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Confusion on Independent Current Source and Independent Voltage Source

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Given the this circuit calculate the voltage through the 3amp source? attached is a jpeg of the circuit. I wasn't able to get a good score on this one question quiz since this boggled me. Is true if I put current source that it will dictate the current of the battery?



    2. Relevant equations

    Hi, I am having some trouble analyzing this circuit.

    Kirshhoff's Voltage Law.

    3. The attempt at a solution

    There is only one mesh for this circuit(square box). I have three components a 12 volt battery, a resistor and a 3amp current source. I want to find the voltage of the 3 amp current source. I found the equation for the loop (clockwise) to be 0 = 12 -i2 - (voltage of the 3 amp current source). How can I solve this with 2 unknowns. My professor said that the key to solving this was ohms law, the i through the 2 ohm resistor is 3 amps and thus the voltage for the 2ohm resistor is 6V and thus we can see that i2 = 6V so 0 = 12 -6 -(Voltage of the 3 amp current source) would then yield 18 as the voltage through the 3amp source. I know however from very basic circuits that the voltage source would produce a current so why can he just say that the 3 amp source is equal to the current produced by the battery?
     

    Attached Files:

  2. jcsd
  3. Oct 14, 2011 #2

    lewando

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    Gold Member

    For this circuit, yes.

    The job of the voltage source is to produce and maintain its specified voltage across its terminals without regard for the current going through it.

    The job of the current source is to produce and maintain its specified current through its terminals without regard for the voltage appearing across it.

    The key is to realize that if the current source is doing it's job, then 3A will be flowing through all the components of this circuit (in a counter-clockwise direction). So your I1 is known (-3A).
     
    Last edited: Oct 14, 2011
  4. Oct 14, 2011 #3

    NascentOxygen

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    Staff: Mentor

    The current source has to produce a voltage = the voltage across the resistor + voltage of the voltage source.

    An ideal voltage source can sink current as well as source it.
     
  5. Oct 14, 2011 #4
    okay, I didnt know that. Thank you lewando.Your explanation makes sense, but lets say that its just a resistor, the 2ohm, the current would then be 6 amps by ohms law but thats because there is a current coming from that battery. you see my dilemna? please explain further. What happened to that current? How do i remember that the current source would take precedence over the current source of the battery?
     
    Last edited: Oct 14, 2011
  6. Oct 14, 2011 #5

    lewando

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    Gold Member

    For a battery in series with the resistor, this is true. The voltage source is still acting like a voltage source. The thing that determines the current is the voltage of the battery and the value of the resistor. If the value of the resistor changes, so will the current--so in that sense the voltage source is providing current but it is not acting like an actual "current source".

    If your dillemma is a battery is not a current source, but can be source of current, I can see how the wording can be confusing. Remember, a battery can also be a sink of current (as NascentOxygen pointed out--like when it is being charged). What happened to the current is that a current source was installed which has the job of making sure that a fixed amount of current is going through it. For amusement purposes let's say a 0A current source was installed (also known as an open circuit) The 6A prior to the installation simply becomes 0A. This is due entirely to a change in the configuration of the circuit.

    Just remember that a battery is not a current source (per the strict definition of a current source).

    Edit: Perhaps it will be more clear to use the terms constant current source and constant voltage source. So a battery is not a constant current source, although it can be a source or sink of current, depending on the circuit in which it is installed.
     
    Last edited: Oct 14, 2011
  7. Oct 14, 2011 #6
    Thanks Lewando,

    I just finished my Electrical Engineering 210 class and I'm glad that you cleared it up for me before my quiz.You made my day, your explanation made a whole lot of sense. I asked my professor after the class and he said it all depends on the circuit your looking at.

    So is it okay for me to say only for this particular case that the current source would "normalize" the whole current going around to be circuit to 3 amps.

    I'm actually going to copy and paste this whole post into a word document for future references. Thanks a whole lot, I would have continued to wrestle with this circuit if it weren't for you.

    Edit:
    I also realized that I wouldn't be able to find the voltage source of the 2 ohm resistor using the voltage source 12 v since that voltage difference would be the whole branch 2ohm resister and amp source. So I can't use voltage divider rule. Is my logic here alright?
     
    Last edited: Oct 14, 2011
  8. Oct 14, 2011 #7

    lewando

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    You could say that. It does whatever it has to do to make 3A go through it. Since the circuit is a series circuit, all elements will have the same current.

    The voltage divider rule is used for a voltage applied across 2 resistors in series, so since the constant current source is not a resistor you can't use it.
     
  9. Oct 14, 2011 #8

    NascentOxygen

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    Staff: Mentor

    The current source sets the loop current to be 3 amps. End of story.
    "voltage source of the 2 ohm resistor"? :confused: Did you mean to type "voltage across the 2 ohm resistor"?

    Use ohms law. It's a 2 ohm resistor and has 3 amps flowing through it, so the voltage across the resistor alone is 6 volts.

    As I pointed out earlier:
    Code (Text):
    The current source has to produce a voltage =
     the voltage across the resistor + voltage of the voltage source.
     
  10. Oct 14, 2011 #9
    Okay, Thanks for the help. I got it.
     
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