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Calculating the EMF

  1. Sep 5, 2015 #1
    1. The problem statement, all variables and given/known data
    607a8abd-b0e5-3f8f-87f9-7f4249e1ff46___5a690bab-031d-3536-8c6b-ec2bc1ad2291.gif
    In the circuit shown in the figure above, the ammeter reads 3.4 A and the voltmeter reads 6 V. Find the emf ɛ and the resistance R.

    2. Relevant equations

    Ohm's law; V= I*R
    EMF equations: ε=I*r+I*R
    ε=I*r+V

    3. The attempt at a solution

    I got the resistance of R right: It's 5.29Ω. To do that, I found the current in the 2Ω resistor (6V/2Ω=3A).
    Then, I know that since the 2Ω and the 4Ω resistors in are in series, the current in the two of them are equal. (4Ω*3A=12V).
    Then, I add up the two voltages (=18V), and solve for R using V=I*R, which gives me 18/3.4 = 5.2941Ω.

    However, when it comes to finding the emf, none of my answers are correct. I've tried "reducing" the circuit so it becomes 3 resistors in parallel (4Ω,2Ω and R on one side, 1Ω and the battery in the middle, and the 2 3Ω ones on the other side). However, this doesn't seem to give me the right answer. At first I thought it might be 6 volts since circuits in series are supposed to have the same voltage everywhere, but then I tried rationalizing it and it didn't work.

    Any push in the right direction would be greatly appreciated!
     
  2. jcsd
  3. Sep 5, 2015 #2

    TSny

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    What is the potential difference across each of the 3Ω resistors?
     
  4. Sep 5, 2015 #3
    Is it 4A? Because 12V/3Ω?
     
  5. Sep 5, 2015 #4

    SammyS

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    Hello Draconifors. Welcome to PF !

    Potential difference can't be 4A . That has units for a current.
     
  6. Sep 5, 2015 #5
    Thank you!

    And ohh. That's true. The only thing I can think of is just ignoring my A and saying it's 4V but I wouldn't know why. :sorry:
     
  7. Sep 5, 2015 #6

    TSny

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    Yikes!:wideeyed:

    How does the potential difference across one of the 3Ω resistors compare to the potential difference across the resistor R?
     
  8. Sep 5, 2015 #7
    Yikes is pretty much my own reaction, honestly.

    And they're all in parallel, so it has to be the same, right?
     
  9. Sep 5, 2015 #8

    TSny

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    Yes, the potential difference across each parallel branch is the same.
     
    Last edited: Sep 5, 2015
  10. Sep 5, 2015 #9
    Ok so if each of the 3Ω resistors has 12V going through it, then I should be able to find all the current running through the circuit and then using ε=I*r+V, no?
    I find that I = 14.4, r=1 and V=12, but the answer is not 26.4. :confused:
     
  11. Sep 5, 2015 #10

    TSny

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    No, not 12 V. What did you use for the potential difference across the resistor R when you calculated R?

    Nit-picky side note: "voltage of a resistor" represents potential difference between a point on one side of the resistor and a point on the other side. "Voltage" does not "go through" something.
     
  12. Sep 5, 2015 #11
    I used 18V, as that was the sum of the current in my 4Ω and 2Ω + the voltmeter.
     
  13. Sep 5, 2015 #12

    TSny

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    OK. If the resistor R is in parallel with one of the 3 Ω resistors, what is the potential difference across the 3 Ω resistor?
     
  14. Sep 5, 2015 #13
    It's also 18V, as potential differences are the same throughout a circuit with resistors in parallel.
     
  15. Sep 5, 2015 #14

    TSny

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    Yes, each 3 Ω resistor will have a potential difference that's the same as the potential difference across R.
    When two branches of a circuit are in parallel, then the potential difference across one branch will equal the potential difference across the other branch.
     
  16. Sep 5, 2015 #15
    Ok so there's the same potential difference going through the branch with the 1Ω resistor and the battery, right? Does this mean I need to calculate the I of the whole circuit and then use ε=I*r+V, with 1Ω=r and V=18V?
     
  17. Sep 5, 2015 #16

    TSny

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    Yes, although I cringed a bit again when I read your phrase "potential difference going through the branch". Charge flows through the branch, but potential difference (or voltage) isn't going anywhere.
    Not sure what you mean here. What do you mean by "I of the whole circuit"?
    Is the I in ε=I*r+V the same as the I of the whole circuit?
     
    Last edited: Sep 5, 2015
  18. Sep 5, 2015 #17
    That was what I understood of my teacher's explanations of the formula, yes. Is it only the I going through each branch?
     
  19. Sep 5, 2015 #18

    TSny

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    There are different ways you could approach the problem. For example, you could consider the junction point J shown in the figure on the left below. Use the "junction rule" that the total current leaving a junction must equal the total current entering the junction.

    Or, you could approach the problem by using the rules for combining resistors to get a simpler circuit as shown on the right.
     

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