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Electric circuits-2 sources (batteries)

  1. Apr 23, 2014 #1
    1. The problem statement, all variables and given/known data
    http://prntscr.com/3cmnbx

    Va=20V R1=3kilo ohm R2=9kilo ohm R3=6kilo ohm R4=7kilo ohm
    Vb=5V

    It asks for the electric current for every resistance(R1-R4)
    2. Relevant equations



    3. The attempt at a solution

    To be honest the thing i want most is to ask some questions rather that getting the solution.
    The questions are the following:
    a)The battery source has 0 resistance.As a result a short circuit is created and so the second battery(Vb) source goes out?
    And now we have a simple circuit that has 4 resistances in a series?And we solve with the Ix=Va/Rx???

    Or maybe the 2 electric circuits from the 2 batteries are added and then flow through resistance R2 and then it goes only through R3??(because R4 and Second battery Vb are out)

    Well i really tried but i couldnt find an answer to my questions or i thought i couldnt.Only this way i can get some help for now.And i dont know if i met the requirements you asked.If you have time,please give an analysis over this.I prefer getting to understand how the whole thing with the 2 batteries work,how the electric current goes and how it is affected by the existence of the 2 batteries.

    Thanks in advance
    Sol
     
    Last edited by a moderator: Apr 23, 2014
  2. jcsd
  3. Apr 23, 2014 #2

    gneill

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    Not sure I understand what you mean. Are you attempting to use superposition to analyze the circuit?

    If Va is shorted the resistances won't be all in series... try sketching the resulting circuit and simplify it step by step. Your diagram doesn't identify Ix or Rx. Can you explain what they are?

    If battery Vb is suppressed (replaced by a short) then the circuit is changed. R2 disappears with the short as you've stated, but you need to consider how the resistor pair R3 and R4 are connected. Can you simplify the circuit?

    Tell us what circuit theorems and analysis methods you're learning about and what it is you want to try.
     
  4. Apr 23, 2014 #3
    First of all dear Gneill i have to tell you that i really appreciate your help.
    Now let me get to the point.So far i have been taught about Kirchhoff's circuit laws and about Ohm's law.However in order to get my questions answered i searched on the net a lot and i encountered some things such as the theories of Norton and Thevenin(from which to be honest i didnt get much) and about the loop method and about node-voltage analysis.I also have no idea about the superposition.

    As asked i tried to make the circuit after the short replacements and i found it really hard imagining how it should be.However let me share with you my guesses.

    If Va is suppressed i reached somehow to the conclusion that R1 and R3 will be in series.And if we add these 2,then R5(R5=R1+R3) will be in parallel with R2 and R4.This conclusion may be silly but that is all i got

    Now if Vb is suppressed,R2 would disappear because the electric current would prefer the "road" with no resistance,right?And the resistances wont be the same with above but with the exception that we wont have R2?

    Well to be honest i doubt i got them right.

    It would be nice if you explained me how the circuit would be if we put out Va and Vb.
    Also if there are 2 "roads" a electric current has encountered.One road has resistance and the other has either a battery or an independent current source,wouldnt the electric current go only to road with the battery-indepedent current source,as there wouldnt be any resistance there?

    Again sorry for all these questions but i need to understand everything.Feel free to answer whenever you can people.
     
  5. Apr 23, 2014 #4

    gneill

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    Staff: Mentor

    Superposition is a circuit analysis method that takes advantage of the fact that circuits with passive components form linear systems, and for linear systems the effects of separate stimuli (i.e. voltage or current sources) add linearly. So you can take one source at a time, suppressing the rest, and determine the currents and voltages that it produces throughout the circuit. If you add up all the contributions of the individual sources to each of those currents and voltages then you will have the their values when all of the sources are active at the same time.

    The problem there is that R1 and R3 won't be in series. You see that where R1 and R3 join there is another connection, too. That is, R4 connects to that junction (node) also, giving another path for the current through R1 to follow.

    attachment.php?attachmentid=68986&stc=1&d=1398269815.gif

    If you identify the individual nodes of your circuit (contiguous conductor "islands" where components attach) then you can see when there are more than two components connected to a given node. I've colored the nodes in the diagram above for you to see how they are identified.

