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What is rms value?

M

Matthew Ling

Jan 1, 1970
0
Hi, I am confused over the term rms.
From an AC waveform, the average should be zero.
how is rms value derived from the AC waveform?
I know that in order to obtain rms, it is just dividing
the max value by square root of 2, BUT WHY??!
Any help is appreciated.
 
Matthew Ling said:
Hi, I am confused over the term rms.
From an AC waveform, the average should be zero.
how is rms value derived from the AC waveform?
I know that in order to obtain rms, it is just dividing
the max value by square root of 2, BUT WHY??!
Any help is appreciated.

the average value of an AC current waveform is NOT the same as the RMS value
RMS is the effective value of the AC current...

so... it's quite simple... say we have two circuit:
1) an AC source connected with a resistor R in series
2) an equivelant DC source connected with a resistor R in series

the key point is
the average power dissipated in R in the AC circuit should be the same as in
an equivelant DC circuit...
^^^^^^^^^^^^^
key point
hint: <P> = <R*i^2> and i=Io*sin(wt+PS) in AC

after hving done some math you will get something like this:
in AC:
<P> = (R*Io^2)/2 <-- still in AC
so Io^2=<P>*2/R <--- AC

in DC: ( current does not alternate in DC so the power is a constant )
I^2 = P/R
because Pac = Pdc
you divide the first equation by the seconde euqation, and u get
(Io/I)^2 = 2
I = Io / 2^0.5
here, I is the current in an equivelant DC circuit... it's also called the
RMS value...
 
J

John Popelish

Jan 1, 1970
0
Matthew said:
Hi, I am confused over the term rms.
From an AC waveform, the average should be zero.
how is rms value derived from the AC waveform?
I know that in order to obtain rms, it is just dividing
the max value by square root of 2, BUT WHY??!
Any help is appreciated.

RMS stands for the three mathematical operations (Root Mean Square)
needed to calculate the equivalent DC voltage (or current) that would
produce the same power in a resistor as the AC waveform. So, first,
you square the waveform, then you take the average (mean) of that
squared waveform over any time that includes an integer number of
cycles (or long enough time to capture a representative average of all
possible variation if the waveform is not cyclic) an then you take the
square root of that mean.

It is just a mathematical coincidence that if the wave is a sine the
RMS result happens to be the peak value divided by square root of 2.
For other wave shapes there are other results. For instance, for a
symmetrical square wave (equal amplitude positive and negative half
cycles) the RMS is the same as the peak.
 
B

Bill Vajk

Jan 1, 1970
0
Matthew said:
Hi, I am confused over the term rms.
From an AC waveform, the average should be zero.

Yes. The time averaged value of a symmetrical sine wave
AC is zero.

Power, however, depends on the absolute value of voltage
and current.
how is rms value derived from the AC waveform?

John has written a succinct explanation to you.
I know that in order to obtain rms, it is just dividing
the max value by square root of 2, BUT WHY??!

For extra points work out why instead of dividing by
the square root of two you can get the same result by
multiplying by 1/2 the square root of two (aproximately
..707--close enough for most work.)
 
J

JeffM

Jan 1, 1970
0
John,
Have you ever been paid for teaching by an institution of (higher?) learning?
 
J

John Popelish

Jan 1, 1970
0
JeffM said:
John,
Have you ever been paid for teaching by an institution of (higher?) learning?

Only by the U. S. Army.
 
B

Bob Penoyer

Jan 1, 1970
0
Hi, I am confused over the term rms.
From an AC waveform, the average should be zero.
how is rms value derived from the AC waveform?
I know that in order to obtain rms, it is just dividing
the max value by square root of 2, BUT WHY??!
Any help is appreciated.

It's not simple to look at an AC waveform (like a sine wave, for
example) and "see" how it compares to DC. It's clear, though, that DC
will heat a resistor by delivering power to it. It's also clear that
AC, too, will heat a resistor. POWER is what heats a resistor. So, if
a given AC waveform heats a resistor exactly as much as a particular
DC voltage, then we can say that their "heating effect" is the same.
An AC voltage's RMS value is often referred to as its heating effect.

When a DC voltage heats a resistor to some temperature, it does it by
delivering power to the resistor continuously. When an AC voltage
heats the same resistor to the same temperature, it does it by
delivering the same AVERAGE POWER to the resistor. (This is true
because power is what heats the resistor. An AC waveform's
"instanteous" power--the power delivered at a particular instant of
time--varies with time because its voltage varies with time. But the
resistor's temperature tends to vary slowly, much slower than the
variation of the AC waveform. So it's AVERAGE POWER from the AC
waveform that we care about.)

