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Internal resistance of each cell+energy dissipated in 1 min

  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A bulb is used in a torch which is powered by two identical cells in series each of EMF 1.5 V. The bulb then dissipates power at the rate of 625 mW and the PD across the bulb is 2.5 V. Calculate (i) the internal reistance of each cell and (ii) the energy dissipated in each cell in one minute.

    Answers: (i) 1 Ω, (ii) 3.75 J.

    2. The attempt at a solution
    (i) I looked for current. P = V I → I = P / V = 625 * 10-3 / 2.5 = 0.25 A. I found the resistance of the lamp: P = V2 / R → R = V2 / P = 2.52 / 625 * 10-3 = 10 Ω. Then plug in E = I (r + R) → 2.5 = 0.25 (r + 10) → r = 0 Ω. I also tried (2.5 + 1.5 + 1.5) = 0.25 (r + 10) → r = 12 Ω / 3 = 4 Ω.

    (ii) W = V I t → W = (2.5 + 1.5 + 1.5) * 0.25 * 60 - 82.5 J.

    What's wrong with every part?
     
  2. jcsd
  3. Oct 4, 2016 #2

    andrevdh

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    (i) E = I(r + R) : to start with the emf of the cells are 1.5 V, so if you want to treat them as a single cell (which is maybe not a good idea) you should use 3V for their combined emf.
     
  4. Oct 4, 2016 #3

    gneill

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    upload_2016-10-4_10-47-12.png

    Just look at the portion of the circuit that comprises the batteries. You have correctly found the current flowing in the circuit: ##I = 0.25~A## . So what is the potential change across just the batteries? Start at the left end of the batteries and do a "KVL walk" until you get to the other end of the batteries. Write down the changes in potential along the way.
     
  5. Oct 4, 2016 #4
    (1.5 + 1.5) = 0.25 (r + 10) → r = 2 Ω so per one cell it is 1 Ω. Right?

    You mentioned 10 Ω on the graph, that means that my lamp resistance is also correct?

    KVL like this -1.5 V - 1.5 V = 0?

    ---

    I applied a different formula W = I2 R t = 0.252 * 1 * 60 = 3.75 J (ii). Why does this formula get the correct answer? W = V I t and W = (V2 / R) * t both of these don't.
     
  6. Oct 4, 2016 #5

    gneill

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    Yes your lamp resistance is fine.

    Note that if you only "walk" partway around a closed path that you will sum up the potential change from your starting location to your stopping point. This lets you find the potential difference between to locations in a circuit. Simply "walk" from one point to the other summing the potential changes. Let's do that for the battery assembly:

    Write out your KVL as you do the walk! Don't try to balance anything, just write the sum. The last thing you should write down is the equals sign. Fill in what it should be equal to afterwords. In this case it should be the same as the measured PD right?

    So in this case (ignoring units for clarity): ##+1.5 - I \cdot r + 1.5 - I \cdot r = PD##

    and since the given PD is 2.5 V,

    Collecting terms: ##3.0 - 2I \cdot r = 2.5##
     
  7. Oct 4, 2016 #6
    1.5 + I r + 1.5 + I r = 2.5
    1.5 + 0.25 r + 1.5 + 0.25 r = 2.5

    How do you determine the signs though?
     
  8. Oct 4, 2016 #7

    cnh1995

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    Using the "KVL walk". Imagine yourself walking along the circuit in the direction of current, starting from any point. Note the "rises in potential" and "drops in potential" along the way and write the KVL equation as "sum of all the potential changes encountered=0". You'll get the correct signs. Refer gneill's KVL equation in the earlier post.
     
  9. Oct 4, 2016 #8

    gneill

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    By noting the whether there's a potential rise or a potential fall as you "walk over" each component in the direction you're doing your KVL walk. When you walk through a voltage source it's simple: you can tell if you're walking from the - terminal to the + terminal or the reverse. For resistors, if you are walking in the same direction as the current flows then you'll see a drop in potential according to Ohm's law. If you're walking against the current, then you'll see a potential rise instead.
     
  10. Oct 4, 2016 #9
    So if we start with 1.5 V, any number or whatever we'll encounter will be minus? Like 1.5 - 0.25 r - 2.5 - 0.25 r = 2.5? (if we had 2.5 V instead of 1.5 V)
     
  11. Oct 4, 2016 #10

    cnh1995

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    RHS should be 0. Sum of all the changes in potential is equal to zero.
     
  12. Oct 4, 2016 #11
    From what you explained I got an understanding that we need to plus everything together. 1.5 + 0.25 r + 1.5 + 0.25 r + 2.5 = 0.

    Current flows in one direction on this picture:

    upload_2016-10-4_10-47-12-png.106961.png
     
  13. Oct 4, 2016 #12

    gneill

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    That's true if you walk ALL THE WAY around the loop. If you only walk partway, the partial sum is the PD between the starting location and where you end up.
     
  14. Oct 4, 2016 #13

    gneill

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    No! You are walking in the same direction as the current, so every resistor you traverse will have a potential DROP.
     
  15. Oct 4, 2016 #14

    cnh1995

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    Right. My apologies..I misread the question.
     
  16. Oct 4, 2016 #15
    I see the current on the graph in clock-wise direction. I see the batteries with their pluses towards the current. I don't see a way to determine in which direction are the resistors (r, r and 10 Ohm).
     
  17. Oct 4, 2016 #16

    gneill

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    Resistors have no direction. The change in potential follows the current direction. Where the current enters a resistor has a higher potential than where it exits.
     
  18. Oct 4, 2016 #17
    In other words just always subtract resistors?
     
  19. Oct 4, 2016 #18

    cnh1995

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    Start from the -ve terminal of the left battery, clockwise.
    First, you'll encounter a "rise" in potential, so it is +1.5V. Next, you'll encounter a "drop" in potential across the resistor, hence it is -I*r. So finally you'll get,
    +1.5V-I*r+1.5-I*r-2.5=0.
     
  20. Oct 4, 2016 #19
    Concerning (ii) I applied a different formula W = I2 R t = 0.252 * 1 * 60 = 3.75 J (ii). Why does this formula get the correct answer? W = V I t and W = (V2 / R) * t both of these don't.
     
  21. Oct 4, 2016 #20

    gneill

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    No, you must pay attention to the current direction. If you're "walking" against the current then you'll see potential rises. Sometimes you don't have a choice in a complex circuit where there are several different currents to account for along your path. You may be "walking" with the current sometimes, and against it other times.

    You pick a direction to walk the path and then as you pass resistors note if you are walking with or against the current. Set your sign accordingly.
     
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