    Note that R3 and R4 are both connected to the same pair of nodes (Blue and Red). That means they are in parallel, and the current I1 will be divided between them as currents I3 and I4 as shown.

    With Vb suppressed R2 gets bypassed (shorted) as you say. Then the currents are driven by Va. R3 and R4 are still in parallel.

    attachment.php?attachmentid=68987&stc=1&d=1398270585.gif

    In both cases you can write KCL to sum the currents at the nodes and KVL around the loops to form the equations you'll need in order to solve for the individual currents.

    I'm not sure what you mean by "put out Va and Vb". It's probably a language translation issue. Can you describe what you mean in a few words?

    Current sources always drive exactly their designated currents no matter what. They will produce any voltage required in order to maintain that current. Voltage sources will produce or consume any amount of current required to maintain their designated potential difference (voltage) across the circuit where they are connected. You shouldn't think of ideal sources in terms of having resistance or not; they simply provide current or potential difference as specified.

     

    Attached Files:

  6. Apr 23, 2014 #5
    Yes sorry about that.By putting out i meant suppression.However you did answer that specific question few lines above.

    Thanks to you sir i went a step forward.Your explanation is good.However what it is said in the last paragraph seems a little strange,but i think i understand it.

    Now to get a little better i have to ask 1-2 questions as well.
    At first i would like to know if there is any information you can give me about the sources.I mean if i have more than one source in my circuit,what happens next?Does this affect my current?
    Can my current meet the negative pole of a source?
    can my current meet the negative pole of a resistance(which was first passed by an oppossite current?
    why cant 2 currents pass through each other?(with opposite directions)

    Sorry if i write it in a way that cannot be understood but this is the best i can do.
    Also i know it is tiring for you to read and i shouldnt do it but thank you again.It will be the last time because i dont want to be repetitive
     
  7. Apr 23, 2014 #6

    gneill

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    Yes, all sources can contribute to the currents. That's the idea behind Superposition. You can calculate the individual contributions of the sources. The contributions add linearly (be sure to account for the directions of currents when summing).

    Sure. Currents flow through voltage sources. In fact, voltage sources are often the sources of the currents in a circuit. The amount of current flowing out of a source must be equal to the amount flowing into its other end.

    They can. This is what you are doing when you sum the contributions of different sources using Superposition. If one of the currents happens to flow in the opposite direction to another, then one of them is designated positive and the other negative. The choice of which sign to assign to each is arbitrary, but the choice should be made ahead of time by placing an arrow on the circuit to designate the positive direction of the current, and this choice once made should not be changed in order to keep the equations and sums consistent.

    Take a look at the two drawings I gave above and note the directions indicated for the currents. Note that the currents that flow through R1 R3 and R4 as driven by the two sources have opposite directions. You would choose one of the sets to represent the positive direction for the currents.
     
  8. Apr 24, 2014 #7
    Just to make this more clear as i dont think i understood pretty well the answer.Here is a cycle i saw a lot on the net.It wasnt for homework.I can post a link to this exact one if any1 wants :P

    This is something i saw that confused me a lot due to the fact that couldnt get how the currents would go.An analysis would be great.Here it is: http://prntscr.com/3d0ci9

    So here are my questions:
    Till now i knew that the current goes from + to -(as it is generally accepted).

    1)So now that i have 2 batteries,2 currents will start?One from V1(from +) and one from V2(from +)??
    2)If i have n batteries,n currents will start and eventually add up or something?

    Now lets leave this and go to another question:

    3)Lets say a current starts from V1 and it goes from A to B.When it meats B,it cant go to E becaus e the arrow indicates that the current only goes from E to B.Right?
    4)If the current goes from B to C how can it passes the battery?I mean at the beginning we take it that the current goes only from + to -.It is something that all batteries should follow.What happens in that case?Can the current pass the other battery and go to F??

    Now some other general questions.

    I read on the net that if batteries are in series then i add them(depending on polarity).Now:
    5)If they are in parallel and they have different numbers,which will be?The one with the smaller number?I know that if they both have the same number(for example 5 volt) then the result will be 5 V.
    6)Also to be able to add batteries etc their joints must end to the same nods?


    Moreover:
    7)Is there a way to determine the number of currents i will have?
    I mean if i have n loops then i will have n currents,one for every loop?
    And if i have n nods then i will have n+1 currents?


