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Thinking about Power as P=IV

  1. May 8, 2012 #1
    Hello Forum,

    electrical power is current times voltage.

    Current is charge per unit time, crossing an hypothetical surface: the more people cross a door per unit time the higher the current.

    Voltage is energy per unit charge: the more energy each person crossing the door, the more the power.

    So to have large power, we need lots of people crossing the door, each with a high (kinetic energy), correct?

    In some cases we have high voltage, low current, hence low power: that means few charges are moving across a surface but those few charges have high energy...

    The max power theorem tells us that in the connection between a source and a load, the max power is transferred at a certain optimal situation, when there is a large number of changes (not the largest) with large energy (but not the largest)...correct?

    All devices need power: some need high current and low voltage or vice versa...why?
    thanks,
    fisico30
     
  2. jcsd
  3. May 8, 2012 #2

    psparky

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    You car needs super low voltage ( 12 volts so you don't get shocked) and relatively high current to work.

    Blender needs low voltage and low amps.

    AC unit at your house needs slightly higher...but still low voltage with higher amps. (needs decent of amount of power to run)

    Factories motors need low voltage (480 V) you could say medium...but I wouldn't....and low to medium current.

    Power lines need super high voltage and super low current to save customers (you) money. With massive amount of power....lower current always has lower power losses.

    Quick answer enough or you want more?
     
  4. May 8, 2012 #3

    sophiecentaur

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    The relationship is very much on a par with torque, rotary speed and Power. You can choose many combinations of two variables to achieve your wanted Power. It's the practical circumstances that dictate the best choice of values.
     
  5. May 8, 2012 #4

    vk6kro

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    All devices need power: some need high current and low voltage or vice versa...why?

    It isn't just a choice of one or the other. Generally, the voltage will be decided first. This may be your mains supply voltage or a 12 V battery. Something fixed that you want to use.

    Then a device is designed to give the required result using that voltage as a power source.

    Generally, you don't decide how much power you want to use. You might select a pump that will pump a certain amount of water per minute. Then, you look at the label or data sheet and see how much current it is going to use and how much power.

    If it is going to switch off your circuit breakers or flatten a 12 volt battery in a few minutes, this is where you start to earn your salary.

    You may be able to compromise and pump the water a bit slower and use a less powerful pump.
     
  6. May 9, 2012 #5

    jim hardy

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    For any given power, available voltage sets the current that will be required.

    When the size of the wire required to carry that required current gets cumbersome, one goes to next higher voltage to reduce wire size.

    Such are the tradeoffs.

    Now that's one you need to figure out by working some examples on paper, for there's a mild subtlety to that theorem, at least the way they worded it when i learned it ca 1961 there was..

    So take say, a one volt source, and place in series with it several values of external and internal resistance. Maybe 0.1, 1.0 and 10 ohms that'd be only nine combinations.
    Tabulate how much power is dissipated internally and externally in each case.. with those numbers the arithmetic is trivial and you'll quickly see the rule..
     
  7. May 9, 2012 #6

    sophiecentaur

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    That voltage would have been selected by someone, in the first place, according to the circumstances they foresaw for the application. A good example would be in the choice of 12V OR 24V for a vehicle electrics - based on the likely size of engine and the starter motor that would be required. The poor devil who finds himself with a 24V system then has to grin and bear the fact that all his replacement electrical items will cost him an arm and a leg.
     
  8. May 9, 2012 #7

    psparky

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    The power lines are another fine example.

    Way back when they litterally needed to run millions of miles of electric cable.

    This has a cost for material and labor. Using the smallest cable possible would be the obvious choice. Small cable.....smallish current. To make the massive amount of power necessesary......a super high voltage must be used which is the case in all power lines.

    The receptacles in your house are another good example. They wanted to have good power with a voltage that wouldn't kill you. In USA they decided on 120 volts.....with a 20 amp breaker in general. Seems to work just fine.

    In factories.....for long distances across super large factories....they will sometimes use 11,500 voltage cables to deliver power from one end to the other....then step it down in a transformer to make it 480 volts for distribution in that area. High voltage is great for avoiding voltage drops.

    As said above....application is everything.
     
