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Internal resistance question

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi I've tried to solve this question but I keep getting stuck at a certain point, it probably very simple but any tips would be greatly appreciated

    The EMF of a cell is measured using a potentiometer and is found to be 1.51V. When a voltmeter, which has a resistance of 3.5k Ohms is connected across the cell, it reads 1.49V. What is the internal resistance of the cell. What current flows through the voltmeter?
    If a resistor of resistance 50 Ohms is connected across the cell, what current flows through it?

    2. Relevant equations
    Kirchhoff's rules
    V=RI


    3. The attempt at a solution

    I said so far:
    (internal resistance)*I = 1.51
    1.49= I1*(3.5*10^3) - 1.51
    & found I1 current flowing through voltmeter = 8.6*10^-4A but i dont know what to do from here and i think this might be wrong anyway...:frown:
     
  2. jcsd
  3. Apr 13, 2007 #2
    Are you sure its "3.5k" ohms?
    Because that would be 3500ohms.
     
  4. Apr 13, 2007 #3
    If you use kirchoff rule and consider the two resistances: where v is the ideal voltage (that is under no load of the emf)

    V-(R(int)-R(ext))i=0
    and V-R(int)i=0.02 solve for R, is that what you got?
     
  5. Apr 13, 2007 #4
    i dont understand,how do you solve that equation when you dontknow what V is, do i take 1.51V as the value for V in that equation?
     
  6. Apr 13, 2007 #5
    yes thats the ideal voltage measured under no current. The 1.49 is with the Ir drop due to current across int resistance.
     
  7. Apr 13, 2007 #6
    and i just use the current value i have already calculated?
    ok i think i follow now,thanks for the help
     
  8. Apr 13, 2007 #7
    I'm a little unclear on your first eqn, if those don't work use the two I posted, remember its 2 unknowns and two equations tho the error from using just the 3500 Ohm resistor to compute i will likely be negligible.
     
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