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Current through ammeter with two batteries

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    For the values shown in the figure below, calculate (a) the current through the ammeter A1, (b) the current through the ammeter A2, (c) the PD across R2.

    242b0846387a.jpg

    Assume that the internal resistance of each cell and the resistance of each ammeter is negligible.

    Answers: (a) 0.05 A, (b) 0.0125 A, (c) 0.5 V

    2. The attempt at a solution
    (c) V = IR = 0.05 * 10 = 0.5 V

    (a-b) We use Kirchhoff's rules: current flowing from E1 is I1, current entering R1 is I1 - I3 and current entering E2 is I3.

    So we get 2 = 40 (I1 - I3) + 10 (1 - I3)
    1.5 = - 40 (I1 - I3), minus since current is moving from left to right while we are moving in an anti-clockwise direction.

    2 = 50 1 - 50 I3
    1.5 = -40 I1 + 40 I3 (multiply by 1.25 and sum up)
    3.875 = 0 which is uncorrect.

    What am I doing wrong?
     
  2. jcsd
  3. Sep 29, 2016 #2

    cnh1995

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    By observing the circuit, what can you say about the voltage across the 40 ohm resistor? What will be the voltage across the 10 ohm resistor?
     
  4. Sep 29, 2016 #3
    Don't we need to find the current first? The voltage across the 10 Ω resistor is easily calculated when we know the current: (c) V = IR = 0.05 * 10 = 0.5 V, where I = 0.05 A is taken from the answers.
     
  5. Sep 29, 2016 #4

    gneill

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    What component is connected directly across that resistor? What do you know about parallel-connected components?
     
  6. Sep 29, 2016 #5
    Current is connected directly across the 10 Ohm resistor.

    Parallel-connected components have the same voltage.
     
  7. Sep 29, 2016 #6

    cnh1995

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    Right.
    So what is the voltage across the 10 ohm resistor?
     
  8. Sep 29, 2016 #7
    Are you implying that we can just 2 V - 1.5 V = 0.5 V? But it is in (c), so I guess we need to find current in (a-b) first.
     
  9. Sep 29, 2016 #8

    cnh1995

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    Sorry I misread the post. Actually, voltage source E2 is directly connected across the 40 ohm resistor.
     
  10. Sep 29, 2016 #9
    Since we have two batteries therefore we need to use Kirchhoff's rules, right?
     
  11. Sep 29, 2016 #10

    gneill

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    Kirchhoff's rules always apply, so it's a pretty good bet they'll come in handy here!
     
  12. Sep 29, 2016 #11

    cnh1995

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    The voltage source E2 is directly connected across the 40 ohm resistor. And you yourself said correctly that
    What is the voltage across the 40 ohm resistor then?
     
  13. Sep 29, 2016 #12
    Ah, now I got it what you imply. Since we have a battery E2 = 1.5 V, then V40 Ω = E2 = 1.5 V. We can then find (c) 2 - 1.5 = 0.5 V.

    But how do we find the currents? I = V / R = 2 / (40 + 10) = 0.04 A, not 0.05 A.
     
  14. Sep 29, 2016 #13

    cnh1995

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    Right.
    Ohm's law gives I=V/R where I is the current through R and V is the voltage across R.

     
  15. Sep 29, 2016 #14
    We can use I = V / R for the R2 resistor. We'll get 0.5 / 10 = 0.05 A.

    Then we find 1.5 / 40 = 0.0375 A. 0.05 - 0.0375 = 0.0125 A.
     
  16. Sep 29, 2016 #15

    cnh1995

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    Right.
     
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