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Circuit consists only of voltage source and current source

  1. Oct 10, 2014 #1
    • Warning! Posting template must be used for homework questions.
    This is more of a theory question than a homework question.....but could a circuit consist solely of a voltage source and a current source?

    What I mean is a very simple circuit with a voltage source of Vx and a current source of Ix, connected in series. No resistor anywhere.

    Could this work? And if so, how could one source supply energy and the other absorb energy?
     
  2. jcsd
  3. Oct 10, 2014 #2

    gneill

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    Hi needhelp171, Welcome to physics Forums.

    Be sure to use the formatting template when posting in the homework areas!

    Draw the circuit. If you apply the basic circuit laws (what are they?) are there any contradictions? If you apply the rules for calculating power delivered/absorbed (what are they?) what can you conclude?
     
  4. Oct 10, 2014 #3
    I don't know how to draw a circuit on this site....

    But when I apply the circuit laws I think there'd be a contradiction because there's no resistor to get rid of the energy (like how you can't just have a voltage source with no resistor because voltage would change without any energy being dissipated, which is a contradiction), but then my friend said the two sources (the voltage and current sources) can somehow cancel each other out or something and now I'm confused.

    I know that a voltage source delivers whatever current it has to in order to maintain its voltage....and that a current source creates whatever voltage difference it needs to in order to deliver its current.

    I'd imagine there's a contradiction when the voltage source delivers a current that doesn't match what the current source wants to supply.....but my friend said they can somehow cancel out or something and I don't get it.

    I didn't include any equations or anything because this isn't really a problem with a numerical answer, it's more something I was just wondering about. I guess a relevant equation would be:

    V = IR
    P = VI
     
  5. Oct 10, 2014 #4

    gneill

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    I imagined that you would draw the circuit on a scratchpad at your desk and ponder the circuit laws, but if you wish to show your picture you can draw a circuit on your own machine and upload it. Use the "UPLOAD A FILE" button at the bottom right of the editing pane.

    Voltage sources and current sources are not arbitrary about the current or voltage they supply in order to accomplish their goals. If the current source must provide, say 1 amp, the voltage source will happily absorb or deliver 1 amp according to the direction of the current. Similarly, if the voltage source insists that it must maintain 10 V, then the current source will happily fix its own potential difference to 10V with the requisite polarity and still insist on its own current value and direction.

    Your second equation, P = VI, is appropriate. The first one doesn't apply because there's no resistance in our imagined circuitry. You also want to throw in KVL and KCL. If there are no contradictions with those two when applied to the circuit then you can safely conclude that all is good.

    Do you know how to ascertain whether a source is supplying or absorbing power?
     
  6. Oct 10, 2014 #5
    You can resume to a simpler model and still run into some problems. Imagine a voltage source only, with terminal connected by a wire.
    If you neglect the resistance of the wire, who absorbs the energy provided by the source?

    There are cases when neglecting something becomes unreasonable.
     
  7. Oct 10, 2014 #6
    Ok I drew it out and I see what you're saying. Thanks!

    So then if the current source is positive, and it's going into the negative terminal of the voltage source, then the voltage source must be supplying energy, which then means that the current source is absorbing energy.
     
  8. Oct 10, 2014 #7

    gneill

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    Right.
     
  9. Oct 10, 2014 #8
    But I think the situation gets really sticky when you throw in a resistor.

    So let's say you have, in series, voltage source --> current source --> resistor --> original voltage source (completing the loop)

    Let the voltage source be 10v, the current be 6 amps, and the resistor be 2 ohms.

    Then, from what I see, they (the voltage and current sources) both supply energy!!!!

    1.) The voltage source takes the ground (0v) and adds 10 volts to it ==> voltage between voltage source and current source = 10v
    2.) The current source adds 2 volts to this ==> voltage between current source and resistor is 12v
    3.) The resistor consumes (6amps * 2ohms) = 12volts of energy. ===> voltage between resistor and voltage is 0v (ground, as it must be)
    4.) Back to step 1.....

