Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How correct are these concepts about circuits?

  1. Oct 26, 2015 #1
    Untitled.png

    first of all sorry for bad drawing, is all of the following correct :
    * number 1 : the current will flow normally
    * number 2 : the voltage source will not give current but Vb-Va = V of the voltage source (10 for example)
    * number 3 : no current will flow
    * number 4 : the voltage source will be considered as a wire Va-Vb=0
     
  2. jcsd
  3. Oct 26, 2015 #2

    berkeman

    User Avatar

    Staff: Mentor

    Looks correct to me. :smile:

    (keep in mind that shorting out a voltage source like in #4 is generally a bad thing...)
     
  4. Oct 26, 2015 #3
    I am not sure what statement number 4 means. If the wire is an ideal wire, with zero resistance, that is the end of the voltage source, and if the wire has sufficient resistance not to damage the voltage source, then statement number 4 is wrong.
     
  5. Oct 26, 2015 #4

    phinds

    User Avatar
    Gold Member
    2016 Award

    Couldn't have said it better myself, except to add that in the first condition, the wire is ideal, it still doesn't really make sense to say "the voltage source will be considered as a wire" but rather to say "this is an impossible situation"
     
  6. Oct 27, 2015 #5
    yes i am talking about ideal wires
    the voltage of the source is just v and we conclude since it is shorted that v is 0 therefore there is no battery primarily and there is just wire, it is like we are testing for the possibility of existence of a battery under this situation, but if we know in advance the existence of a battery with a specific voltage (10 for example), this voltage rise will not meet the required voltage drop to make the net voltage 0 so it may explode producing an open circuit not a wire
     
  7. Oct 27, 2015 #6

    phinds

    User Avatar
    Gold Member
    2016 Award

    I disagree and stand by my original statement.

    Your png does not load

    EDIT: I think the problem is that you do not understand the normal conventions of drawing circuits. Your #4 is just not valid. You asked a question but seem to want to argue about the answer in this case.
     
    Last edited: Oct 27, 2015
  8. Oct 27, 2015 #7

    berkeman

    User Avatar

    Staff: Mentor

    An ideal voltage source has zero output resistance, so shorting it results in infinite current. A real voltage source has a finite output resistance, so shorting it results in a large (and potentially destructive) current. Please do not ever do this in real life.
     
  9. Oct 27, 2015 #8
    Mohamed el teir,
    i was looking at your circuits and i do not understand the logic of the four drawings-1. you have shortcircuited perhaps the cell/which we do not do or forbid people in the lab- it may lead to burns/accidents.
    if one wants to draw a circuit he puts up a load- a bulb and puts in an ammeter and a voltmeter as by looking at the source and wires you can not say which way the current is moving and what is the state of voltage source. so if you have a multimeter you can say definetly about the current etc - so my request will be to pick up a school science text book a nd draw electrical circuits by their 'rule' otherwise people will be just guessing what your aim is!
     
  10. Oct 27, 2015 #9
    sorry the attachment was by mistake, what do you think will happen if a battery is shorted other than damage ? or you think that the battery primarily can't be shorted ?

    yes i agree, i want to say that in case 4 when we apply KVL to the loop : v-0i=0, so we have two possibilities, that v is 0 (the volt of battery is 0 so it is just like a conducting wire not actual battery) and also i is 0 therefore it is just wire loop, and the 2nd possibility if we know there is a battery 10 v for example therefore 10-0i=0 leads to i = 10/0 = infinite (in ideal case) or 10/(very small resistance) (in real case) therefore damage occurs

    ofcourse my purpose is the implementation of these parts in larger circuits not dealing with the drawn circuits alone, in norton theorem for example to get norton current you make a short circuit instead of the resistance you are working on, if this SC is connected in parallel with current source or battery i just wanted to know the case
     
    Last edited: Oct 27, 2015
  11. Oct 27, 2015 #10

    phinds

    User Avatar
    Gold Member
    2016 Award

    You are missing the point. The assumption for such drawings unless otherwise stated, and you did not state otherwise, is that the elements are ideal. You can't have an ideal voltage source with an ideal short circuit across it. Of course you can short circuit a battery, but it would either explode or overheat enough to possibly start a fire, unless it was a low-power capable battery.

