Re: Power Measurent: Watts Vs. Volts

On 11 Aug 2005 07:35:02 -0700, "Kaimbridge" <Kaimbridge@xxxxxxxxx>
wrote:

>Power = Watts = Volts * Amps
>
>If the unit measure of power is Watts, how come voltage is usually
>expressed.
>For instance, "Danger, Keep out: 50,000 volts!" (or just "High
>Voltage").

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
distance.

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.

> Static electricity can zap you with 50,000 volts, and
>you'll just feel a snap--Why?: Because there is very little amperage,

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.

>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?

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.

.

Relevant Pages

  • Re: Intensity and what else affects a single ligth beams temperature?
    ... now and the voltage now to compute the power now. ... you still measure or compute the current NOW and the voltage NOW. ... Then remove phi from the calculation and demonstate ... That is going to give 100 Watts at that instant in time. ...
    (sci.physics.relativity)
  • Re: Power Inverters
    ... The "designer" copied the circuit ... When using a square wave or modified sinewave ... Newer power supplies have built-in power factor correction to make ... the slightly higher voltage won't matter. ...
    (comp.sys.laptops)
  • Re: AC power measurement
    ... voltage and current will give the power dissipated. ... A simple way for mains powered devices is to measure the ... RMS current and multiply by the nominal/actual RMS voltage. ...
    (uk.comp.homebuilt)
  • Re: AC power measurement
    ... voltage and current will give the power dissipated. ... RMS current and multiply by the nominal/actual RMS voltage. ... Metrawatt analogue wattmeter using 'simple' analogue multiplier ...
    (uk.comp.homebuilt)
  • Re: UPS Advice
    ... many periods of low voltage. ... and power outages aren't as much of an issue. ... PC repair technicians aren't usually well-versed in either electricity or electronics. ... I was an electrician and an electronic technician long before the PC showed up. ...
    (alt.comp.hardware.pc-homebuilt)
Loading