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Reading of the voltmeter with closed switch

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) In the circuit shown below E is a cell of source (internal) resistance r and the resistance of R is 4.0 Ω. With the switch S open, the high resistance voltmeter reads 10.0 V and with S closed the voltmeter reads 8.0 V. Show that r = 1.0 Ω.

    (b) If R were replaced by a cell of e. m. f. 4.0 V and source resistance 1.0 Ω with its negative terminal connected to B, what would be the reading of the voltmeter with S closed?

    b2ff33ce1773.jpg

    Answer: (b) 7.0 V.

    2. The attempt at a solution
    (a) E = I (r + R) → 10 = I (1 + 4) → I = 2 A. Then we find E = I r + V → E - V = I r → 10 - 8 = 2 r → r = 1 Ω. Looks correct.

    (b) Maybe something like 10 - 4 = 1 I3 + 1 I1, 10 - V = 1 I1?
     
  2. jcsd
  3. Oct 3, 2016 #2

    gneill

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    Yes, that looks okay.
    There's only one current circulating in the loop, so no need to introduce more current variables.

    Have you learned about Kirchhoff's Voltage Law (KVL)? It would really help you to write these equations since it can be done in a formulaic manner. It would also help you to find these voltages in a methodical way.
     
  4. Oct 4, 2016 #3
    In my book I have Kirchhoff's rule 1: the algebraic sum of the currents flowing into a junction is zero, i. e. Σ I = 0, I1 = I2 + I3.

    And rule 2: in any closed loop, the algebraic sum of the EMFs is equal to the algebraic sum of the products of current and resistance, i. e. Σ E = Σ I R. Using this rule each resistor within a particular loop must be traversed in the same sense (either clockwise or anticlockwise).

    KVL is a different name for the second rule?
     
    Last edited: Oct 4, 2016
  5. Oct 4, 2016 #4

    gneill

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    Yes, KVL is the second rule. In practical terms the rule states that the sum of the changes in PD that occur when traversing the loop is zero. So if you "walk" around the loop in either direction, summing up any rises and falls of PD on the components as you go, when you arrive back at your starting location that sum must be zero.

    It's analogous to walking a closed path on a terrain and counting the changes in elevation. You expect the net change to be zero when you get back to the starting point.
     
  6. Oct 4, 2016 #5
    In our case it would be -4 V + 8 V = 0?
     
  7. Oct 4, 2016 #6

    gneill

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    No. First of all it makes no mathematical sense, since 8-4 is clearly not zero. Secondly, you haven't accounted for all the potential changes. When I look at the circuit for part b I see two EMFs of 10 V, one of 10 V and one of 4 V, and two internal resistances of 1 Ohm. I don't know the current yet, but I can write the KVL sum around the loop using a variable ##I## for that current.

    So, write the KVL equation for the loop with assumed current ##I##. The sum, since your'e arriving back where you started having completed a closed loop, will be zero.

    [edit: Fixed the EMFs. Too many threads ongoing to keep track! :confused:]
     
    Last edited: Oct 4, 2016
  8. Oct 4, 2016 #7
    8 - 1 I - 4 - 1 I = 0
    4 - 2 I = 0
    I = 2 A
     
  9. Oct 4, 2016 #8

    gneill

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    That's the idea, only one of the EMF's is 10 V, not 8 V.
     
  10. Oct 4, 2016 #9
    It's 10 when it's open. We have a situation when it's closed in (b).
     
  11. Oct 4, 2016 #10

    gneill

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    The EMF is inviolate. It doesn't change. It's the voltage drop across the internal resistance that makes the PD drop at the battery terminals. You measure the EMF of a battery by insuring that no current is flowing --- put a good voltmeter across the battery terminals with no other load. That was the purpose of part (a), to determine the EMF of the battery. It is 10 V. It stays 10 V no matter what.
     
  12. Oct 4, 2016 #11
    10 - 1 I - 4 - 1 I = 0
    6 - 2 I = 0
    I = 3 A

    V = I R = 3 * (1 + 1) = 6 V?
     
