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question about a car alternator

Discussion in 'General Electronics Discussion' started by dietermoreno, Mar 5, 2013.

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  1. dietermoreno

    dietermoreno

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    Dec 30, 2012
    N00b question about an alternator.

    In my study of electricity, I remember that I have in the past charged my car battery with a plug in charger and the greater the current the faster the car battery charged; also I remember that in the past when I have had my car jumped, the faster the RPM of the car charging my car, the faster my car battery was charged.

    So I have come to the conclusion that the faster that an alternator spins, the greater the current output, and its voltage output is determined by Ohm's Law.

    So I have also come to the conclusion that in a car, alternating current does not power the car radio, rather the car radio is DC powered from the battery, which the battery is charged from alternating current from the car alternator.

    but what if you wanted to use the alternating current without being connected to a battery? like in a power plant.

    Why has 60 Hz been chosen for power transmission when wouldn't an alternator spinning as fast as possible for the greatest current output, like spinning at 10,000 Hz, emit the greatest current?
     
  2. duke37

    duke37

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    Jan 9, 2011
    Not all power transmission is 60Hz, were it is 50Hz.

    Car alternators have a varible frequency output depending on the position of the loud pedal. They run at high frequency to make them small and they have leakage inductance built in. The inductance has a reactance proportional to frequency which compensates for the increased voltage at high frequencies.

    The alternators are three phase and are full wave rectified to give DC. The current output is limited by the wire thickness.

    Alternators have replaced dynamos partly because they will spin faster without bursting apart.
     
  3. dietermoreno

    dietermoreno

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    Dec 30, 2012
    Wouldn't the wire thickness not limit the current output of the alternator, rather wouldn't increasing the wire resistance increase the voltage and do nothing to the current, according to Ohm's Law?

    Since V=IR, increasing the resistance increases the voltage if the current is constant.

    A creek that has been redirected to flow through a garden hose has much more potential energy in the garden hose at an instant of time compared to if the creek had been redirected to flow throw a 6 foot diameter pipe, but the same amount of water will flow through the garden hose, just it will take more time to flow through and the potential energy at an instant of time will be greatly increased.

    So voltage is potential energy.

    So a car battery of 12 volts with 4 ohms of internal resistance so with 3 amps of current, when it is connected to resistance wire, the 3 amps of current will still flow through, just it will take more time for the 3 amps of current to flow through, and the voltage will be greatly increased (and the resistance wire will glow red hot before lighting your house on fire, unless it is in a vacuum tube so that it doesn't burn).

    So if it takes more time for the 3 amps of current to flow through the resistance wire, wouldn't that keep the current constant but the power is what is effected because current has no time domain component but power is equal to potential energy divided by time, so that for more work in less time the power is greater, and for the same work in less time the power is greater?

    P=W/T

    So then for electricity the equation would be P=V/T

    but that doesn't make any sense because Ohm's Law for power is P=IV

    wait, so does that mean that current has a time component, and ampere is not a base unit?

    If we had P=V/T, in order to get to P=IV, that would mean that current is equal to 1/T, so increasing the time for the current to flow decreases the current, so greater resistance decreases the current, but that would contradict Ohm's Law of V=IR which says that voltage increases as resistance increases?

    oh wait, I think I get it now: Ohm's Law of V=IR says that if EITHER the current OR the resistance increases, then the voltage increases, and if both current AND resistance increase, then the voltage increases. So since V/R = I, that means that as resistance increases and voltage is constant, current decreases, which makes sense from the time component in current.

    So not only does it take more time for the 3 amps of current to flow through the resistance wire, but also the current output is decreased, even though the current of 3 amps will still flow through EVENTUALLY.

    So if you connect a 3 amp 36 watt guitar amp (P=IV, V= 12 volts, I = 3 amps, P=36 watts) to a 3 amp car battery, then the battery will drain MUCH FASTER than if you connect a 500mAmp 6 watt cell phone charger (P=IV, V=12 volts, 0.001 amps are in a milliamp, I=0.5 amps, P= 6 watt).





    So the conclusion is that yes, the resistance of wire will limit the current output of an alternator, as well as determine the voltage output using Ohm's Law.

    but still, wouldn't the frequency of the alternator determine the current output of the alternator IN AN IDEALIZED WIRE WITH NO RESISTANCE?
     
    Last edited: Mar 5, 2013
  4. davenn

    davenn Moderator

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    Sep 5, 2009
    No, the thicker the wire, the more current it can handle
    But the important parts is .... the thicker the wire the less resistance it has and therefore the lower the voltage drop over its length

    Dave
     
  5. dietermoreno

    dietermoreno

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    Dec 30, 2012
    So do you mean that wire resistance is not static, like resistor resistance is static, rather wire resistance is dynamic changing with the current? So a thick gauge wire and a high current will have a low resistance, but a thick gauge wire and a low current will have a high resistance?

    I know that resistance in wire can change in liquid such as in my car's gas tank.
     
  6. davenn

    davenn Moderator

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    Sep 5, 2009
    no thats wrong and thats not what I said

    read again what I said :)

    Dave
     
  7. dietermoreno

    dietermoreno

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    Dec 30, 2012
    Oh, so you mean that the resistance of a wire is a static value, that will not change no matter what the current is, but is a function of the wire gauge.

    The thicker the wire gauge, the less resistance of the wire; therefore the more current the wire can handle because of lower resistance.

    The thinner the wire gauge, the greater resistance of the wire; therefore the less current the wire can handle because of higher resistance.
     
  8. davenn

    davenn Moderator

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    1,950
    Sep 5, 2009
    Exactly :)

    also, power loss over that wire is a function of current and resistance

    Power (loss) = I squared x R .... reduce R and you reduce your power loss over that wire

    (grrrr cant do super scripted characters on here)

    Dave

    PS EDIT ....
    not so much because of the lower resistance but the more current that can flow
    think of a 1 cm water pipe and 10cm water pipe .... which can have a larger flow ?
     
    Last edited: Mar 5, 2013
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