    So the voltage supplies energy by increasing voltage to 10 volts.....the current supplies energy by increasing voltage from 10 to 12 volts.....12 volts consumed in resistor.

    I know this is wrong just because it's wrong, but there's nothing wrong, that I can see, with my logic....
     
  10. Oct 10, 2014 #9

    gneill

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    A word on terminology. The Volt is a unit of potential difference, not energy. Although you could say that it is equivalent to the energy per unit charge given to or taken from that charge if it traverses that potential : V = J/Coul. Volts are not something that is "consumed" (unless you're invoking literary license and rendering the statement somewhat unscientific). Now Power is something that can be consumed or produced.

    Look at each component in your circuit and state whether it is producing or consuming power, and how much. Do the powers all add to zero?

    Fig1.gif
     
  11. Oct 10, 2014 #10
    Well in my drawing I had the resistor and current source flipped from your drawing, with the current from the current source flowing directly into the resistor, though I think it doesn't matter?

    Anyways, going off your drawing, the voltage source is a supplier because it raises the voltage from 0 to 10 (so it produces power).

    The current source is a supplier because it raises the voltage from -2 to to 0 ( so it, too, produces power).

    10 + 2 - 12 = 0, so everything adds up, despite the fact that they both somehow supply energy.....
     
  12. Oct 10, 2014 #11

    gneill

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    Right, the order makes no difference.
    It is a supplier because it raises the potential in the same direction that the current is flowing. So if current is coming out of a voltage source's + terminal, it is supplying power to the circuit.
    Right. Same deal, if a current source is raising the potential in the direction of current flow then it is supplying power. So if the terminal that the current is coming out from is positive with respect to the other terminal, it's supplying power to the circuit.
    By KVL the sum of the potential changes around the loop MUST be zero, so no surprise there. But potential changes are not power consumed or produced. Add up the powers!
     
  13. Oct 10, 2014 #12
    So......I guess I should use p = vi

    Power for voltage source : 10 * 6 = 60
    Power for current source : 2 * 6 = 12
    Power for resistor: -12 * 6 = -72

    60 + 12 - 72 = 0 ........
     
  14. Oct 10, 2014 #13

    CWatters

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    I have a circuit here that's similar to this. It involves a constant current source in series with a constant voltage source. The current source is actually a NiMH battery charger and the voltage source is an NiMH rechargeable battery. Ok so neither are true ideal sources but they are close enough for this discussion. In this case the constant current source emits power and the constant voltage source absorbs it. There is some resistance in the wires and in the battery.
     
  15. Oct 10, 2014 #14
    How does the voltage source absorb power when it increases the potential difference across it?
     
  16. Oct 10, 2014 #15

    gneill

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    If the current was entering the + terminal of the voltage source then the charges will be falling through that potential difference as they pass through the component rather than being raised up to the higher potential. So the charges each "deposit" energy in the source. That is, the source is absorbing energy.
     
  17. Oct 10, 2014 #16

    CWatters

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    Gneil beat me to it.

    In the example I gave the current is indeed flowing into the +ve terminal of the voltage source/battery.

    How does it absorb energy? In the case of a battery it's stored chemically.
     
  18. Oct 10, 2014 #17
    Yes, but in our example (the one of the posted diagram), the current is entering the negative terminal, hence the source is supplying energy.....
     
  19. Oct 10, 2014 #18

    gneill

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    If the chemical reaction that supplies the energy for charge separation is designed to be reversible then the energy "undoes" the chemical changes that the battery undergoes when it's powering your iPod while you're listening to the entire discography of The Moody Blues :smile: If the reaction is not so designed then the battery gets hot through the usual resistive (Ohmic) heating and may explode :nb)

    This is why some batteries are designated as "rechargeable" and others are not.
     
  20. Oct 10, 2014 #19
    So was what I said in post #12 correct?

    I suppose I had assumed that resistors can't produce or consume power.....but they can??
     
  21. Oct 10, 2014 #20

    gneill

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    Yes what you said is true, assuming the convention that a negative power value implies absorbing power from the circuit.

    Resistors can ONLY absorb power; they turn electrical energy into heat.
     
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