Voltage Divider Circuit question

Discussion in 'Homework Help' started by Curls, Jan 6, 2010.

  1. Curls

    Thread Starter New Member

    Jan 6, 2010
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    Hi all,

    I'm new to electronics and I'm trying to learn some stuff on my own by reading the courses on all about circuits and trying to solve the worksheets. 'Till so far, I managed to find and explain most things myself, but now I'm stuck on question 36 on this page (all the way at the bottom): http://www.allaboutcircuits.com/worksheets/e_divide.html

    Strange thing is: I think I found a way to make the calculations, because they work out for R4 and R3, but when I try to do the same thing for R2, I get a different result.

    So here's what I do:

    I start with the bottom resistance, R4, which isn't very hard. You've got to have a voltage of 45V and a current bleed of 5mA, so the resistance there should be 9k (correct according to the answers you can see by clicking "reveal answers"). The equivalent resistance with the bottom load connected (a load that draws 10mA of current at 45V, so which has a resistance of 4k5), would then be 3k (9k and 4k5 in parallel).
    Then I look up and take R3. Right above R3 there should be a total voltage of 100V, so the voltage over R3 should be 100V-45V (which is the voltage over R4), so 55V, correct? The equivalent resistance for R4, at 45V is 3k (as calculated above), so I calculate R3 using the voltage divider formula: E(R3)=Etot*R3/Rtot, Rtot being nothing more than R3+R4, so R3+3k. With E(R3) being 55V and Etot being 100V, a little bit of algebra gives me 45R=300, so R3=3,67k... which is also correct according to the anwers.
    So to me this way of working seems ok and I go on with the next resitance, R2. Total voltage over R2, R3 and R4 should be 320V, voltage over R3 and R4 is 100V, so voltage over R2 should be 220V. I already calculated the equivalent resistance at R4 (3k). The equivalent resistance at R3 (with a load connected that draws 5mA of current at 100V, so which has a resistance of 20k) would then be 3.1k (20k and 3k67 resistance in parallel). SO, total resistance of R3 and R4 is 3k+3k1=6k1, with a 100V over it. Now I can calculate the resistance at R2, using again the voltage divider formule: 220V=320V*R2/6k1+R2, and when I calculate this, I get 1342=100R2, so R2=13,42... While according to the answers, R2 should be 11k...

    Now, this could be a completely wrong way of thinking and calculating (I tried to start the calculations at R1, but couldn't find a way to do it), BUT since it worked for R3 and it gave me the exact result, I can't figure out what I'm doing wrong to calculate resistance at R2... I do the exact same thing (I think) as I did for R3, but the answer is incorrect.

    Can someone help me out here? :)
    Thanks for your time!

    S.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Hmm,
    R1=U/I=(450V-320V)/(20mA+5mA+10mA+5mA)=130V/40mA=3.25K
    Don't forget about I Kirchhoff law
     
  3. Curls

    Thread Starter New Member

    Jan 6, 2010
    18
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    Ok, makes sense :) But that doesn't of course explain what's wrong in my calculations? Thanks anyway, this seems like a more logical way to solve the circuit... But still curious though!
     
  4. thyristor

    Active Member

    Dec 27, 2009
    94
    0
    You've made an error in that you have taken the 20K resistance (100v @ 5mA) as being across the 3.67K resistor (R3) when actually all the loads are to ground, so the 20K resistor is across the total of R3 and the bottom parallel combination which is 3K.

    So the 20K resistance is in parallel with (3 + 3.67)K which gives 5K.

    It doesn't explicitly state that the loads are to ground but the implication is that they must be otherwise we coudn't get 450v across the top load.

    Otherwise you have the right idea.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    1,097
    But, you couldn't find more complicated way to solve this :D
     
  6. Curls

    Thread Starter New Member

    Jan 6, 2010
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    Allright! Thanks a lot! :)
     
  7. Curls

    Thread Starter New Member

    Jan 6, 2010
    18
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    Yeah well, since the whole chapter and all the exercises where completely and only about voltage and resistors, it didn't occur to me I could also involve current in a better way :) Thanks anyhow for the tip!
     
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