    As I said in a previous post, the problem here is that you are refusing to understand that there are conventions for such drawings.
     
  12. Oct 27, 2015 #11
    yes elements are ideal, why it is refused to short circuit ideal battery ? mathematically it would give infinite current (which translates to very large current in reality due to the very small resistance causing damage) so is it refused only by convention to short circuit an ideal battery?
    take a look at this:
    Untitled.png
    if you want the equivalent norton circuit at terminals a-b (in the 2 ohm resistor area),to get R norton you will remove the 2 ohm resistor, put SC instead of battery and OC instead of current source so the equivalent R will be 1 ohm connected in parallel with 0 giving 0 resistance, to get i norton you will put SC instead of 2 ohm resistor, here we see that the battery is shorted, will we say i norton is infinite ? and since v thevenin (= 10) divided by R thevenin (= R norton = 0) equals i norton we will get infinite also, is norton inapplicable here ?
     
  13. Oct 27, 2015 #12

    jbriggs444

    User Avatar
    Science Advisor

    Mathematically, there is no real-valued current that can satisfy the requirement that current times resistance = potential difference. Infinity is not a real number. If you insist on using an ideal power source in conjunction with an ideal zero resistance then mathematics does not define the resulting current at all. You are not allowed to divide by zero.
     
  14. Oct 27, 2015 #13
    so why it is commonly said that in open circuit resistance is infinite (v/i if v = 10 for ex so it will be 10/0) ? aren't the cases so close ? in open circuit infinite resistance means no current is flowing and in short circuit infinite current means no resistance is present ?
     
  15. Oct 27, 2015 #14

    phinds

    User Avatar
    Gold Member
    2016 Award

    @mohamed el teir, an ideal short on an ideal voltage source is impossible because it is an inherent contraction in terms. An ideal short has zero volts across it and an ideal voltage source has a non-zero voltage output. Do you really think you can have both at the same time?
     
  16. Oct 27, 2015 #15

    Dale

    Staff: Mentor

    Both 1 and 2 are fine. Both 3 and 4 are impossible for ideal components.
     
    Last edited: Oct 27, 2015
  17. Oct 27, 2015 #16

    phinds

    User Avatar
    Gold Member
    2016 Award

    Gads. I can't believe I overlooked that 3 is also impossible. I blame it on the fact that I just barely believe in current sources :smile:

    Good catch, Dale.
     
  18. Oct 27, 2015 #17

    jbriggs444

    User Avatar
    Science Advisor

    With no real disrepect intended, one reason is that physicists and engineers are more casual about infinities than mathematicians.
     
  19. Oct 27, 2015 #18

    Dale

    Staff: Mentor

    The problem with 3 and 4 is not about whether you want to allow infinity as a valid value. The problem is that both circuit 3 and circuit 4 are not self-consistent.

    For circuit 3 the current source says "the current is I" while the open circuit says "the current is 0". The current cannot be both I and 0, so the circuit is not self consistent.

    For circuit 4 the voltage source says "the voltage is V" while the short says "the voltage is 0". The voltage cannot be both V and 0, so the circuit is not self consistent.
     
  20. Oct 27, 2015 #19

    phinds

    User Avatar
    Gold Member
    2016 Award

    You are carrying Ohm's Law too far when you try to take it to the realm of infinity.
     
  21. Oct 27, 2015 #20

    phinds

    User Avatar
    Gold Member
    2016 Award

    This is a perfectly acceptable circuit. The current through the 1 ohm resistor is 10 amps, the current through the 2 ohm resistor is 5 amps. That's 15 amps, 3 of which is provided by the current source and 12 of which is provided by the voltage source. The current source has 10 volts across it, which is fine, and the voltage source has 12 amps through it, which is fine.

    This floundering around with specific cases really isn't the best way to learn this stuff. Clearly, you need to go back and study the basics of voltage sources and current sources. It's all very simple, it just takes a bit of getting used to.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How correct are these concepts about circuits?
Loading...