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Homework Help: Voltage divider with added resistance

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure.


    2. Relevant equations
    N/A.


    3. The attempt at a solution

    First I did KVL in the large loop and the small loop so,

    [tex]KVL_{large}: i_2R_1 + (i_1+i_2)R_2 - V_s = 0[/tex]

    [tex]KVL_{small}: i_1R_m - i_2R_1 = 0[/tex]

    Using KVL of the small loop one can say the following,

    [tex]i_1 = \frac{i_2R_1}{R_m}[/tex]

    Plugging that into the KVL equation of the larger loop and simplify I found that,

    [tex] i_2 = \frac{V_s}{R_1 + \frac{R_1R_2}{R_m} + R_2}[/tex]

    and for [tex]i_1[/tex] simply plug and chug.

    Now this is a pretty messy equation for the current I2 which leads me to believe I've done something wrong... Is this the case or am I just worrying myself over nothing?

    If so, what would be the better approach to solving this problem?

    Thanks again!
     

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  3. Sep 13, 2010 #2

    berkeman

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    I personally prefer to use KCL equations, but that's up to personal preference.

    Your equation looks very reasonable (I didn't check the detailed math). Look at the limits as Rm = 0 (short circuit) and Rm = infinity (open circuit). Your equation appears to give the right values. What about the test case when Rm = R1?
     
  4. Sep 13, 2010 #3
    I'm not entirely sure what you're getting at here. If Rm = R1 then those resistors in parallel will become a R1/2 in series with R2, and we have ourselves another simple voltage divider.

    How can I use this to further check my answer?
     
    Last edited: Sep 13, 2010
  5. Sep 13, 2010 #4

    cepheid

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    I prefer using KCL as well. Look at the node that is at voltage V2'. The current going into this node has to be equal to the current coming out of it. The current coming out of it is just V2' / R2. The current going into it is just I1 + I2, where I2 = (V1' - V2') / R1 and I1 = (V1' - V2') / Rm. Simple.

    You can use it to further check your answer by seeing if this simple voltage divider gives you the same answers for V1' and V2' as your equations for the more general case do. :rolleyes:
     
  6. Sep 13, 2010 #5

    berkeman

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    Only that I2 goes up, and by what looks like a reasonable amount. I didn't have any quantitative magic in mind. You could also probably just plug in some real values and see if the equation you've come up with gives the right answers. Or re-do the analysis with KCL -- that's different enough that it should show up any errors.
     
  7. Sep 13, 2010 #6
    I'm going to give this a shot when I get home. I'll update this thread with my results.

    I'd love it for you to come back and check my work again if possible! :wink:
     
  8. Sep 13, 2010 #7
    Okay so using KCL,

    [tex]\frac{V_2^'}{R_2} = \left( \frac{V_1^{'} - V_2^' }{R_1} \right) + \left( \frac{V_1^{'}- V_2^' }{R_m} \right)[/tex]

    I'm kinda lost as to what to do now, I need another equation right?
     
    Last edited: Sep 13, 2010
  9. Sep 13, 2010 #8

    cepheid

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    Do you? Don't you already know V1'? Furthermore, I1 and I2 are known in terms of the two voltages. So how many unknowns does that leave?
     
  10. Sep 13, 2010 #9
    Hmm... Is V1 defined as follows,

    [tex]V_{1}^{'} = V_{s} - V_{2}^{'}[/tex]

    ?

    If so, then I'd only have one unknown, [tex]V_{2}^{'}[/tex], and I could solve it!
     
  11. Sep 13, 2010 #10

    cepheid

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    Hmm? The node labelled V1' and the + terminal of the voltage source are connected by an ideal wire, are they not? Therefore, they should be at the same electric potential.
     
  12. Sep 13, 2010 #11
    Sorry, as you can tell this is where I have lots of my difficulties. I often get confused with node voltages.

    So,

    [tex]V_{1}^{'} = V_{s}[/tex]

    I'm so used to seeing voltage across an element, as soon as someone writes it on a node I get confused.

    Any ideas for some good exercises/readings/tips/suggestions to help clarify things?

    EDIT:

    After thinking it over, that wouldn't make any sense either. If V1' = Vs then when we put KVL on the larger loop we'd find that V2' = 0 but there is current passing through the R2 so is must have some voltage across it.

    What am I misunderstanding?
     
    Last edited: Sep 13, 2010
  13. Sep 13, 2010 #12

    cepheid

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    Indeed, a voltage is a difference between electric potentials, so I can see why labelling a single point as having a "voltage" might be confusing. The labelling of the node with V1' is intended as a statement that the electric potential at that node is V1'. But, you might protest that electric potential is kind of like altitude -- its value is meaningless unless if you specify a reference point, and the choice of reference point is arbitrary -- only differences in potential really matter. So what is meant by "the electric potential" at that point?

