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Using Kirchhof's laws to solve a circuit

  1. Jun 6, 2017 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure (attached with the solution). Determine:
    1. the reading of the ammeter when the key (K) is open.
    2. The reading of the ammeter when the key (K) is closed.
    2. Relevant equations


    3. The attempt at a solution

    syhUR.jpg

    What confuses me in the circuit above is that when the key is open, I will have only one loop to work on, and that will make only one equation (equation no.2) and then I will have two unknown values, and I will not be able to find i₂ (the current passing through the ammeter), so what should I do here?
     
  2. jcsd
  3. Jun 6, 2017 #2

    gneill

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    Staff: Mentor

    How many independent currents can you have if there is a single loop?
     
  4. Jun 6, 2017 #3
    When I have only loop 1, I will have two currents: i₁ and i₂.
     
  5. Jun 6, 2017 #4

    cnh1995

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    Homework Helper

    Are they two different currents?
     
  6. Jun 7, 2017 #5
    Oh, they aren't.
    So there will be only one current, and the equation will go like this:
    20-5 = 15i -----> i = 1 A
    So in the first case the ammeter reads 1 A.
     
  7. Jun 7, 2017 #6

    cnh1995

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    Homework Helper

    Right.
    What about the second case?
     
  8. Jun 7, 2017 #7
    OK, I will solve the three equation I had after applying Kirchhoff's laws.
    The value of i2 is 2.57 A.
     
  9. Jun 7, 2017 #8

    cnh1995

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    Homework Helper

    Sorry for relpying so late! I was travelling.

    Your first equation is incorrect seeing the directions of the currents.

    For second case, you have two independent loops. So use two loop currents.

    See this example.
    download (1).png
     
  10. Jun 8, 2017 #9
    Oh, yes, I can't believe I still do these silly mistakes with Kirchhoff's laws until now.
    Anyway, it will be like this:
    i₁ = i₂ + i₃ ----> 1
    15 = 10 i₂ + 5i₁ -----> 2
    15 = 5i₁ + 5 i₃ -----> 3

    Ok, and I will use two loop currents. It will be like this:
    15 = 10i₂ + 5i₁ ------> 2
    15 = − 5i₂ + 10 i₁ -----> 4

    Then i₂ = 0.6 A
     
  11. Jun 8, 2017 #10

    cnh1995

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    I am getting the same answer.
     
  12. Jun 8, 2017 #11
    Thank you, thank you, thank you!
     
  13. Jun 8, 2017 #12

    cnh1995

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    Homework Helper

    No probs!:smile:
     
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