  13. Oct 4, 2016 #12

    gneill

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    The current value looks good.

    What are you calculating with that last line?
     
  14. Oct 4, 2016 #13
    Voltmeter voltage (b) part.
     
  15. Oct 4, 2016 #14

    gneill

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    Then your calculation is not correct. The voltmeter is connected across EMFs and resistors. You've only assumed resistors and that they are connected in series between A and B. They are not.

    Instead, do a KVL walk from A to B. You can choose either direction (via the upper path through the 10 V EMF, or the lower path via the 4 V EMF).
     
  16. Oct 4, 2016 #15
    10 V - 3 A * 1 Ω = V
    V = 7 V

    I used 10 V as you said, not 8 V.

    4 + 3 * 1 = V
    V = 7 V

    Honestly, I'm still struggling to understand where a plus or a minus should go. For example why is it not 4 - 3 * 1 = V? I know we can just play with the numbers to get my answer 10 - 3 * 1 = V and 4 + 3 * 1 = V and get V = 7 V in both formulas but I want to understand the logic behind : (.
     
  17. Oct 4, 2016 #16

    gneill

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    Make a drawing of the circuit. Draw in the current direction. For every resistor that the current passes through place a small + sign where it enters and a - sign where it exits the resistor. That's it. When you do your walk along a path of the circuit just account for the change in potentials using those signs. + to - is a drop so it's negative. - to + is a rise so that's positive.

    Some people sum the drops and treat them as positive, while rises are accounted as negative. There's aesthetic reasons for this having to do with making the math look nice for techniques they use that you haven't learned yet. So for now just make drops negative, rises positive.
     
  18. Oct 4, 2016 #17
    vnfmkg.jpg

    Question 1: how do we determine the current direction? At AF I looked at the 10 V and took the direction from the + side.

    Question 2: voltmeter is like a resistor? Where current enters is a plus and where it goes out it's a minus?

    ABEF: - V - (1 * 3) + 10 = 0. V = 7 V

    BEDL: - V - 4 - (1 * 3) = 0. V = - 7 V. On LD both are minus since we are moving clockwise while the current is anti-clockwise as on the graph, therefore the signs are in the other direction.
    http://tinypic.com/r/vnfmkg/9
     
  19. Oct 4, 2016 #18

    gneill

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    If you can determine the current direction just by noting or even estimating the net EMF driving it, that's fine. But sometimes a circuit is complex and you can't determine ahead of time which way the current is going to flow in any given loop or along a given path if there are separate currents. That's fine! Just pick a direction and pencil it in. From then on just be consistent and use the same assumed direction when writing your equations. The math will sort it out for you! When you solve the equation for the current and the result turns out to be negative, then it just means that your initial guess was not right: the current actually flows in the opposite direction from your guess. The magnitude of the current you find will still be correct, and you can still plug the negative value into any equations you wrote and get correct results.
    Yes. Although good voltmeters generally have very high resistance and pass negligible current. Unless you are specifically analyzing a voltmeter or how it affects a particular circuit, their current is taken to be zero. Ideal voltmeters shown on a circuit to show you where to "read" a potential difference are taken to have infinite resistance and so pass no current at all (effectively they are an open circuit).
    In your diagram the 3 A you've shown for the voltmeter branch is not correct. It should be zero amps. If you do KCL (Kirchhoff's Current Law) at node B then you you would see that its sum would not be zero if you had two paths of 3 A leaving and only one entering.
     
  20. Oct 4, 2016 #19

    gneill

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    You're going clockwise around the loop, so when you pass through the 4.0 V source you're going from - to + through it. That makes it a potential rise. You're also moving against the current when you go through the resistor, so again going from - to + and that's another rise. Both those terms should be positive.
     
  21. Oct 4, 2016 #20
    Ah yes, since it's a voltmeter there's no current. And since we got V = -7 V then the current is going in the opposite direction to the one we put on the graph. But we can still say that the voltmeter says V = 7 V.

    This part was particularly helpful, I didn't get it before what is a drop and what is a raise.

    ---

    I think this should be it concerning this question : ). Thank you!
     
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