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Power Measurent: Watts Vs. Volts

Discussion in 'Electronic Basics' started by Kaimbridge, Aug 11, 2005.

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  1. Kaimbridge

    Kaimbridge Guest

    Power = Watts = Volts * Amps

    If the unit measure of power is Watts, how come voltage is usually
    For instance, "Danger, Keep out: 50,000 volts!" (or just "High
    Voltage"). Static electricity can zap you with 50,000 volts, and
    you'll just feel a snap--Why?: Because there is very little amperage,
    so the 50,000 volts may be with only .1^10 amps, giving
    ..000005 watts, whereas the "Danger, Keep out: 50,000 volts!" may be at
    10 amps, providing 500,000 watts! So why don't they say "Danger, Keep
    out: Up to (if the amperage varies) 500,000 watts!"?
    Similarly, product "letter" batteries (A, AA, AAA, C, D) are all
    measured as "1.5 volts" (I believe 2 equal 3 volts, 4 equal 6,
    etc.--?). Don't the batteries have a defined amperage (maybe that's
    the difference between A and AA and C, etc.?), thus why aren't they
    defined as "___watt" batteries (if the amperage--and therefore the
    wattage-- decreases as the battery weakens, its defined value would be
    its full, maximum strength)?
    Finally, radio signal strength: Why is it measured as "microvolts" and
    not "microwatts", especially since a station's transmitter output power
    *is* measured in wattage?

  2. JeffM

    JeffM Guest

  3. tlbs

    tlbs Guest

    Batteries are rated by their terminal voltage AND by their capacity,
    measured in Amp-Hours. Many AAA, AA, C, D cells have their capacity
    printed on the side, as well as their terminal voltage. Since the
    terminal voltage is known, and the capacity is known, the total
    available energy is also known (Volts x Amps x seconds = energy, in
    Joules). This is a simplistic answer -- the actual available energy
    is different from V x Capacity x time, because the terminal voltage
    drops with: current draw, rate of current draw, time, temperature, and
    other environmental factors.

    The rate at which you can draw that energy (Watts) from a battery is
    also more complex -- if you need to know that you should consult the
    manufacturer's data sheets. Automobile batteries list their "cold
    cranking Amps" -- that is a pretty good indicator of how much power you
    can draw from it (CCA x V).

    In measuring radio emmision energy, the measurement is the electric
    field strength. The units are Volts/meter (or microVolts/meter).
    Power in free space is measured in Watts/meter^2.

    Only when the RF energy is converted to RF current through an antenna,
    can direct power be measured. Then other factors must be taken into
    account: antenna impedance, antenna gain, antenna type. For a given
    field strength different power levels may be obtained depending on the
  4. Don Bowey

    Don Bowey Guest

    Are you just ranting or responding to a post? It's hard to tell when you
    don't quote anything (such as above).

    Go away and read up on board etiquette.
  5. Kaimbridge wrote:

    No, the current is determined from the voltage and resistance (body
    resistance and ground resistance in series) by Ohms law R = U/I. But in
    the case of static electricity the current will stop immediately,
    because the capacity of the "power source" is extremly low.
    Again current is determined by Ohms law. There are two main differences
    between a C battery and an AAA: Capacity and internal resistance. A
    battery converts chemical into electrical energy, a C battery has more
    of the chemicals and hence can sustain a given current for a longer
    time. In addition, its electrodes are bigger, hence internal resistance
    is lower. The voltage across a battery is determined by the kind of
    chemicals used inside, for example Zn/C = 1.5 V.
    The field strength is V/m, this will result in a certain voltage across
    the poles of an antenna. Given the antennas resistance (usually 220 or
    75 Ohm) this will result in a certain power, since R = U/I and P = U*I
    the power is P = U^2/R.

    A good high school physics textbook will cover these fundamentals in
    more detail.
  6. tlbs

    tlbs Guest

    In the context of the thread, I was the 3rd poster, answering the
    unanswered questions in the primary post. I would think in context my
    post was obvious. I guess I was wrong. Yes, I can paste text from
    that to which I am responding (as you can see, above), but in this case
    I didn't think it was necessary.

    I don't know any other way to read these boards, other than by
    thread-context. If there is a way to read my post as an isolated
    entity, then what you say makes sense. So my question to you is: how
    is it that you can read my post as an isolated entity, yet I read them
    as whole threads (10-posts-at-a-time)?

    PS: I have no obvious way to change the color of the quoted text (as I
    do on other boards), I had to add the ">" characters manually, and I
    had to add, "Dr. Bowey wrote:", manually. Would Google Groups Beta
    allow some simple bulletin board functions (like those of phpBB, etc.),
    I might be more prone to follow better etiquette in these groups.
  7. JeffM

    JeffM Guest

  8. Rich Grise

    Rich Grise Guest

    Because you're using google. Google wants to pretend that USENET is
    some kind of chatroom, which it isn't.

    Real newsreaders quote context by default. But you need a real newsserver,
    which should be provided by your ISP.

    Good Luck!
  9. Kitchen Man

    Kitchen Man Guest

    Reason number one: because the voltage is the known quantity.
    Capacity and amperage will vary with usage and time, so there is
    little point in stating these quantities, except as maximum values.
    Imagine: "DANGER! Possible maximum capacity of 120,000 amp-seconds!"
    Would that terminology be meaningful to the most casual observer?
    Would it even be mathematically justifiable?

    Reason number two: it is important to know what the voltage quantity
    is when dealing with high voltage. There is an immense difference
    between 500KV and 50KV as far as minimum approach distance (MAD) is
    concerned. 500KV will "reach out and touch you" from a greater

    As far as how many watts will be used in the frying of your body when
    you come in contact with the high voltage, that is a quantity that can
    be extremely variable, even when the voltage is known.
    Actually, with static electricity, it is the capacity, not the
    current, that is the limiting factor. The initial instantaneous
    current may be quite large, dampening quickly.
    Transmitter output power is known and measured, because the
    characteristics of the transmitter are known. The received power is
    not known until a particular receiver picks up the signal, and the
    calculation for the received power will be dependent upon the field
    strength in microvolts at its particular location.

    Hope this all made sense.
  10. John Fields

    John Fields Guest

    and its antenna
    No, the received power is _only_ dependent on the field strength of
    the incident signal and the impedance presented to that signal by
    the receiving antenna and its load.
  11. Kitchen Man

    Kitchen Man Guest

    Of course. I was counting the antenna as part of the transmitter.
    Not completely technically correct, I suppose, but then, what
    transmitter is any good without an antenna?

    Well, that's what I meant to say. Perhaps I could have phrased it

    Again, I meant to speak of one particular receiver, antenna included.
    The answer to your hypothetical situation is that the size of the
    antennas help determine gain, and thus, directly affect received
    power. The receiver with higher gain will get more power from the
    same field strength, and yes, the field strength may be measured in
    units other than microvolts, if the measurerer is so disposed.

  12. John Fields

    John Fields Guest

    On Sat, 13 Aug 2005 18:01:59 -0700, Kitchen Man

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