Calculating Vout/Vin (voltage division)

Discussion in 'Homework Help' started by TsAmE, Jun 16, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    I am not sure how to attempt this question as Vin isnt given. If it was I would:

    *Work out the first Vout after the current has gone through 1kΩ
    *I would use the first Vout to work out the final Vout after the current has gone through the second 1kΩ resistor
    *For these calculations I would use the voltage division formula:
    Vout = Vin x R2 / R1 + R2.

    How would I work out Vin, in order to do the calculations above?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well assume that Vin=1V

    Or we can write nodal equation an solve unknowns.

    [​IMG]


    \frac{(Vin-Va)}{R2}=\frac{Va}{R4}+\frac{(Va-Vout)}{R2}

    \frac{(Va-Vout)}{R2}=\frac{Vout}{R1}

    But Va can be found by inspection (because we have here two voltage divider)
    First
    Va = ( R4||(R2+R1)/ ( R2 + R4||(R2+R1) ) * Vin

     Va=Vin *\frac{\frac{R4*(R2+R1)}{R1+R2+R4}}{R2+\frac{R4*(R2+R1)}{R1+R2+R4}}

    second

    Vout = Va * R1/(R2+R1)
     
    Last edited: Jun 16, 2010
  3. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    Oh ok I see. Something else I am curious about is how the various currents flow in this circuit. Would it be like this:

    *Current starts flowing at Vin (I1)
    *It then goes through R2, splits at Va (the one goes through R4 to earth (I2), and the other goes through R2 (I3) therefore I1 = I2 + I3 )
    *At the next common point, spilts again (one goes through R1 to earth (I4), and the other one goes through Vout (I5) (I3 = I4 + I5).
     
  4. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    You have to assume that no current flows out of Vout. Imagine it as a test probe point. If current flowed out and we had no way of knowing what it was, the system would be unsolvable.
     
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  5. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    Here is the Millman's Theorem analysis technique applied to the circuit.

    I have co-opted jony130's diagram to permit comparison between the two approaches.

    Begin by forming (by inspection) the Millman's Theorem equation for Vout in terms of Va:

    V_{out} \,=\, \Large \frac{\frac{V_{a}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_1}}\normalsize \, \, (1)

    V_{out} \,=\, \frac{R_1V_{a}}{R_1+R_2}\normalsize \, \, \, \, (2)

    Rearrange equation 2 to isolate Va on the left side of the equal sign:

    V_{a} \,=\, \frac{(R_1+R_2)V_{out \.}}{R_1}\normalsize \, \, (3)

    Next, form the Millman's Theorem equation (by inspection) for Va in terms of Vin:

    V_{a} \,=\, \Large \frac{\frac{V_{in}}{R_2}}{\frac{1}{R_2}+\frac{1}{R_4}+\frac{1}{(R_2+R_1)}}\normalsize \, \, \, \, (4)


    V_{a} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (5)

    Set equation (3) equal to equation (5) and solve for Vout/Vin:

    \frac{(R_1+R_2)V_{out}}{R_1} \,=\, \large \frac{R_4(R_2+R_1)V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (6)

    \frac{V_{out}}{R_1} \,=\, \large \frac{R_4V_{in}}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4}\normalsize \, \, (7)

    \frac{V_{out}}{V_{in}} \,=\, \large \frac{R_1R_4}{R_4(R_2+R_1)+R_2(R_2+R_1)+R_2R_4} \normalsize \, \, (8)

    Done.

    hgmjr
     
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