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[Long Post] Different batteries - lamp problem

  1. Sep 16, 2014 #1

    Rectifier

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    Gold Member

    Hey there! Can somone please help me with this problem?

    The problem
    Translated from Swedish (sorry for possible grammatical errors):
    L(from Europe) travels on the Trans-Siberian railway from Moscow to Vladivostok. His lamp battery dies when he reaches Irkutsk. He buys a Russian battery and that battery works as good as the European battery he had. After closer inspection of the battery, he finds that it is labeled 3,7 V. But he doesnt understand how it could work as good as his European one wich was labeled 4,5 V.

    He asked a Russian engineer that he met on the train and he explained that the Russian battery uses GOST classification. This means that 3,7 is the voltage across a 10 Ohm resistor that is attached to the battery and the voltage of the European battery describes the voltage of the source and does not say anything about the internal resistance.

    What is the internal resistance of the Russian battery that L bought?

    The attempt at a solution
    I started with drawing a picture to represent the information given in the problem:
    X2Tdp1O.png
    Russian battery on the left and European battery on the right.

    We can draw following conclusion from the picture:
    [itex] V_s=4.5 [/itex] and we want to know what [itex]R_i[/itex] is.

    One more important thing to notice is that the battery worked as good as the old one. Which means that the lamps brightness was the same in both cases. Since the lamp has a fixed resistance value (we can call it [itex]R_L[/itex]) and the voltage-drop ([itex]V_x[/itex]) was the same in both cases (when he had the 4,5V battery and the Russian one). Since resistance is fixed and voltage-drop is the same - the current through the lamp (and circuit) must be the same too.

    These statements lead to following equations:

    [itex]V=RI \\ I=\frac{4,5}{R} \\ I=\frac{4,5}{R_L+R_a} [/itex]

    for the other battery it is

    [itex] I=\frac{V_o}{R} \\ I=\frac{V_o}{R_L+R_i} [/itex]

    Inserting one equation into another gives us

    [itex] \frac{4,5}{R_L+R_a} = \frac{V_o}{R_L+R_i} \\ \frac{4,5(R_L+R_i)}{R_L+R_a} = V_o [/itex]

    I have a few unknown variables here. In the next step I write an equation for the top picture of the Russian battery in terms of the voltage source.

    [itex] 3,7=\frac{10}{10+R_i}V_o \\ \frac{3,7(10+R_i)}{10}=V_o [/itex]

    After inserting one equation into the other, we get :

    [itex] \frac{3,7(10+R_i)}{10} = \frac{4,5(R_L+R_i)}{R_L+R_a} [/itex]

    And here where I get stuck :uhh: . I have got several unknown values left and I have no idea how to calculate them.
    I have been sitting with this problem for two days now. I would really appreciate if somone could help me.
     
    Last edited: Sep 16, 2014
  2. jcsd
  3. Sep 16, 2014 #2

    NascentOxygen

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    With one battery working as well as the other, it follows you'll have to say Vo = Vs, and Ri = Ra.

    In other words, the batteries can be regarded as identical. I can't see any other way.
     
  4. Sep 16, 2014 #3

    CWatters

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    I agree. Treat both batteries as 4.5V when measured open circuit.
     
  5. Sep 16, 2014 #4

    Rectifier

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    :thumbs:
    Thank you for your comments! I really appreciate that you took your time to help me.

    Exactly, I was thinking of this too at some point but I dont have a good way to prove it. The only way I can prove it, is by looking at the last formula.

    [itex] \frac{3,7(10+R_i)}{10} = \frac{4,5(R_L+R_i)}{R_L+R_a} [/itex]

    We have no value for RL given, thus it must be canceled out.

    [itex]\frac{(R_L+R_i)}{R_L+R_a} [/itex]

    and the only way of doing it is by saying that Ri = Ra.

    To be honest, I am not convinced by my own conclusion :uhh:
     
  6. Sep 16, 2014 #5

    NascentOxygen

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    It's much simpler to establish them as identical right at the start, and then deal with just the one circuit. You did it the hard way.
     
  7. Sep 16, 2014 #6

    andrevdh

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    Note that the current for the two batteries is the same. This enables you to solve for the internal resistance of the European battery and the current.
     
  8. Sep 16, 2014 #7

    Rectifier

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    That is the thing I dont understand here. How do I do that and why? I think that I am missing some small but crucial bit of theory behind circuits probably.

    I know that the current is the same but we have the internal resistance left. Its possible that the internal resistance is for some reason very is small(although very unlikely) [itex]\rightarrow[/itex] almost 0V drop over the internal resistance [itex]\rightarrow[/itex] [itex] V_o = 3,7 [/itex]

    If I try to apply KVL without assuming that the resistance i am looking for is very small, it gives me:
    [itex]V_o=V_{R_i}+V_x \\ 4.5=V_{R_a}+V_x[/itex]
    I is the same wich gives me
    [itex]V_o=R_iI+R_LI \\ 4.5=R_aI+R_LI \\ \frac{V_o}{R_i+R_L}=I \\ \frac{4.5}{R_a+R_L}=I \\ \frac{V_o}{R_i+R_L}=\frac{4.5}{R_a+R_L} \\ \frac{R_a+R_L}{R_i+R_L}=\frac{4.5}{V_o}[/itex]
    I keep getting back to the similar formula all the time...

    Sorry for being wierd here but my mind is a total mess right now.
    Maybe there is a smarter way of doing this. Maybe a worth to mension is that this problem was in the chapter about Thévenin and Norton Equivalents.
     
  9. Sep 16, 2014 #8

    Rectifier

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    Yeah.


    [itex] \frac{4.5}{R_a+R_L}=I [/itex]

    This is what I get using KVL from what you wrote. If I write a similar one for the other I get [itex] \frac{R_a+R_L}{R_i+R_L}=\frac{4.5}{V_o}[/itex] in the end.
     
  10. Sep 16, 2014 #9

    NascentOxygen

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    But you won't assume internal R is negligible, because you know it's causing 0.8V loss. A 10Ω load causes the 4.5V to drop to 3.7V, so write

    [itex]V_o=V_{R_i} + V_x \\ 4.5=I.R_a + 3.7[/itex]

    and you know what I is under these conditions, don't you.
     
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