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Figuring out why experimental results are different

  1. Nov 11, 2016 #1
    1. The problem statement, all variables and given/known data
    question.jpg
    Hello everyone, I am trying to figure out how to approach part c, parts i and ii
    2. Relevant equations

    OK. I know that we have the current equation in milliamps as shown above
    3. The attempt at a solution

    My problem is giving one possible explanation for the finding and explaining an implication that the answer to part i has for the predicted capacitance.

    One attempted solution of mine is that the battery has a smaller internal resistance than expected.
    This would create a greater overall battery voltage and thus, the capacitor would be charged to a greater voltage to meet the battery's voltage. That explains part1, but I do not know how that would affect the capacitance in part2.

    Could anyone please direct me and help me think of at least one (or even a few possible solutions) if possible in how to approach this problem? This would be useful to me as a learning exercise.

    Thanks in advance, and make it a great day!
     
  2. jcsd
  3. Nov 11, 2016 #2

    gneill

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    Is it possible that there's some missing information in the problem statement regarding the value of the resistor in the circuit?

    What value did you initially assume for the battery's internal resistance when you worked parts (a) and (b)?
    I'd think that the battery voltage would be measured with a very high resistance voltmeter so that essentially no current would flow in the process. The battery's internal resistance wouldn't noticeably affect the meter's reading of the EMF.

    The potential at the battery's terminals would only drop when current is flowing, causing a potential drop across the internal resistance. When no current is flowing the potential across the battery terminal would reflect that of its cell's EMF.
     
  4. Nov 13, 2016 #3
    No, I am afraid not. This is a complete question from an AP Physics E&M exam, and they make sure that their questions are accurate.

    I initially assumed that the battery's internal resistance is zero that it was an ideal battery. It is an ideal assumption that the test makers want us to make unless they say explicitly otherwise.

    OK. I see what you said there. If the battery is charging then there is obviously going to be a current going through so the charging voltage of the battery would be less than its ideal voltage (EMF).

    How would that be related to part c of the question that says that the charging voltage is GREATER than predicted?

    Thanks for the help!
     
  5. Nov 13, 2016 #4

    gneill

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    And no resistor value was trimmed off from above the resistor in the diagram when it was cropped for posting?
    Right.
    The predicted battery voltage comes from when the battery is in operation, supplying current.
     
  6. Nov 13, 2016 #5
    No. This is the whole package.

    I am confused. It would be the actual battery voltage which would be EMF - Iinternal * R that would be there. And that would make it less than the ideal battery's voltage EMF, not greater, right?
     
  7. Nov 13, 2016 #6

    gneill

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    Since my last post I did a web search and found the same question in a pdf document. The circuit diagram there showed a resistor value R = 50 kΩ. So I think the copy of the problem that you received was incorrectly trimmed. The value of the resistor is needed in order to answer the question with numerical values.
    Right. It would be less while the circuit is operating, or at least the effects on the circuit would be such that a value of battery voltage lower than its actual EMF would be predicted from the given data. When measured separately while no current is being drawn it would show a larger value (the cell's actual EMF).
     
  8. Nov 18, 2016 #7
    Yes, I now realize that I trimmed that off. My apologies.

    So, now that we know the R= 50 kilo ohms, how can I approach part c parts i and ii?
    Can you hint or direct me at some possible approaches? Thanks!
     
  9. Nov 18, 2016 #8

    gneill

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    Knowing the resistor value doesn't change or help with part c. It does allow you to make numerical calculations to answer parts a and b.

    Part c relies on understanding the effect that ignoring the battery's internal resistance had on the "prediction" of the battery's EMF, which we've been discussing in the above posts.
     
  10. Nov 27, 2016 #9
    Thank you for being willing to assist me in figuring out this problem. Unfortunately, I am still stuck, but am trying to figure the approach out.

    Yes, I realized that and was able to solve parts (a) and (b) correctly.

    So we have one possible explanation for part (c) i. that the EMF was measured with an ideal voltmeter with no current being drawn, with no internal resistance causing a voltage drop. However, in part (a), I got the predicted charging voltage (correctly) to be 260 volts by applying Kirchoff's circuit loop voltage rule and assuming that the battery's internal resistance is zero. So thus, this explanation would yield a measured charging voltage that is equivalent to the calculated voltage in part (a) and not greater.

    Right? If what I am saying is right, then I believe that this approach would not work in trying to solve both parts of part c.

    Thanks again for the help, and I am looking forward to hearing your reply!
     
  11. Nov 27, 2016 #10

    gneill

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    Yes, that's the whole point. You found the predicted value in that manner. Then the battery was removed from the circuit and its terminal voltage measured with a good voltmeter and found to be larger than that predicted value.
    No, it's the information needed for answering all of part c. Write the expression for the circuit's time constant knowing that the battery internal resistance exists.
     
  12. Nov 27, 2016 #11
    No no. I did not calculate part a accounting for the internal voltage. I assumed that the battery was ideal and I set the internal resistance to zero, as needed.
    IF and only IF I had calculated part (a) saying that the internal resistance is not zero, then I would agree with you and say that the charging voltage measured in the laboratory is greater than the one predicted in part (a). Do you see and possibly agree with what I am saying here?

    Time constant = ( C ) * (50*10^3 + RInternal)

    O.K. I am curious. Why did you mention the time constant here? Is this another line of thinking?

    Thanks again.
     
  13. Nov 27, 2016 #12

    gneill

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    No, I strongly disagree. Let's review how you found the predicted voltage. The initial current specified by the current function was Io = 5.20 mA, and the given resistor value was R = 50 kΩ. That would be the only resistance in the circuit if the battery was ideal. Then you calculated the predicted battery voltage as: Vp = Io R = 260 V. That would also be the actual potential difference across the battery terminals for the circuit for that initial current.

    Now, it turns out that the battery is not ideal and there was some internal resistance. So there must have been some potential drop across that internal resistance. Even so, the battery terminal voltage was 260 V to produce that initial current. So the battery's actual EMF must be higher than 260 V, the drop across the internal resistance lowering the external voltage to 260 V when the initial current is being drawn.

    When the battery's voltage is measured using a good voltmeter it shows the actual EMF, and it's higher than 260 V. In fact its 260 V plus the drop across the internal resistance.
    Go back to how you found the capacitor value in part (b). If the total resistance is actually greater than just R, what does that do to a value for C calculated from the observed time constant?
     
  14. Dec 4, 2016 #13
    I apologize for my late reply, but I must thank you for your willingness to help me decipher this problem.

    Yes, I see what you are now saying. Yes, the total EMF will be greater and the Rtotal will be greater due to h

    For part b, the value for C will decrease if the Rtotal will increase due to the added internal resistance. Right?

    Thanks again for the help!

    I
     
  15. Dec 4, 2016 #14

    gneill

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    Right. And you're very welcome.
     
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