1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation with a voltage divider

  1. Sep 12, 2007 #1
    The problem seems quite easy, and it probably is, however i'm having trouble understanding how I can put up an equation to solve this at all considering the examples in the book are like ten thousand times easier than the equation we're given here:

    [​IMG]

    I am supposed to find v0, the only problem is, I really don't know what v0 is. The notion used at v0 is different from anything i've seen (And this seems to be the only time it's used in the entire book). If I were to have a guess I'd say it's supposed to indicate there's some kind of voltage divider. However, if that is the case, I can't figure out how to solve the equation considering all the examples in the book only have an R1 and an R2.

    What I would really like is just a hint to what i'm supposed to do to solve this circuit. I've tried switching around the circuits to make this look more like your standard voltage divider, but to no avail. I've also tried kirkhoff's laws just for the kick of it but there simply doesn't seem to be enough information to go by. So i'm left rather stumped.

    In general my biggest problem at the moment is defining v0.

    Any help would be greatly appreciated, as i'm sincerely stuck here.
     
    Last edited: Sep 12, 2007
  2. jcsd
  3. Sep 12, 2007 #2
    Ugh.. this problem would be difficult with voltage divider, did you learn how to do Nodal Analysis? Because this problem would be surprisingly easy if you did. But I'm assuming you haven't.

    Is the value given for that voltage supply?

    Well at any rate, I think the way to tackle this problem is to write KVL equations, and whatever voltage you get on the bottom-right loop, is the answer.
     
    Last edited: Sep 12, 2007
  4. Sep 12, 2007 #3
    I do know Nodal Analysis, the problem is this excersise is presented in the chapter before the one you learn nodal analysis, and hence it's obviously supposed to be solved without it. (For that matter I tried solving it with nodal analysis and it didn't help much. I put the bottom node as the reference node, and ended up with 3 essential nodes and 4 unknowns (v1, v2, v3 and v0), however, since the two currents on the left and right-hand side of v0 are, as I see it, supposed to be equal, you can turn that into another equation, it still got way messy, though).

    I did try putting up KVL and KCL equations (and use of ohms law) and ended up with 6 different equations with 6 unknowns, but it just got way way too messy to solve.

    I still haven't completely understood what v0 is supposed to symbolize though. It's not a voltage source or a current source, or for that matter a resistance, what is it?
    Anywho thanks for the comment, it was very helpful, i'm probably just being a bit stupid here but i'm pretty much 100% new to this and this is among the first excercises they gave us.
     
  5. Sep 13, 2007 #4

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    If V0 is just the difference between the two points marked + and - (i.e. what you would measure if you connected an ideal voltmeter between those points) the question gets a lot easier. There is no current through the 2K and 4K resistors, for a start.

    So you have an 18ma current source driving 25k and 15k resistors in parallel...
     
  6. Sep 13, 2007 #5
    Aah so that's what v0 is, it's just an opening in the circuit, which tells us that no current flows through the 2k and 4k resistors. That helps A LOT to know, I was really churning my noggin at that. I was treating v0 as a voltage supply, for some reason, so I got another unknown.

    I got the correct answer by using current division with the paralell-couplings to find what the drop in voltage was, then I could calculate the current and go back to the original circuit to check what the voltage drop across v0 was in the bottom right loop and it became correct.

    Thanks A LOT for the help! It was incredibly appreciated!
     
  7. Sep 14, 2007 #6
    Can you tell me what you got?

    I got Vout = 162.165V

    Seems really high power, but an ideal current source of 18 mA going through a high resistance load is going to result in high voltage i guess.
     
  8. Sep 15, 2007 #7
    The equivalent impedance of the circuit is
    [tex]R_{equiv}={10}^3\cdot\left(\frac{1}{10+15}+\frac{1}{3+12}\right)^{-1}=9.375\cdot{10}^3\ \Omega[/tex]

    The voltage of the voltage source, in order to get 18mA flowing, is
    [tex]U=I\cdot R_{equiv}=18\ \mathrm{A}\cdot9.375\ \Omega\cdot{10}^{-3+3}=168.75\ \mathrm{V}[/tex]
     
    Last edited by a moderator: Sep 15, 2007
  9. Sep 15, 2007 #8
    So then, to get Vout, the voltage on the Vout nodes is

    [tex]V_+=168.75\ \mathrm{V}\cdot\frac{15}{10+15}=101.25\ \mathrm{V}[/tex]
    [tex]V_-=168.75\ \mathrm{V}\cdot\frac{12}{3+12}=135\ \mathrm{V}[/tex]

    Then,

    [tex]V_{out}=101.25\ \mathrm{V}-135\ \mathrm{V}=-33.75\ \mathrm{V}[/tex]
     
  10. Sep 17, 2007 #9
    hmm.. i reworked it, and basically got the same values, except I was using the wrong direction for one of the currents, and got a much different answer of Vout = 7.714V. The thing is I used the current divider method last time, and this time the KVL mesh analysis and got the same currents and voltages, but I did the current direction right. So how do you end up with -33.75V? I don't quite follow your analysis.


    ugh, nevermind I used a wrong resistor value both times >.<

    I put in the new value into my equations, and now I get POSITIVE 33.75. Maybe I am using the wrong convention.
     
    Last edited: Sep 17, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Equation with a voltage divider
  1. Voltage divider (Replies: 11)

  2. Voltage Divider (Replies: 2)

Loading...