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Math Help - Physics Circuitry

  1. #1
    Senior Member DivideBy0's Avatar
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    Physics Circuitry

    Uh... ever since we switched teachers everything's gotten more confusing... anyway,
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    What is the difference between a half-wave and full-wave rectifier? I saw what the difference was on the CRO reading, but, for instance, why would you use one instead the other?
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    What is this 'line loss' thing? As electricity passes through a wire it will lose voltage... or was that power...? Anyway, for some reason there is a formula P=I^2R where P is power loss, I is current, and R is wire resistance. How was the formula derived? How, by stepping up the voltage at the start of the transmission and stepping it down at the end will the energy loss be decreased?
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    What does the term 'Phase' mean?
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    Thanks... more I might think of more questions as time goes by...
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  2. #2
    Super Member malaygoel's Avatar
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    1)you have seen the result on CRO.
    Full wave rectifer gives continuos pulses while in half wave rectifier there is a constant break(0V) two consecutive pulses.
    It depends on the requirement which type of pulse train you want.

    3) In a wave, every point vibrates about its mean postion.Phase is used to denote the relative displacement from a postion taken as standard(usually it is mean position). If two points always have the same displacement, they are in the same phase.If at some instant, two points have the same displacement, they are not necessarily in same phase because they may be moving in opposite directions, hence will not be in the same phase at the next instant.

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    Malay
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    What is this 'line loss' thing? As electricity passes through a wire it will lose voltage... or was that power...? Anyway, for some reason there is a formula P=I^2R where P is power loss, I is current, and R is wire resistance. How was the formula derived? How, by stepping up the voltage at the start of the transmission and stepping it down at the end will the energy loss be decreased?
    V = IR <-- Ohm's Law, take this as a given.
    This describes the drop in potential across a resistor.

    P = VI
    You might wish to also take this as a given: you can derive it, but the derivation deals with charges moving across a potential difference and using Coulomb's law, etc. (There may be a simpler way, but it's not coming to mind right now.) This equation represents the power required to move those charges across a potential difference.

    Then
    P = (V)I = (IR)I = I^2R

    Now, as far as reducing losses are concerned note that no matter what you do to the voltage in the line the resistance stays the same. So the power loss over a line depends directly on the size of the current: the larger the current the larger the power loss.

    So in a step-up or step-down transformer the power in the line is conserved (actually, it's energy that's conserved). So if we ramp up the voltage we are reducing the current flow. Thus I drops and we lose less power when transmitting over long distances (large R.)

    -Dan
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