    With these answered my remaining ones wont be many i believe.

    Thnx in advance once again
     
  9. Apr 24, 2014 #8

    gneill

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    Staff: Mentor

    That's not a realizable circuit if V1 is different from V2. You NEVER connect in parallel two ideal batteries with different voltages; it leads to (theoretically) infinite currents, and for real-life batteries, can result in fire or explosion.

    Not necessarily. It is possible for current to flow into the + terminal of a battery if it is being driven by an even higher potential difference than the battery's potential difference. After all, this is how rechargeable batteries get recharged.

    Batteries in series all share the same current (all components in series share the same current). So no, you don't necessarily have n different currents in the complete circuit.

    However, you can use superposition and consider the contributions of each source individually. But the individual contributions are not recognized as different currents; they are components of a single net current that all the components sum to.

    Keeping in mind that your example circuit is physically unrealizable (you might put resistors in branches AB and BC to make it valid),

    Right. An ideal current source defines the current that can flow in its branch, both magnitude and direction.

    Batteries pass current just fine. They will pass or generate any current required in order for them to impress their designated voltage across the circuit they are attached to. The only thing you need to know about an ideal voltage source is that it represents an enforced potential difference between its terminals.

    Batteries of different voltage in parallel will be a mess, possibly an explosion. Do not do it. If you see a circuit diagram where two different ideal voltage sources are in parallel (with no resistance between them) then the circuit is wrong. Period.

    Battery potentials add when they are stacked end-to-end.
    Battery currents add when they are connected in parallel (Exactly the same battery voltages only!).

    Loop currents are a minimum count for currents; where two loops are adjacent both loop current pass through the shared component(s) and sum to a new current.

    Node counts also give you a minimum count for currents, since there may be several parallel branches between any two nodes.

    It's probably easier simply to sketch in currents for all branches then count them. Keep in mind that series-connected components carry one current.
     
  10. Apr 24, 2014 #9
    Ok now with your help i can see some light.Even though i dont know if i am ready i will try to make an attempt solving the initial exercise i posted here.I will try doing it step by step showing my way of thinking to see what is that i didnt understand good.

    Well here is the problem again:
    http://prntscr.com/3cmnbx
    and here is the data:
    Va=20V R1=3kilo ohm R2=9kilo ohm R3=6kilo ohm R4=7kilo ohm
    Vb=5V
    and It asks for the electric current for every resistance(R1-R4)

    ok lets give a try:

    (gneill scrolls down and goes to the attached file :P)



    Ok this is my honest try.I have zero confidence about it though.Would be nice if you could tell me once again my mistakes and give the right analysis explaining your thoughts.

    As always i have another question:

    It is about an independent current source.

    1)if in a circuit there is a battery and an independent current souce both elements will "create" currents?The battery because of the potential difference and the independent current source because it is made that way(to produce current)???

    2)also if in a route because of this independent source only xmA are allowed,every other current wont go through that route but to another one?

    3)can we have a circuit with independent current sources and with no voltage and why?

    4)what i should know about the independent current source :P?About their function i mean and how it affects the circuit?


    Lol ok.I know i ask many things but well...cant really help it.I need answers to my infinite questions :P
     

    Attached Files:

  11. Apr 24, 2014 #10

    gneill

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    On the first page of your PDF you claim that R3 is "out" because the path from the right end of R3 to R4 has no resistance. This is not true. All wires are taken to have no resistance and this doesn't "cancel" any components unless the wire shorts them out (is in parallel with the component). Because you've ignored R3 the loop equations that you've written are not be correct.

    In this circuit the fact that R3 and R4 share exactly two nodes with their connections means that the two components are in parallel. They will both conduct some current.

    For a beginning student who has probably just learned about Ohm's Law, KVL, and KCL, I would suggest labeling all the currents (much like I did in my diagrams above) and then writing KVL loop equations for the three obvious loops and KCL equations for the nodes (junctions). Then solve the simultaneous equations for the individual currents. Trying to guess the paths of currents is not the way to go. Just write the loop and node equations and solve.

    attachment.php?attachmentid=69038&stc=1&d=1398380146.gif

    I will give you a helpful hint: Note that Vb connects directly across R2. What does that tell you about the current through R2 no matter what else is going on elsewhere in the circuit?

