Voltage divider loaded

Discussion in 'Homework Help' started by Random3s, Jul 30, 2014.

  1. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Hi all,

    There is something which I am simply not grasping regarding loaded voltage dividers.

    I am working through the voltage divider circuits worksheet here on "all about circuits" (linked below), question 17 has a divider circuit with 2 disconnected loads.

    I understand how the principle to calculate a divider circuit (Vo=Vi x Rn / Rt), therefore for the unloaded example its pretty easy, my issues occur for the loaded examples.

    As I see it, a ground is common, therefore load 1 and the 700ohm resistor are in parallel, likewise for load 2 and resistor 3kohm and 700ohm... however when I calculate these in parallel, complete with the load, I do not arrive at the correct answer...

    Could someone please quickly explain how its possible to arrive at the correct answer for the loaded examples?

    http://http://www.allaboutcircuits.com/worksheets/e_divide.html
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Can you show as your work?
     
  3. mitko89

    Member

    Sep 20, 2012
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    Draw the circuit in proteus. Calculate the paralel resistance and exchange the 2 resistors for one with the calculated resistance. Run the simulation and check the answer. In other words, draw the equivalent schematics.
     
  4. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    There are multiple ways to check the answer without letting the computer do it for him. I'd hate for him to start using a computer as a crutch before he has mastered the techniques himself.
     
  5. mitko89

    Member

    Sep 20, 2012
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    I never said to let the PC do it for him. This way provides just answers. I mentioned the Proteus, because it is a good way to check your calculations and extremely easy to experiment with.
     
  6. joeyd999

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    Jun 6, 2011
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    A better way, at least for someone trying to educate himself in the topic, is to learn (or discover!) different methods of analysis and cross-check his answers manually.
     
  7. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Thanks for your replies. Firstly, new to electronics, and not accustomed to simulation software... I would really like to battle through this on paper first, in order to gain a proper understanding of the principles.

    I am a little pushed for time now, but I just cracked load 1 as follows:

    Total R= 1200 + 3000 + (1/1/700 + 1/5000) = 4823

    So Vi = 35v 635ohm/4825 = 4.5v (point b)

    and

    35v 3000/4823 = 21.7v + 4.5v =26.2v (point a)

    I am not sure why this wasnt working out for me previously, I guess I am getting a little chaotic with my calculations at times and confusing myself.

    I will try load 2 and then both loads later on and post results for feedback if no one minds commenting further?

    Id appreciate any feedback reference my workings, or how I can simplify or streamline this process?

    cheers
     
  8. shteii01

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    Feb 19, 2010
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    The link does not open. Is the site effed up or the link?
     
  9. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    This is the correct link:
    http://www.allaboutcircuits.com/worksheets/e_divide.html

    About half way down the page is question 17.


    In general when a load to ground is placed across the lower resistor in the voltage divider you can remove the lower resistor and replace it with a 'new' resistor made up of the old resistor in parallel with the load resistor.

    The formula for parallel resistors is:
    Rp=1/((1/R1)+(1/R2))

    or a little more clearly written:
    Rp=[1/R1+1/R2]^-1

    and in simpler form for just two resistors:
    Rp=R1*R2/(R1+R2)

    Note the one with all the reciprocals is easy to mess up when using a calculator so this last one works better. The fastest way is probably like this:
    1. Calculate R1+R2.
    2. Take the reciprocal of that (1/x).
    3. Multiply the result times R1.
    4. Multiple the result by R2.
    5. The Rp shows up on the calculator, done.

    If you have a second load added later in parallel then you have to do the same thing again. For Q17 however there is another resistor in series with the newly calculated Rp so you have to add that to Rp first before you apply this procedure a second time for the second load.
     
  10. Jony130

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    You did some small error in calculations.
    Rz = 1/(1/700 + 1/5000) = 614.035Ω

    So Rtot = 1200 + 3000 + 614 = 4814Ω

    And VB = 35V * 614/4814 = 4.464V
    VA = 35V * (3000 + 614)/4814 = 26.27V
     
  11. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    http://www.allaboutcircuits.com/worksheets/e_divide.html

    The OP's link had too many http://

    I did the ol' standby of

    Va = (35V * 700) / (1200 + 3000 + 700) and Vb = (35V * (3000 + 700)) / (1200 + 3000 + 700)

    Then I expanded that formula to account for the parallel resistors when needed.

    It would be nice if all the calculations specified to a significant digit or scientific notation to two decimal places.
     
    Last edited: Jul 30, 2014
  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Here is what LTSpice shows. Note the way switches are implemented.
     
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  13. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Hi guys,

    Thanks for all of the responses so far.

    I calculated the same questions for load 2 and both loads, with the following results.

    Load 2:

    Va = 35 * (1/ 1/700 + 1/3000 + 1/ 5000) / (1/ 1/700 + 1/3000 + 1/ 5000) + 1200

    = 35 * 2126 / 3326 = 22.4V

    (this is where my problem starts!)

    Vb= 35 * (1 / 1/700 + 1/5000) / (1/ 1/700 + 1/3000 + 1/ 5000) + 1200

    = 35 * 614 / 3326 = 6.4V

    I know the above is incorrect, what have I included which shouldn't be there?

    Both loads:

    I have tried varying methods to get to the correct voltage output.... my latest attempt...


