Voltage Divider Circuit with two unknowns?

Discussion in 'Homework Help' started by TripleDeuce, Sep 20, 2010.

  1. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    Analysis with window not broken:

    When the window is not broken, the voltage
    divider circuit is as shown on the right. For
    clarity and for comparison with the next case,
    we have denoted the output as vo1. Derive a
    relationship between the output vo1 and the
    unknown variables E and R. We note that
    one of the design constraints is that vo1 should
    be less than 6 V. This yields one constraint
    equation. Write this equation in the space
    below.​

    [​IMG]

    I know that the 400 Ω resistor and 200 Ω resistor are in parallel so their R_equivalent can be written as follows

    R_eq = 400(200) / (400+200) = 800/6 Ω

    Using the Voltage Divider formula: I get V_01 = E(R_eq/R+R_eq)

    but since I have two unknowns, I don't know what to do. That 6V they inform me about plays a role but I don't know how to use it.

    Substitute V_01 with 6 which would yield 6V = E(R_eq/R+R_eq)? This is also a problem I encounter while doing the second part of the problem.

    Analysis with window broken:

    When the window is broken, the voltage di-
    vider circuit simplifies to the one shown on
    the right. In this case, we have denoted the
    output as vo2. Derive a relationship between
    the output vo2 and the unknown variables E
    and R. We note that one of the design con-
    straints is that vo2 should be greater than 6 V
    but less than 24 V. This yields another con-
    straint equation. Write this equation in the
    space below.

    [​IMG]

    The resistors are connected in series so I can use the Voltage Divider formula

    V_02 = E(400Ω/R+400Ω)

    but then I am stuck again because E and R are not known.

    Select the values for E and R to satisfy the above two constraint equations, and verify that the chosen values indeed will work. Do all your work in the space below, and backside if needed.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    You may want to rerun the equation:
     
  3. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
    26
    0
    Whoops. 400/3
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    You have the initial formula correct -
    The formula is usually referred to as Product over Sum.

    What is 400 * 200?
    What is 400 + 200?
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Hi beenthere,

    Not sure what your objection relates to ...

    Req=400*200/(400+200)=400*200/600=800/6=400/3=133.333 ...

    I think the OP's maths is OK.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    TripleDeuce - this looks rather like a linear programming exercise. Are you familiar with that method?
     
  7. beenthere

    Retired Moderator

    Apr 20, 2004
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    Missed the dropped 0's.
     
  8. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    @ t_n_k

    No, I don't think I've done any linear programming before. This is for my electrical engineering class.
     
  9. Georacer

    Moderator

    Nov 25, 2009
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    Both of the constraints make two equations, suitable for calculating both quantities. But the exercise seems to ask for them to be calculated separately. Wierd...
     
  10. TripleDeuce

    Thread Starter New Member

    Sep 20, 2010
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    0
    Could you show me how to use the constraints?

    What do I do with the
    ?

    Do I substitute 6V for vo1 and then try to solve?
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The problem lacks any constraints on either E or R and therefore one can presumably calculate a conditional but infinite set of values for E and R which satisfy the vo1 & vo2 constraints. Something seems to be missing from the problem statement.
     
  12. Georacer

    Moderator

    Nov 25, 2009
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    Maybe we misunderstand the question, and in reality it just asks at first for an expression, not a solution and in the end it wants an area of values.

    With that direction in mind we solve as follows:

    You correctly extracted the two inequalities from each constraint
    \left\  \begin{array}{l} 22.2 \cdot E < R +133.33 \\<br />
 R+400<66.67 \cdot E<4 \cdot R+1600 \end{array} \right}

    Now, my skills in solving inequalities are a bit rusty, so I recommend the intuitive method. Break the above system into 3 equalities. Plot them in a graph with R and E as axis. Mark the areas that correspond to the inequalities. The area that is marked by all 3 inequalities is the one that contains all the pairs of solutions.

    Or have WolframAlpha do it for you! :p
    Inequality Plot.gif
     
  13. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Nice outcome - you have an infinity of choices!

    As Georacer's analysis indicates, you may select any pair of (E,R) values whose co-ordinates lie at any point within the permissible area.
     
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