Understanding grounding - very basic question

Discussion in 'Homework Help' started by phenohm, Nov 12, 2006.

  1. phenohm

    Thread Starter Member

    Nov 12, 2006
    Hello! I'm self-taught, not taking a class, and I am just starting out. I was hoping someone could give me a hand with these two questions I don't understand from the Socratic Electronics Project. Someone let me know if my picture upload doesn't work...

    q1 - Is V(out) not electrically common with the ground? How do you measure the voltage drop without a loop circuit? V(total)-V(pot)? Is the added resistor there in parallel with V(out)? How do you calculate a total unknown (V(out)) in parallel with a known resistor?
    q2 - In question 7, why is the ground connection in the loop on the right just after the source, pre-load? How will that circuit function? And, if it can't function, i.e. if it's shorted, how are we measuring voltage across it?

    Thanks a lot!
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  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    Greetings phenohm,

    As regards the problem concerning the potentiometer, it is a straightforward voltage divider with the some parallel resistance calculation thrown in. You can simply replace the potentiometer with two resistance whose value depends on the position of the wiper.

    Another way to tackle the problem would involve calculating the Thevenin's equivalent resistance and voltage at the wiper of the pot without the external resistor attached. If you want to take this route, I suggest you review the material in the AAC Tutorial section on Thevenin's Method.

    Since you have taken it upon yourself to learn Electronic Theory on your own, you may want to take advantage of the full range of tutorial material that is available at www.allaboutcitcuits.com.

  3. kubeek


    Sep 20, 2005
    In q7, the position of the ground in the circuit is irrelevant, it is just a common point for measuring voltage. This means that by moving the ground to any point of the circuit (still talking about one halve of the of q7 circuit separately) won´t change anything, the circuit will also function with no ground at all.

    the second part is just substracting the two results, like in no.6. Notice that voltage on point A is positive from ground and B is negative from ground, bacause in B part the ground is connected to the + pole of the battery.
  4. phenohm

    Thread Starter Member

    Nov 12, 2006
    hi guys, thanks for your responses. i've been out of town for a week or so and away from the board. i'll respond by individual question:

    q1 - i didn't understand that Vout was floating - i.e. that there wasn't more circuit that i wasn't seeing, and that it wasn't connected directly to ground. it was just an error being pretty new to reading diagrams i think.
    q2 - i understand the math involved, i was just confused about whether or not you can connect a ground to a circuit at any spot. but if again that's just another diagram issue, and that's not an actual indication that we're not asking the current to choose to move through those resistors instead of a resistance-free path to ground then i think i understand that too.

    thanks again! i was all the way through the "lessons in electronic circuits" book 1 before i started doing the "socratic" worksheets , so i feel like i generally understand some stuff that's well beyond the point where i start getting actual circuit analysis wrong. i'm catching up, slowly...