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Homework Help: Find the value of resistance in the circuit

  1. Jul 28, 2017 #1
    1. The problem statement, all variables and given/known data
    ?temp_hash=f10be61cac4499f1c867d406a9a8e332.png

    2. Relevant equations


    3. The attempt at a solution

    I don't understand how can current in ammeter be zero .Why is 2V battery not sending current in the right loop ?
     

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  3. Jul 28, 2017 #2

    cnh1995

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    What is the voltage across R?

    A specific value of R will make the ammeter read zero. What should be that value?
     
  4. Jul 28, 2017 #3

    cnh1995

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    In other words, what should be the current through R so that the ammeter reads zero?
     
  5. Jul 28, 2017 #4
    I think what you are suggesting is that if the potential drop across R is 2V , then there would be no current in the ammeter .

    Right ?
     
  6. Jul 28, 2017 #5

    cnh1995

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    No.
    Potential drop across R is 2V no matter what the value of R is. Do you see why?

    How would you apply Kirchhoff's current law at the juntion of R , the 5000 ohm resistor and the left terminal of the ammeter?
    What should be the current through R if you know the current through the 5000 ohm resistor?
     
  7. Jul 28, 2017 #6
    Yes.

    Will the current across 5000 ohm be 12/(5000+R) irrespective of E2 ? Doesn't E2 affect this current ?
     
  8. Jul 28, 2017 #7

    cnh1995

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    What is the voltage drop across the 5000 ohm resistance?
    What is the current through the 5000 ohm resistance?
     
  9. Jul 28, 2017 #8
    Please explain your point . A question for a question is not helping much :smile: .

    I get the correct answer if I equate [12/(5000+R)]R = 2
     
  10. Jul 28, 2017 #9

    cnh1995

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    That's the correct equation.
    Voltage across R is 2V.

    I am afraid if I started explaining, I might just give out the answer (and get a warning from some mod:-p). I am not good at the art of explaining something precisely.

    How about some reverse engineering?
    You know the correct value for R. What can you say about the relation between the currents through R and 5000 ohm resistance? What would it be if R had some other value?
     
  11. Jul 28, 2017 #10

    cnh1995

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    The 12V battery is driving a current through the 5000 ohm resistor. Say that current is I1. The current through R is I2. How are I1 and I2 related if the ammeter reads zero?
    Have you studied Kirchhoff's current law?

    Should the current I1 split at the junction of R, ammeter and 5000 ohm resistance?
     
  12. Jul 28, 2017 #11
    It is because 2V battery is placed across R .Right ?

    Suppose there is no 12V battery and no 5000 ohm resistor i.e only the right loop is present .

    What is the difference between this case and the setup given in the problem ? Why does current flow in the former but not in the latter despite potential difference across R being 2V in both the cases ?
     
  13. Jul 28, 2017 #12
    No . It should not split .I1 =I2 . I know KCL :smile:
     
  14. Jul 28, 2017 #13

    cnh1995

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    In the former case, the current is driven by the 2V source.
    In the latter case, the current is supplied by the 12V source and the 2V source only maintains a voltage of 2V across R. It does not supply any power. If it did, it would violate KCL and KVL.
     
  15. Jul 28, 2017 #14

    cnh1995

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    Yes.
    Now given that the voltage across R is 2V, what is the voltage across 5000 ohm resistance? Do you know KVL?
     
  16. Jul 28, 2017 #15
    OK. This is something new and interesting . Thanks .

    But why is only 12V battery driving the current and not 2V battery ? Why doesn't it supply any power ? What if there was a current in the ammeter , wouldn't 2V battery supply current in that case ?
     
  17. Jul 28, 2017 #16

    cnh1995

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    Yes. But for all values of R except the one which is the answer to this problem.
    This specific value of R prevents any current from flowing through the ammeter.
    Have you studied Wheatstone's bridge? This is exactly like a balanced Wheatstone's bridge.
     
  18. Jul 28, 2017 #17
    10 V . Current through 5000 ohm will be 10/5000 ohm .This same current will flow through R i.e (1/500)R = 2 gives R .
     
  19. Jul 28, 2017 #18

    cnh1995

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    That's correct.
     
  20. Jul 28, 2017 #19
    So role of 2V battery is to maintain a potential difference of 2V across R irrespective of whether any current flows through Ammeter or not .

    Is that correct ?
     
  21. Jul 28, 2017 #20

    cnh1995

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    Yes.

    You can see for any other value of R, there will be a current through the ammeter and the 2V source will be supplying or absorbing power depending on the direction of the current through the ammeter.
     
  22. Jul 28, 2017 #21
    I am not sure about Wheatstone bridge , but I think this same mechanism happens in potentiometer . Right ?
     
  23. Jul 28, 2017 #22

    cnh1995

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    Yes.

    In this problem, suppose you removed the 2V source. Now because of the 12V source only, there will be a current in the leftmost loop. The voltage across R is still 2V.
    This means the 12V source establishes a 2V potential difference across R. Now when you connect the 2V source back in the circuit, you are connecting a 2V source between two points where the potential difference is already 2V.
    Hence, adding the 2V source doesn't make any difference.
     
  24. Jul 28, 2017 #23
    Thanks a lot :smile:
     
  25. Jul 28, 2017 #24

    cnh1995

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    No probs!
    Good night!
     
  26. Jul 28, 2017 #25
    Everything else remaining same in the circuit , if I remove the 5000 ohm resistor and try to write KVL in the two loops I end up with an absurd result 10 =0 .

    Is it because now both the batteries are trying to maintain their respective potential differences across resistor R ?

    Now if I put a resistor somewhere in the left or right loop then again things fall in place .

    Why does this happen ?
     
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