As John Popelish explained, "RMS stands for the three mathematical
operations (Root Mean Square) needed to calculate the" AC voltage that
is equivalent to a DC voltage. R, M, and S are the mathematical
operations in reverse. First, square the voltage. This makes sense
because voltage squared is used to calculate power.

Next, take the mean (average) of the squared voltage. That is,
calculate the average power.

Finally, take the square root. This makes sense because the square
root of power can be used to calculate voltage.

So, by operating on an AC waveform, first by squaring, then averaging,
then by square-rooting, you convert the AC voltage to an equivalent DC
voltage.

Here's an example:

A sine wave has a peak value of 10 volts. What is its RMS (or
equivalent DC) value?

V = 10 * sin(2 * pi * f * t)

SQUARE:
V^2 = 100 [sin(2 * pi * f * t)]^2

but [sin(x)]^2 = 0.5 [1 - cos(2x)] (Check a trig reference if you're
not familiar with trigonometry.)

so V^2 = 50 - 50 * cos(4 * pi * f * t)

MEAN:
Ave(V^2) = 50 - 50 * Ave[cos(4 * pi * f * t)]

Since the average of the cosine function over a full period is zero,
then
Ave(V^2) = 50 - 0 = 50

ROOT:
SquareRoot[Ave(V^2)] = SquareRoot[50] = 7.07, approximately.

So, a sine wave with a peak value of 10 has an RMS value of 7.07. This
means, of course, that a sine wave AC voltage with a peak value of 10
volts will heat a given resistor exactly as much as a DC voltage of
7.07 volts. For a sine wave, then, to find the RMS value, just divide
the peak value by the square root of 2: 10 / sqrt(2) = 7.07.
 
B

Bob Myers

Jan 1, 1970
0
John Popelish said:
learning?

Only by the U. S. Army.

John, I am SO tempted at this point, but I'm going to let
this one pass by...it's just too much of an easy target...:)

Bob M.
 
R

Ratch

Jan 1, 1970
0
POWER is what heats a resistor.

ENERGY is what heats a resistor. Power determines the energy dissipation
and ultimately the temperature rise of the resistor. Ratch
 
B

Bob Penoyer

Jan 1, 1970
0
ENERGY is what heats a resistor. Power determines the energy dissipation
and ultimately the temperature rise of the resistor. Ratch

Well, let's see. If I dump 1000 joules into a 1-watt resistor but I
take all of February to do it, the resistor won't get very hot. But,
if I do it over a period of 100 seconds, the resistor will get damn
hot. Rate! Rate is important. It's power that heats the resistor.
Hence, terms like 1-WATT resistor.
 
Bob Penoyer said:
Well, let's see. If I dump 1000 joules into a 1-watt resistor but I
take all of February to do it, the resistor won't get very hot. But,
if I do it over a period of 100 seconds, the resistor will get damn
hot. Rate! Rate is important. It's power that heats the resistor.
Hence, terms like 1-WATT resistor.

i wonder if you did your lil dumping-joul experiment in VACUMM AND there is
no heat RADIATING from that resistor!!!

joul is the unit of energy, also the unti of heat!!!
which means... no matter how many jouls you dump into a object...
as long as there's no heat lost... the temp of that object will go up by a
certain degrees...

for the case you were talking about... it's becuase a partial heat that
CURRENT generated is LOST in air and radiation... that's why you dont feel
it really hot if you take a whole month to dump that 1000J into the
resistor...
for only 100s... you feel it much hotter becuase not much heat is running
away~~~

as the current thru a R goes up, the montion of free electrons becomes more
active in order to keep that current.( remember the definition of current I?
it's I = Q/t... )therefore, the molecules are moving in a faster rate in
high current than in low current... thus, the collision among molecures
occurs in a higher rate which generate more HEAT...
so it's CURRENT that heats up the resistor... not POWER...

furthermore, power is defined as POWER = ENERGY / T
doesn't this look similar to VELOCITY = DISPLACEMENT / T ??

:) have a nice day~~~
 
B

Bob Penoyer

Jan 1, 1970
0
i wonder if you did your lil dumping-joul experiment in VACUMM AND there is
no heat RADIATING from that resistor!!!