    In circuits, by convention, all potentials are measured with respect to the point that is labelled as ground. In other words, ground is the reference point -- we define the electric potential to be 0 V there. Therefore, when we say that the "voltage" at that node is V1', what we really mean is that the difference in electric potential between that node and ground is V1'. Does that alleviate your confusion?
     
  14. Sep 13, 2010 #13

    cepheid

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    How do you figure that, exactly?
     
  15. Sep 13, 2010 #14
    Your explaination definetly clears some confusion up with my understanding of node voltages, but I'm still kinda confused about what V1' is equal to. (See the EDIT in my post above).

    When you say the following,

    This makes me want to write the following voltage for V1',

    [tex]V_{1} = V_{s} - V_{2}^{'}[/tex]

    Because these are the voltages across the two endpoints or nodes of our element, and one branches down to connect to ground. It feels sort of intuitive but I'm still not entirely convinced it's correct.(We determined it wasn't previously)

    I might still be a little confused but maybe once I see how V1' is determined I'll have a better feel of what's going on.
     
  16. Sep 13, 2010 #15
    [tex]KVL_{largeloop}: V_{1}^{'} + V_{2}^{'} - V_{S} = 0[/tex]

    If,

    [tex]V_{1}^{'} = V_{S} [/tex]

    then,

    [tex] V_{2}^{'} = 0[/tex]
     
  17. Sep 13, 2010 #16

    cepheid

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    I see what you are trying to do, so let me explain further. Let's call the node that has been labelled V1', "node 1." The potential difference between node 1 and ground is V1' volts. In other words, if I take a voltmeter and touch one lead to node 1 and the other lead to ground, the voltmeter should read V1' volts.

    Now, my claim is that this potential difference between node 1 and ground is equal to Vs. If that's true, then when I "travel" from ground to node 1, no matter what route I take through the circuit, I should gain an electric potential of Vs volts. How can I travel from ground to node 1? There are two possible routes. I could go through the branch that has the voltage source. If I do so, then clearly I WILL gain an electric potential of Vs volts, since I have moved across the voltage source. This serves to illustrate that the difference in potential between node 1 and ground is equal to the difference in electric potential between the terminals of the source (since the + terminal is connected to node 1 and the - terminal is connected to ground).

    The second route I could take between ground and node 1 is across the branch with the two resistors R1 and R2 in it. When I move across this branch, I gain an electric potential equal to VR2 + VR1, where VR2 is the voltage across R2, and VR1 is the voltage across R1. Agreed? We know what these voltages across the resistors are:

    [tex] V_{R2} = V_2^\prime - 0 [/tex]

    [tex] V_{R1} = V_1^\prime - V_2^\prime [/tex]

    Agreed?

    Therefore, the total electric potential I gain by going from ground to node 1 by travelling across the two resistors is just:

    [tex] V_{R1} + V_{R2} = (V_1^\prime - V_2^\prime) + (V_2^\prime) = V_1^\prime [/tex]

    So in summary, going across the branch with the source in it, I gain an electric potential of Vs. Going across the branch with the two resistors in it, I gain an electric potential of VR1 + VR2 = V1'. Since both routes involve travelling between the same two points, the potential difference must be the same for both routes, meaning that it must be true that:

    [tex] V_s = V_{R1} + V_{R2} = V_1^\prime [/tex]

    Do you see why this equation is right and the one you wrote down is not?
     
    Last edited: Sep 13, 2010
  18. Sep 13, 2010 #17

    cepheid

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    This KVL equation is simply wrong, because the voltage across the right side of the loop is not V1 + V2. As I explained in my previous post, KVL demands that the voltage across the source is equal to the sum of the voltages across the two resistors R1 and R2, so that the KVL equation should read:

    Vs = VR1 + VR2

    I won't elaborate further, since I would just be rehashing my previous post.
     
  19. Sep 14, 2010 #18
    Thank you for the excellent post cepheid this clears up alot of confusions I had concerning node voltage. I can even refer back to this post if I ever get confused again!

    Now back to the original question.

    Since we determined that,

    [tex]V_{1}^{'} = V_{s}[/tex]

    Then we should find that,

    [tex]\frac{V_2^'}{R_2} = \left( \frac{V_s^{'} - V_2^' }{R_1} \right) + \left( \frac{V_s^{'}- V_2^' }{R_m} \right)[/tex]

    Simplifying I find that,

    [tex]V_{2}^{'} = \frac{(\frac{1}{R_1} + \frac{1}{R_m})V_{s}}{1 + (\frac{1}{R_{1}} + \frac{1}{R_m})}[/tex]

    EDIT: The part on the right of the fraction line should be the denominator in the fraction.

    How's this look?
     
    Last edited: Sep 14, 2010
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