    Last edited: May 9, 2012
  9. May 9, 2012 #8

    sophiecentaur

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    Actually, I'm not so sure that they did do "fine" because they need their so-called split phase system to give them 240V for high powered domestic equipment so as to avoid needing immensely thick cables. Too late to change now but the 120V standard was clearly chosen with only lighting and light equipment in mind.
    But is has proved to be a rich vein for discussion of PF!!!
     
  10. May 9, 2012 #9

    psparky

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    I agree with what you are saying, but it actually does work just fine. And it even runs strong circular saws and air compressors at 120 volts. If you need more power.....you just beam up Scotty....or you run a 240 volt line.

    That reminds me of the old school fuse panel I have in my house off of a main breaker panel. I've considered replacing the fuse panel for a breaker panel....but I'm often asked if I'm having any problems with it. The answer is no.....and the reply of the electrician I am speaking with is ussually....."then why change it?". I mean I've never even popped a fuse. Works perfect without exception. If it aint broke....don't fix it.
     
  11. May 9, 2012 #10

    sophiecentaur

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    Copper is getting more and more expensive, though. . . . . . .
     
  12. May 11, 2012 #11
    Simple related question:

    Say device A needs 10 W of power and a voltage of 5V. The current must be 2 A.

    Suppose a source is connect to device A. the source can provide the correct voltage, 5V.
    I would assume that once that is accomplished the device will draw the current it needs, i.e. 2A, correct?

    That is all good as long as the source is able to emit 2A of current, correct? If the max current output of the source is, say, 1 A, then the device will not work properly....

    I guess there must be current continuity:what goes into device A must be what comes out of the source. Figuratively speaking, device A is pulling charge in the conductors to cause a current of a certain size but the inner working of the source don't allow such a current...

    I naively assumed that once a device was provided a certain voltage, the current that enters the device itself would only depend on the inner impedance of the device and nothing else...

    thanks,
    fisico30
     
  13. May 11, 2012 #12

    vk6kro

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    Yes, you have most of that OK.

    Sometimes the supply can actually supply the current for the load, but it would eventually heat up too much and may be destroyed.

    More usually, the supply has some internal resistance that stops it delivering more current at the correct voltage.
    What follows is a bit of simple maths. Try to work through it one line at a time and you will get an idea of what actually happens.

    Suppose you had a 6 volt supply with 0.5 ohms of internal resistance in series with it.
    Now, you connect a 2.5 ohm load on it.
    The combined resistance is 0.5 + 2.5 ohms, or 3 ohms, so the current is (6 volts / 3 ohms) or 2 amps.

    The load will have a voltage of I * R or 2 amps * 2.5 ohms ie 5 volts across it.
    The other volt is lost inside the battery across the internal resistance.
    This is the example you gave.

    Can you see what would happen if you connected a 1 ohm resistor?
    It would make a total of 1.5 ohms with the internal resistance (1 ohm + 0.5 ohms) so the current would be (6 volts / 1.5 ohms) or 4 amps but it would only have 4 volts across the 1 ohm resistor.

    In fact this supply could give 12 amps into a short circuit, limited only by the internal resistance, but it can only give 2 amps at 5 volts and only if the load is 2.5 ohms.

    This sort of behaviour of power supplies and batteries was the reason that regulated supplies were developed. These give a steady voltage out, almost regardless of load (up to a maximum current, of course).
     
  14. May 13, 2012 #13
    Hello vk6kro,

    thanks a lot. what I am trying to do is charge a small electronic device with a solar cell.
    The solar cell has a certain output voltage. I don't know the impedance of the electronic device. Based on its impedance and the impedance of the solar cell, a certain output current will be generated.....

    The goal is to be able to transfer all the power generated by the cell to the device.

    The device probably operates at a DC voltage that is different from the DC voltage offered by the solar cell. Do I need a DC to DC converter to match the two voltages? I would think so. If the device voltage is less than the solar cell voltage, I can use a diode to prevent power to go from the device to the cell...but my objective is really to charge the electronic device....

    Once the voltages are the same, the problem that the impedances of the solar cell and electronic device are different remains....What should I do about that?

    In summary, what do I need to buy/build to transfer the most power from the cell to the device?

    thanks
    fisico30
     
  15. May 15, 2012 #14

    psparky

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    Yes....and even more important.....

    Higher voltage means lower current....which means reduced cable size.

    Multiplied by millions of miles of overhead lines....this means millions and millions of dollars saved in materials and labor for installing smaller diameter overhead lines.

    Money....money....money.....
     
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