Heat radiates just fine from a resistor in a vacuum. If heat couldn't
radiate through a vacuum, the sun wouldn't do us much good.
joul is the unit of energy, also the unti of heat!!!
which means... no matter how many jouls you dump into a object...
as long as there's no heat lost... the temp of that object will go up by a
certain degrees...

A resistor won't be insulated under normal circumstances. So heat will
be lost by conduction, convection, and/or radiation.
for the case you were talking about... it's becuase a partial heat that
CURRENT generated is LOST in air and radiation... that's why you dont feel
it really hot if you take a whole month to dump that 1000J into the
resistor...
for only 100s... you feel it much hotter becuase not much heat is running
away~~~

Nonsense. As I said in my earlier post, it is rate that effects the
heating of the resistor. If you dump energy into the resistor faster
(i.e., apply more power), the resistor will get hotter.
as the current thru a R goes up, the montion of free electrons becomes more
active in order to keep that current.( remember the definition of current I?
it's I = Q/t... )therefore, the molecules are moving in a faster rate in
high current than in low current... thus, the collision among molecures
occurs in a higher rate which generate more HEAT...
so it's CURRENT that heats up the resistor... not POWER...

It's pretty tough to pump current through a resistor without
encountering P = (I^2)R. Now, if I apply the same current to a 10-ohm
resistor as I do to a 20-ohm resistor, the 20-ohm resistor will get
hotter--even though I use the same current in both cases. It's power
that heats the resistor, not current. That's why resistors are rated
in watts, not amps.
furthermore, power is defined as POWER = ENERGY / T
doesn't this look similar to VELOCITY = DISPLACEMENT / T ??

Only because they are both RATES. Other than that, I don't see a
connection.
:) have a nice day~~~

You, too.
 
Bob Penoyer said:
Heat radiates just fine from a resistor in a vacuum. If heat couldn't
radiate through a vacuum, the sun wouldn't do us much good.

in vacuum and radiation are different statements... they are not a bit
related...
A resistor won't be insulated under normal circumstances. So heat will
be lost by conduction, convection, and/or radiation.


Nonsense. As I said in my earlier post, it is rate that effects the
heating of the resistor. If you dump energy into the resistor faster
(i.e., apply more power), the resistor will get hotter.

alrite... one simple Q
do u measure heat in Jouls or in Watts?
 
R

Ratch

Jan 1, 1970
0
alrite... one simple Q
do u measure heat in Jouls or in Watts?

Heat is the lowest form of energy. Heat/energy is measured in joules for
International System units. Heat/energy rate of gain/loss is measured in
watts for International System units. Ratch
 
R

Robert C Monsen

Jan 1, 1970
0
AND there
is

in vacuum and radiation are different statements... they are not a bit go up by
a dont

alrite... one simple Q
do u measure heat in Jouls or in Watts?

Stop it, you are making my head hurt...

The amount of power radiated from a resistor is related to its
temperature. The higher the temp, in relation to the environment, the
faster the object will radiate energy.

If you add energy by passing current through it, it will heat up until
the rate of radiation = the rate of addition, ie, it will reach a
thermal equilibrium.

If it is unable to radiate energy for some reason (maybe the
surroundings are kept at the same increasing temperature) the resistor
will continue to heat up, because energy is getting added by the
current flow.

Thus, you are both right, it's energy that heats a resistor, but in
the real world, its power that determines the final temperature,
because, given a particular resistor configuration, the resistor will
radiate at a particular rate dependent on temperature. Thus, the
temperature will satisfy a formula something like

I^2*R = P(T)

where P(T) is the function of radiated power for the object in its
environment, given a temperature T.

Regards,
Bob Monsen
 
Ratch said:
Heat is the lowest form of energy. Heat/energy is measured in joules for
International System units. Heat/energy rate of gain/loss is measured in
watts for International System units. Ratch

Thank you...
Bob, i think you should read this...
heat , which is a form of a energy, is measured in Jouls...
 
B

Bob Penoyer

Jan 1, 1970
0
in vacuum and radiation are different statements... they are not a bit
related...

It was your proposition. You must have thought it made sense.
alrite... one simple Q
do u measure heat in Jouls or in Watts?

Joules.

Here's a question: Two resistors of equal power-handing capability are
connected in series. One resistor is twice the value of the other. A
current is passed through the resistors so that they are warm to the
touch. Which of the following choices is correct?

A. The larger resistor gets hotter.
B. The smaller resistor gets hotter.
C. The resistors are equally warm because it is current that
determines the amount of heating.
 
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