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emcrazy14 Group Title

I will be really grateful if someone can explain this to me. A battery connected in series with a resistor R of resistance 5.0 Ω. The electromotive force (e.m.f.) of the battery is 9.0 V and the internal resistance is r. The potential difference (p.d.) across the battery terminals is 6.9 V. Calculate the current in the circuit.

  • 7 months ago
  • 7 months ago

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  1. emcrazy14 Group Title
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    Now the formula I used was E=V+Vr where V= terminal p.d and Vr=p.d across battery 9=6.9+I(5) I= 0.42A But apparently the answer is incorrect. Why? @experimentX @ganeshie8

    • 6 months ago
  2. sidsiddhartha Group Title
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    but it seems correct

    • 6 months ago
  3. sidsiddhartha Group Title
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    E-ir=IR

    • 6 months ago
  4. emcrazy14 Group Title
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    No, the answer is 1.38 A.

    • 6 months ago
  5. emcrazy14 Group Title
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    I know. That's what is confusing me. The formula is right, then why is the answer incorrect?

    • 6 months ago
  6. sidsiddhartha Group Title
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    |dw:1398454706860:dw| that's the ckt is'nt it

    • 6 months ago
  7. emcrazy14 Group Title
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    Here's the diagram that's given with the question.

    • 6 months ago
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  8. emcrazy14 Group Title
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    Yes. O.o

    • 6 months ago
  9. emcrazy14 Group Title
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    They got the answer by simply putting it this way. V/R=6.9/5=1.38 A. I want to know why do it this way? :O

    • 6 months ago
  10. inkyvoyd Group Title
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    another way to look at it is that kirchoff's voltage law dictates that the sum of all voltages in a closed loop must equal zero. But because the voltage across the battery equals -6.9 (when measured in the direction of conventional current flow), the voltage across the resistor must similarly be 6.9. Thus the current in the resistor must equal 6.9/5 or 1.38

    • 6 months ago
  11. theEric Group Title
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    Just a side note, since it wasn't explicit, @inkyvoyd 's reasoning explains why the answer was as simple as 'V/R."

    • 6 months ago
  12. emcrazy14 Group Title
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    I still don't get why the formula I used cannot be applied in this case. :/

    • 6 months ago
  13. theEric Group Title
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    Oh! I'm not sure what your variables mean in \(E=V+V_r\), despite the short explanations. What I can tell from your equation, 9=6.9+I(5) is that you are saying \(9\) volts is going to be the sum of the \(6.9\) and \(I(5)\) voltages. Kirchhoff's law states that the sum of voltages as you trace around the circuit will be zero. That is the sum of all. So, you can say \(\sum\)all voltages=0 \(\Downarrow\) \(\sum\)forward voltages-\(\sum\)backward voltages = 0 Right? My notation might not be proper, but let me know if you agree.

    • 6 months ago
  14. theEric Group Title
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    Forward versus backwards is just a choice.

    • 6 months ago
  15. emcrazy14 Group Title
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    Right.

    • 6 months ago
  16. Vincent-Lyon.Fr Group Title
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    If 6.9V is the potential difference across the battery, then it is also the potential difference across the resistor. Hence current I = 6.9V/ 5Ω = 1.38 A There is no problem with that.

    • 6 months ago
  17. emcrazy14 Group Title
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    Okay, so isn't it that the total emf of the battery is equal to the p.d across the resistor and the p.d across the battery?

    • 6 months ago
  18. theEric Group Title
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    \(\sum\)forward voltages-\(\sum\)backward voltages = 0 \(\Downarrow\) \(\sum\)forward voltages=\(\sum\)backward voltages Does that make sense, too? If there's a 1V forward, then there must be 1V backward. Otherwise, there's some voltage unaccounted for. That's silly. So! The given forward voltages are \(6.9\rm V\) from the battery (which is 9V forward and something internal backwards) The given backward voltages are \(I(5\Omega)\) from the resistor. So \(\rm 6.9V=I(5\Omega)\) ------------------------------- What you were thinking was that the forward voltage is \(9\rm V\) the backward voltages are \(\rm6.9V\) \(I(5\Omega)\) The part that is incorrect is that you suggest that \(\rm6.9V\) is the backwards voltage. That's what it looks like to me. Really, it is given as the forward voltage. The backwards part of it is the difference.

    • 6 months ago
  19. theEric Group Title
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    That's my analysis of your equation! So it was just a slight lapse. You can say that \(\rm9V=(9-6.9)V+I(5)\), and I think that's how your mind wants you to think.

    • 6 months ago
  20. emcrazy14 Group Title
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    I thought 6.9V was the voltage drop in the internal resistance of the battery.

    • 6 months ago
  21. emcrazy14 Group Title
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    I think my concepts are all flawed.

    • 6 months ago
  22. theEric Group Title
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    Oh! Okay! Well that was it, then! It says that the battery (chemistry stuff) produces a \(\rm9V\) potential difference. And then it says that that the voltage across the battery is \(\rm6.9V\). So, that \(\rm6.9V\) is the voltage the battery actually contributes to the circuit when it's hooked up with that \(5\Omega\) resistor.

    • 6 months ago
  23. emcrazy14 Group Title
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    Oh okay. So that means 2.1V is lost in the battery itself, right?

    • 6 months ago
  24. theEric Group Title
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    Don't worry! You actually had the right idea!! You just misinterpreted the meaning of the \(6.9\rm V\)! If you were told that the internal resistance caused a \(6.9\rm V\) drop, then you can apply what you thought. In fact, that would lead to the Kirchhoff's law equation you made exactly. \(\rm 9V=6.9V+I(5\Omega)\) \(\huge\color{lime}{:)}\)

    • 6 months ago
  25. theEric Group Title
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    Right! 2.1V is what is lost in the battery.

    • 6 months ago
  26. emcrazy14 Group Title
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    Okay. Can you please explain what exactly is terminal p.d in context of such questions?

    • 6 months ago
  27. theEric Group Title
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    So, since we both understand that, we can make another equation, this one correct, being \(\rm9V=2.1V+I(5\Omega)\)

    • 6 months ago
  28. theEric Group Title
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    The potential difference across the terminals would be what you could actually measure with a voltmeter on the outside of the battery, if that helps.

    • 6 months ago
  29. emcrazy14 Group Title
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    Yes, I get it now. xD Thank you so much.

    • 6 months ago
  30. theEric Group Title
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    What is the potential difference from one terminal to the other. Cool! You're welcome :)

    • 6 months ago
  31. emcrazy14 Group Title
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    Okay, so here in this question is the terminal p.d 2.1V or 6.9V?

    • 6 months ago
  32. theEric Group Title
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    In this question, the potential difference between the terminals is \(\rm6.9V\). The EMF is the potential difference made by the chemical reaction. That is \(\rm9V\).

    • 6 months ago
  33. emcrazy14 Group Title
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    Okay, thank you. I'm sorry, I took so much of your time.

    • 6 months ago
  34. theEric Group Title
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    Why do I come here if not to do this? Hehe, I like this. I'm glad I can help. Don't apologize and you're welcome! :)

    • 6 months ago
  35. theEric Group Title
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    I have to go. Good timing :) Take care! :)

    • 6 months ago
  36. emcrazy14 Group Title
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    Goodbye. You too. ^_^

    • 6 months ago
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