# Power Measurent: Watts Vs. Volts

K

#### Kaimbridge

Jan 1, 1970
0
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"). 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?

~Kaimbridge~

T

#### tlbs

Jan 1, 1970
0
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
antenna.

D

#### Don Bowey

Jan 1, 1970
0
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
antenna.
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.

D

#### Dr Engelbert Buxbaum

Jan 1, 1970
0
Kaimbridge wrote:

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,

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.
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)?

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.
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?

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.

T

#### tlbs

Jan 1, 1970
0
Dr. Bowey said:
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.

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

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
allow some simple bulletin board functions (like those of phpBB, etc.),
I might be more prone to follow better etiquette in these groups.

R

#### Rich Grise

Jan 1, 1970
0
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

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!
Rich

K

#### Kitchen Man

Jan 1, 1970
0
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.

J

#### John Fields

Jan 1, 1970
0
Transmitter output power is known and measured, because the
characteristics of the transmitter

---
and its antenna
---
are known. The received power is
not known until a particular receiver picks up the signal,

---
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.

K

#### Kitchen Man

Jan 1, 1970
0
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
better.

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.

:-Þ

J

#### John Fields

Jan 1, 1970
0
On Sat, 13 Aug 2005 18:01:59 -0700, Kitchen Man

....

Replies
13
Views
156
Replies
1
Views
102
Replies
2
Views
158
Replies
1
Views
147
Replies
5
Views
256