Hello everybody

Discussion in 'General Electronics Chat' started by gepdiana, Aug 17, 2013.

  1. gepdiana

    Thread Starter New Member

    Aug 17, 2013
    Hi, Im a very noob... but interested in lot of things.. especially science things... Im from Italy so forgive my crappy english... :)
    I decided to register to this forum cause I read some of the lessons about electic circuits here... and i liked them.
    But there is a concept that I can't really understand... so I decided to ask for help...
    I read about other few similar questions.. but I didn't "like" that much the answers I read...
    Suppose I've one simple closed circuit, a battery connected to a bulb : if I measure the voltage before and after the bulb I'll get 9 volts right? But if I test with the two probes AFTER the bulb (or before) I'll get 0. Fine. I don't like this... :) I mean If I think only at tha part of the circuit I m testing (a piece of wire actually) I can see no voltage... BUT CURRENT! (the one comin out from the bulb).
    Am I totally wrong in my reasoning? (maybe it is more like a philophical question...)

    Thank you very much for your patience.

    Have a nice day!
  2. Shagas

    Active Member

    May 13, 2013
    Try reading again this . Maybe a few times so you understand it well.


    Your testing is wrong
    Voltage is a RELATIVE quality. That means that you put 1 probe on the reference voltage (or ground) or wherever , and you put your other probe on the place that you want to measure and you will get that voltage - reference (or ground) voltage.

    If you put a 9 volt battery to a light bulb then you have a complete circuit .
    If you put your probes across the battery before you put the light bulb you will see 9 volts.
    If you put it when the circuit is completed you will still see 8-9 volts (abit less due to internal resistance)
    Read the article that I linked and the ones after that and before and you will start to understand

    Also , go on youtube and write : How to measure voltage and current using multimeter .

    Or something like that , and I guarantee it will start making more sense very fast
  3. gepdiana

    Thread Starter New Member

    Aug 17, 2013
    You are absolutely right, and I think I got what you explained in your message.
    But a thing is still missing to me... I suppose:
    I supposed to put the 2 probes after the bulb, so no difference in potential so no V.
    My battery is 9v and the bulb 3 ohm. In the circuit I should have 3 Amperes.
    By ohm's law, in the piece of wire after the bulb I have 0V=0R*3I which sounds mathematically good.. but confuses me because of current without difference in potential...
    Is it because it's totally wrong trying to measure V in that way? But, anyway, does not any real wire have its own resistance?
    OR is it just because "all the potential left after the bulb is converted to current?" so: ~0 V = ~0 R* 3 A?
    I m thinking to water analogy, if the water falls from 10 meters I can measure a difference in potential anywhere... from 10 to 0 from 5 to 0 and so on...
    Sorry to bother you... but I can't really understand what I am missing.. If you think I really didn't understand anything about it, just let me know... I ll study more.
    (I already read the link you send me... but I couldn't find my answer...)

    Thank you very much.
  4. Shagas

    Active Member

    May 13, 2013
    I'm not understanding everything you are asking but i'll try :
    First of all , normal wire has resistance and when current flows through it , then there is a potential developed across ANY two points in the wire.
    Problem is that the resistance of the wire is very small and therefore even for large currents the potential developed will be small and difficult to measure with the (probably cheap and crappy) multimeter that you are using .

    one of the basic laws: Your total emf (in this case the 9 volts battery voltage) = The sum of the potential drops (in this case the potential drop across the light bulb + the wires + the internal resistance of the battery )

    Second : You are using a 9 volt battery and trying to get 3 amps out of it .
    Not going to happen.....
    The battery has internal resistance so if you try to pull 3 amps out of it then the 9 volts will drop to very low voltages and the battery will start heating up .
    use a light bulb which has a much higher resistance .
    Also how do you know that the bulb has 3 ohms?
    Did you measure it with a multimeter? If yes then your reading is wrong . The resistance rises up to 30,40,50 ohms when it's heated up

    Yes , you probably skimmed through the articles looking for your answer and that's why you didn't find it .

    Read through them In detail and study more for a few days and you will see that the pieces of the puzzle will start coming together.
  5. gepdiana

    Thread Starter New Member

    Aug 17, 2013
    Maybe it's my english, and sometimes I notice my questions are not very clear to other people...
    I think your second answer is much closer to what I'm looking for. Agree about battery internal resistance, and all the other "details" but mine was just a "theoretical" example.
    Thank you very much for the time u spent trying to explain.

    I ll try too study abit more... :)

    Have a nice day.

    Best regards, Igor.