
  2. jcsd
  3. Oct 25, 2008 #2

    Doc Al

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    I'm a bit puzzled by this problem, since the battery EMF is given. Are you being asked to find R and r?

    That's one equation. E and I are given.

    Write another equation describing the given voltage reading across the battery terminals when the switch is closed.
     
  4. Oct 25, 2008 #3
    I have two equations there. Are they right? What would the emf be? the 3.07V or 2.95V? And how would you find the r and R?

    Thanks
    Stephen
     
  5. Oct 25, 2008 #4

    Doc Al

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    You must have snuck the second one in after I had posted. The one I saw is certainly correct. The other is just Ohm's law.
    The terminal voltage across the battery (which is what the voltmeter measures) is the EMF minus the voltage drop across the internal resistance (which is given by Ir). When the switch is open, what's the current through the circuit?
    By combining two equations. The one I quoted from your first post and another that you should write for the terminal voltage when current flows.
     
  6. Oct 25, 2008 #5
    So use 3.07 for E in E=Ir and 1.7 for I thus r=3.07/1.7=1.81? Then to find the emf you do the voltage 3.07 + (1.7)(1.81)?Then to find R would you use 2.95=I(r+R)=(1.7)(1.81+R)????

    Are these right?
     
  7. Oct 25, 2008 #6

    Doc Al

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    No. Reread what I said about EMF, terminal voltage, and internal resistance. Note that 3.07 is measured when the switch is open.
     
  8. Oct 25, 2008 #7
    The terminal voltage across the battery (which is what the voltmeter measures) is the EMF minus the voltage drop across the internal resistance (which is given by Ir).

    Well doesn't this mean that the emf = the voltage measured by voltmeter +the voltage drop, Ir? and to find r you use E=Ir or 3.07=(1.7)*r?

    If this is not right I need a little bit more info? The concept is not sinking in.
     
  9. Oct 26, 2008 #8

    Doc Al

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    No. Note that the terminal voltage (in this setup) is the voltage measured by the voltmeter. So that means: Measured voltage = EMF - Ir.

    And the current (I) depends on whether the switch is open or closed.
     
  10. Oct 26, 2008 #9
    From your equation EMF = Measured voltage +Ir
    right?

    And since the switch is not closed what would the current be when the voltmeter measures 3.07V? And then how would you find r from the equation E=Ir? Would you use the 3.07V for E and the current found above for I? If this is not right, would you please guide me to the right direction. Thanks.
     
  11. Oct 26, 2008 #10

    Doc Al

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    Yep.
    You tell me. When the switch is open you have an open circuit.
     
  12. Oct 26, 2008 #11
    I have no idea my professor only went over closed circuits. Since the circuit is not closed, I do not see how a current exists. But since the voltmeter is giving a measurement there is obviously a current. Need a starting step.

    V=Ir or I = V/r, but since we do not know r, I do not see how we can solve it.
     
  13. Oct 27, 2008 #12

    Doc Al

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    That's just the point: With an open circuit, there is no current. I = 0.
    Not true. The voltmeter measures voltage, not current. You don't need a current to have a voltage.
     
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