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Difference between power 'supplied' and 'absorbed' circuits?

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  1. Sep 11, 2015 #1
    • Member warned to use formatting template in Homework Help sections of PF
    It is my first time learning about power in circuits. From reading my electric circuits text, I understand that when current enters the positive terminal of a circuit, power is being "absorbed" and when it enters the negative, it is "supplied". The book doesn't really go in depth on what those means and was wondering if somebody could give me a brief idea of the concept of supplied and absorbed?

    much appreciated!
     
  2. jcsd
  3. Sep 12, 2015 #2

    phinds

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    Batteries, generators, and so forth supply power and the circuits they power absorb (use) that power. I don't follow what you are saying about positive and negative terminals supplying and absorbing.
     
  4. Sep 12, 2015 #3

    NascentOxygen

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    Yes, that's about it. When current exits the positive terminal of the device, that device is supplying power to an external circuit; and when current enters a device at its positive terminal, then that device is absorbing or being supplied with power.

    You may find, after doing the calculations, that the current exiting the positive terminal has a negative value, so this means that current is actually entering that terminal, therefore the device is in fact absorbing power from some other circuit.

    Note that for rechargeable batteries, such as lead acid cells, current exits the positive terminal while the battery is discharging, but enters the positive terminal while that battery is being recharged, and this accords with your supplying vs. absorbing description.
     
  5. Sep 12, 2015 #4

    Great explanation! But just to clarify on what you said above... The reason why there is a negative value when current enters the negative terminal and leaves the positive terminal is when actually current is being SUPPLIED to it from another device of the circuit--hence the current returning back to the positive terminal? Correct me if I've misquoted you...
     
  6. Sep 12, 2015 #5

    NascentOxygen

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    The reason why your calculations may, for example, give a negative value to the current you initially assumed to be exiting the positive terminal is simply because your initial assumption (or guess) had the direction wrong. There is no need to repeat any calculations, this is a normal occurrence in circuit analysis. You assume a direction for the current, and subsequent calculations will either confirm that and produce its magnitude, or will produce a value for your current that is negative and of the correct magnitude. In the latter case, it just means that your initial assumption that current had a particular direction of flow was 180 degrees out.

    So a negative value of current leaving the positive terminal indicates current flow is into the device, so some external source is supplying it, as you say.
     
  7. Sep 13, 2015 #6

    Zondrina

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    You are referring to the head-to-head convention that engineers use to determine whether power is absorbed/supplied and whether to use a negative or positive sign in many common equations.

    In the case of power, usually we use the equation ##P = \pm IV##.

    The (+) version of this equation is chosen if the head of the current arrow meets the head of the voltage arrow (the head of the voltage being the positive end of the voltage drop). So if the (+) current meets the (+) voltage, we use the (+) sign.

    Lets say the current is now negative, and the head of the current arrow is meeting the tail of voltage drop (aka the negative end of the voltage drop). This is equal to a positive current meeting the positive head of the voltage drop. This is equal to the case outlined above, and therefore we would use the (+) version again.

    Now if we had a positive current arrow meeting the tail of the voltage drop, we would use the (-) version.

    If we had a negative current arrow meeting the head of the voltage drop, this is equal to a positive current arrow meeting the tail of the voltage drop, which again leads to the (-) version of the equation.

    This convention applies to all electrical circuit components. Did you know it was actually ##V = \pm IR## depending on which end of the current meets the resistor?
     
  8. Sep 30, 2015 #7

    CWatters

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    Try drawing a simple circuit comprising a battery and a load/resistor? Mark up the voltages and currents. Apply what the book says and you will see its correct.

    Then perhaps try other simple block circuits such as a battery charger connected to a battery.
     
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