    They can both cause currents to flow.

    A voltage source's only ambition in life is to maintain its specified potential difference. It will source or sink any amount of current in order to force the terminals where it connects to the circuit to maintain that potential difference.

    A current source's only ambition is to maintain a fixed current through itself. It can produce any potential difference across itself necessary to maintain that current flow.

    So:
    Never short-circuit an ideal voltage source (infinite currents will result!)
    Never open-circuit a current source (infinite potential difference will result!)

    If I understand what you're saying, yes that's true. Current sources fix the current in their branch. To see what happens at nodes (junctions) where current sources attach, simply sum the currents at the junctions to find out (KCL).

    Sure. Short circuit a current source. Can you figure it out from there?

    All you need to know is that an independent current source always has a fixed current flowing through it.
     

    Attached Files:

  12. Apr 24, 2014 #11
    Kirchhoff's Loop rule is really all you need to solve this.

    You've got 3 loops, which means three equations, and you've also got 3 linearly independent mesh currents. Also, if you go around the loop on the right, you can immediately solve for the current flowing through that resistor.
     
  13. Apr 24, 2014 #12
    Helpful hint huh? :P
    i have some guesses but the answer to this may be something i am not aware of, even if it is obvious.
    Maybe because it is directly connected to Vb and because you said "no matter what else is going on elsewhere in the circuit",it returns to Vb,instead of splitting into the other routes as well?
    if this is actually true,why does this happen?

    So the logic here is to just "write down" the currents that pass through every branch(not trying to find them like i did) and then through the equations i will know,right?
     
  14. Apr 24, 2014 #13

    gneill

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    What is the potential difference across R2? What does Ohm's law say about the potential difference across a resistor?

    That's the "entry level" approach to KVL and KCL analysis of a circuit, yes.
     
  15. Apr 24, 2014 #14
    it is 5 volt.so I2 will be equal to 5/R2 mA?
    Do you mean that?
     
  16. Apr 24, 2014 #15

    gneill

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    Yup, if R2 is in kΩ. So that takes care of the current through R2.
     
  17. Apr 24, 2014 #16
    Well ok i just thought there was a trap or something,so you gave a hint.surely my bad :P
    Sir,first thing tomorrow i will try again to solve this exercise and hope to do it right and then go a small step further.Good night
     
  18. Apr 25, 2014 #17
    Well unfortunately i got off schedule today and didnt have time to post the answer earlier as i planned.Now before i post the answer i have to ask something to see if i did a mistake...in the circuit you posted above,does
    I1 equals to I3+I4?

    and last but not least if i found a negative current,i am not wrong am i?
     
  19. Apr 25, 2014 #18

    gneill

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    Yes.

    Not necessarily, no. A negative value may simply mean that the assumed direction of the current (the assumption concerning its direction that you made before performing the analysis) was incorrect. The magnitude of the current will be correct.
     
  20. Apr 25, 2014 #19
    ok here it is once again.ehh i dont care about the exact results.i care about the equations and the logic.

    also one question:

    Lets say i have an independent current source in a loop.If i apply kirchhoff's loop law and take the equations about the loop,does the independent current source contribute in any way?


    i forgot to say that in the exercise i found that Ib is negative
     

    Attached Files:

  21. Apr 25, 2014 #20

    gneill

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    Yes. It will fix the current in the branch that it's in. In general it can be tricky to us basic KVL when you have a current source in the circuit because you don't know to begin with what the potential drop (or rise) across the current source will be. You need to introduce a new variable to represent that unknown potential, or use a more advanced circuit analysis method (You'll probably see it in your studies soon when you cover mesh analysis. The concept is that of the "Supermesh"). The alternative is nodal analysis, which you should also cover.

    I don't think it should end up negative. Vb will be putting out about 1.66 mA to drive the current through R2, but I1 from your first loop should be slightly larger, and it goes through Vb in the assumed direction.

    I've looked at your PDF and your approach is good. There's a problem with your determination of I1 though. Re-do your reduction of the line marked (1). Something went wrong in going from
    $$15 = \frac{39 + 43}{13} I_1$$
    to
    $$I_1 = ...$$
     
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