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About electricity and the EMF -- Multiple choice question

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    The cell in the circuit has an emf of 2.0 V. When the variable resistor has a resistance of 4.0 Ω, the potential difference (pd) across the terminals of the cell is 1.0 V.

    tPicture.asp?sub=AA_PA&CT=Q&org=175b675e2ecaeeff89e96ac7b647ab93&folder=QSAS232_files&file=img02.png

    What is the pd across the terminals of the cell when the resistance of the variable resistor is 12 Ω?


    A

    0.25 V



    B

    0.75 V




    C

    1.33 V




    D

    1.50 V





    2. Relevant equations
    The only relevent equation I am thinking of is V=IR

    3. The attempt at a solution

    There are two ways I tried to solve this problem

    1) by using proportion so if I volt = 4 ohms then what would 12 ohms would be.... The answer came out to 3 which is wrong because it can not be more the emf voltage as it is voltage source.

    2) It has to be more that 1 volt because we know that as the voltage increase the resistance also increase, so this means I can eliminate the multiple choice A and B. so now there are two choices, it can be either 1.50 V or or 1.33V.



    Please help me on this question guys. Teach me the proper method to solved this question. Thanks indeed
     
    Last edited by a moderator: Apr 19, 2017
  2. jcsd
  3. Apr 7, 2016 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Solve this by using the first circuit to tell you what the internal resistance of the voltage source is. If the open circuit EMF of the voltage source is 2V, and when you connect an external 4 Ohm resistor it drops to 1V, what does that tell you the internal resistance is?

    And then re-draw the circuit with a 12 Ohm resistor outside. The voltage across the 12 Ohm resistor will be the divided-down voltage that starts at 2V and divides with the internal resistance and the external 12 Ohm resistance. That should get you to the correct answer. :smile:
     
    Last edited by a moderator: Apr 19, 2017
  4. Apr 7, 2016 #3
    I am still stuck on the question, I do not know how solve if there is no current mentioned, because we will have two unknown values.
     
  5. Apr 7, 2016 #4

    berkeman

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    Staff: Mentor

    Nope, it's easy to solve. Just use the 2-step procedure that I outlined. What is your answer to the first step?
     
  6. Apr 7, 2016 #5
    Basically i am thinking of using EMF= I( R+r)
    When i substitute in 2=I(4+r). This is what I mean, I feel like I am over thinking it as it is a only one. Please help me mentor.
     
  7. Apr 7, 2016 #6

    berkeman

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    Staff: Mentor

    Think in terms of the voltage divider equation. Do you know how to calculate the output voltage from a voltage divider?

    https://www.facstaff.bucknell.edu/mastascu/eLessonsHTML/Resist/Resist3A01.gif
    Resist3A01.gif
     
  8. Apr 7, 2016 #7
  9. Apr 7, 2016 #8

    berkeman

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    Staff: Mentor

    Correct. So for the first situation with the 4 Ohm resistor outside (call it R2), what value of R1 do you need to get Vout = 1V when Vin = 2V?
     
  10. Apr 7, 2016 #9
    this gave 4 ohms
     
  11. Apr 7, 2016 #10

    berkeman

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    Staff: Mentor

    Correct! Now use that value for the internal resistance and change the external 4 Ohm resistor to 12 Ohms. Vin is still 2V from the voltage source. What do you get for Vout now? And that just happens to be one of the available multiple-choice answers... :smile:
     
  12. Apr 7, 2016 #11
    Thank you indeed mentor, I got the answer (1.5V)
    but is there any easy way to tackle this question as it is only mark and this is only my first year in the course.
     
  13. Apr 7, 2016 #12

    berkeman

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    Staff: Mentor

    That's the main way to work on these types of questions. The only speed-up was that the first external 4 Ohm resistor cut the source voltage in half to 1V. To do that, the internal resistance had to be equal to the external resistance of 4 Ohms. So that saves you having to use the voltage divider equation. For the second part, you pretty much have to use the voltage divider equation.

    Hang in there -- it gets easier with more practice. :smile:
     
  14. Apr 7, 2016 #13
    Thank you mentor
     
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