    Va = 35 * (1 / 1/3000 + 1/ 5000) / (1 / 1/3000 + 1/ 5000) + 1200

    35 * 1875 / 3075 = 21.3V

    Vb = 35 * 614 / 3326 (Rtot) = 6.4V

    I am uncertain how to deal with the circuit when both loads are on. I dont have simulations software currently (maybe someone can recommend a freebie version?).. if both loads are on, does that mean that 2 parallel circuits are created 1 steming from a and one from b? If so how do you deal with a parallel within a parallel?

    Feedback appreciated.
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    There a many ways to solve this circuit. But let as try to use voltage divider equation only.

    For Load 2 and Va we have this situation

    [​IMG]

    So

    Va = Vsup * Rt/(Rn + Rt)
    and

    Rt = (R2 + R3)||RL2

    Now VB

    [​IMG]


    And Vb = VA * R3/(R2 + R3) = 22.4 * 0.7/3.7 = 4.23V


    Try the same approach for case with both loads

    [​IMG]
     
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  15. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Jony thanks for your reply, much appreciated.

    Everything adds up now... but why?

    I understand the principles behind Va at load 2, this makes sense.

    But the Vb result doesn't make sense, considering what I have so far learned regarding parallel circuits.

    Vb = VA * R3/(R2 + R3) = 22.4 * 0.7/3.7 = 4.23V

    Ok so the Voltage has been reduced to the output at Va. But why is the net resistance only including R2 and R3... and why isn't R3 in Parallel with the load?

    I dont understand the concept changing here.
     
  16. Jony130

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    Feb 17, 2009
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    Because R3 isn't parallel with the R_load_2

    (R2 + R3) is in parallel with R_load_2

    Also don't forget that in a parallel circuit the voltage is the same for all elements.
    This means that Va voltage is present across R_load_2 and also across (R2 + R3).
    http://forum.allaboutcircuits.com/showthread.php?p=537140#post537140
    http://forum.allaboutcircuits.com/showthread.php?t=66769
     
  17. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Jony, thanks for your patience, and assistance.

    I have the following for load 1 and 2 together:

    Rtot= 1200+2097 (equiv R2+R3) = 3297

    Va= 35 * 2097/3297 = 22.3

    Vb= 22.3 614/3700=3.7

    So far so good.. but I am still not fully understanding what's happening here....and why..

    1: Firstly going back to the previous example RL2 on, the answer (provided by your good self) was:

    Vb = VA * R3/(R2 + R3) = 22.4 * 0.7/3.7 = 4.23V

    But the diagram shows that R2 and R3 are in parallel with RL2...leading me to believe that I should use the equivalent resistance.

    Vb= VA * R3/(R2 + R3) = 22.4 * 614/2126 = 6.4V

    I know that the above is incorrect, but I am just trying to show you along my lines of thinking.... what am I lacking here, an understanding of how to analyze a circuit... a misconception of the appropriate laws? I dont know myself, I am really trying to grasp these (basic) concepts, but it just doesnt seem to be coming home..

    I can calculate the circuits following your lead, no problem... but i am afraid that when I get off of the beaten track, and come across a real world example, I wont have a clue how to tackle it.

    My questions is pretty much the same for RL1 and RL2 together..

    Vb is 22.3 * 614/3700

    But why is the total R 3700... why not the equivalent of 3700, 2126? seeing as they are in parallel with RL 1 and RL 2 respectively? or even the Equiv of R2 + equiv of R3 ... 1875+614?

    I hope this isnt too confusing, I guess my mind is kinda confused right now!
     
  18. Jony130

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    (R2 + R3) are in parallel with RL2. Yes?
    And this means that the voltage drop across (R2 + R3) must be equal RL2.

    [​IMG]

    VA = V1 = V2 (KCL) And because of this VB voltage is only determined by VA and R2, R3 voltage divider.
    And don't forget that RL2 don't steal any current from VB node. So how can RL2 has any effect on VB voltage?
    RL2 will steal some current only from VA node. And this is why RL2 will affect VA voltage.
     
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  19. MrAl

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    Jun 17, 2014
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    Hi,


    It appears that you are just having a hard time learning how to lump circuit elements together.

    You already know how to compute resistors in series, and resistors in parallel, so you're half way there already.

    What you should do to start with is once you've calculated two resistors in parallel, REPLACE those two resistors with a single resistor with the computed value and REDRAW the circuit with the new resistor only.
    To redraw usually you can just erase one resistor and relabel the other resistor with the new computed value.
    You can use any drawing program, like Paint.

    I suggest redrawing because you seem to be losing track of what you are doing once you combine a few things. By redrawing you will be starting with a new circuit each time but each new circuit will be simpler than the last one. Until you get used to doing these problem you should use this REPLACE and REDRAW method.
    Try it and see how much easier it makes life.
     
  20. Random3s

    Thread Starter Member

    Jul 30, 2014
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    Ok, but what about using equiv resistances?

    why do we use the 700Ω and 3700Ω resistance values in this formula...? The ACTUAL resistance of R2+R3 in parallel with RL2 must be 2126Ω... this is the part which is confusing me the most...its as if we are calculating a resistance in series instead of in parallel...?

    I guess its the placement of "B" in this case, as it indicates that a meter reading is being taken from within the parallel circuit...

    Thanks again for taking the time to tutor me, this forum is great!
     
    Last edited: Aug 1, 2014
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