Question about AC, frequency, and watts.

[GreenXenon]

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

.... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.


Thanks

 

[Narthix]

"GreenXenon" <glucegen1x@gmail.com> wrote in message
news:ed1eef67-db73-413d-b8ff-a22bb4c60236@c19g2000yqc.googlegroups.com...

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

no.

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

no.

Thanks

sure, anytime

 

[HardySpicer]

On Jun 4, 1:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks

The average power dissipated is I^2R where I is the rms current and R
is the real part of the load (assuming the load is complex ie with
inductance). So if both rms currents are the same (which you said they
were) and R is the same then the power is the same. Average power is
also given by VIcos(phi) where V and I are both rms voltage and
current and cos(phi) is the power factor ie phi is the angle bewteen V
and I. This gives the same answer as previous.

However, if two ac voltages of the same rms value but different
frequencies are applied to a load R+jwL

where R is resistance L is inductance and w is 2pif rads/s then the
current for each case will be different.

rms current I=V/sqrt(R^2+(wL)^2) and hence I will be smaller if w is
larger and hence the power will be smaller.
So a larger freq dissipates smaller power for a given load. The power
for both cases will be

P =R . V^2/(R^2+(wL)^2)



Hardy

 

[zzbunker@netscape.net]

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

It doesn't if you measuring what's called apparent power.
But, with AC lines people usually think in terms of RMS power.
It which case the frequency doesn't have anything to do with power
throughput.

Thanks

 

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks

OK Greenxenon I strongly suggest you get a physics text book on E&M
and read it. Do all the problems. And then come back and ask some
questions. Volts times Amps is Watts. I don't care what the
frequency. However if you want to calculate the number of photons in
the 50 Hz signal you will find that there are more than in the 60 Hz
signal. Same is true for any frequency or wavelength

George Herold

 

[HardySpicer]

On Jun 3, 7:37 pm, ggher...@gmail.com wrote:

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:



Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks

OK Greenxenon I strongly suggest you get a physics text book on E&M
and read it.  Do all the problems.  And then come back and ask some
questions.  Volts times Amps is Watts.  I don't care what the
frequency.  However if you want to calculate the number of photons in
the 50 Hz signal you will find that there are more than in the 60 Hz
signal.  Same is true for any frequency or wavelength

George Herold

The current drawn is frequency dependent if the load is inductive (and
resistive). So therefore the power is also frequency dependent. For a
purely resistive load however you are right.


Hardy

 

[Uncle Al]

GreenXenon wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?
[snip rest of crap]

Idiot

Solve the equations, moron.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

 

[Benj]

On Jun 3, 10:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net>
wrote:

I am wondering if the same analogy applies to electric current.

   It doesn't if you measuring what's called apparent power.
   But, with AC lines people usually think in terms of RMS power.
   It which case the frequency doesn't have anything to do with power
throughput.

Which is why real scientists and engineers spend their time working on
perpetual motion machines, free energy, channeled schematic diagrams,
UFO tracking, debunking relativity, time travel, ghost busting, aether
theory, faster than light communications, warp drives, etc. ...

Idiot.

(Not because your answer is wrong, but because you responded to
GreenXenon at all! )

 

[Benj]

On Jun 4, 10:08 am, Uncle Al <Uncle...@hate.spam.net> wrote:

[snip all crap]

Idiot

Solve the equations, moron.

Idiot

Do not respond to GreenXenon, moron.

 

On Jun 3, 7:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net>
wrote:

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:





Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

   It doesn't if you measuring what's called apparent power.
   But, with AC lines people usually think in terms of RMS power.

Actually, it's 'average', not 'RMS'. RMS power is a useless quantity.
RMS is applicable to voltage & current, but not power.

Paul Cardinale

 

[dave y.]

On Thu, 4 Jun 2009 11:49:40 -0700 (PDT), pcardinale@volcanomail.com
wrote:

On Jun 3, 7:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net
wrote:
On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:





Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

   It doesn't if you measuring what's called apparent power.
   But, with AC lines people usually think in terms of RMS power.

Actually, it's 'average', not 'RMS'. RMS power is a useless quantity.
RMS is applicable to voltage & current, but not power.

Paul Cardinale

Not to disagree with the point you're trying to make, but as an aside,
RMS power can be used to describe temperature rise situations
such as motor duty cycle.

dave y.

 

[Arlowe]

ggherold@gmail.com expressed precisely :

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:
Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks


... Volts times Amps is Watts. I don't care what the
frequency....

George Herold

You are correct if your voltage is ripple free DC.
However DC doesn't have frequency, so your answer doesn't apply.
AC power really can not be explained as simply Volts X Amps (P=VI) due
to the inductive, capacitive components that can be present in an AC
circuit.
You can use P=V(sq)/R or P=VIFp.

 

[John Fields]

On Sat, 06 Jun 2009 09:49:18 +1000, Arlowe <bare.arsed@gmail.com> wrote:

ggherold@gmail.com expressed precisely :
On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:
Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks


... Volts times Amps is Watts. I don't care what the
frequency....

George Herold

You are correct if your voltage is ripple free DC.
However DC doesn't have frequency, so your answer doesn't apply.
AC power really can not be explained as simply Volts X Amps (P=VI) due
to the inductive, capacitive components that can be present in an AC
circuit.
You can use P=V(sq)/R or P=VIFp.

---
You can't use either of those equations to determine the reactance of
the circuit's components, and neither can you use them to determine the
impedance of the full circuit.

Do you want to know why not?

JF

 

On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

Thanks

IT MOST PROBABLY APPLIES ONLY TOO "CURRENT"
THE FREQUENCY CAN BE USED ALTER YOUR CURRENT NOT THE POWER AS ;LIKE;LY

I AM PROTEUS

 

[Don Kelly]

"dave y." <nospam@myhouse.com> wrote in message
news:nl8i25hf5efir9l1eqdh8ajt6km6ni8r5h@4ax.com...

On Thu, 4 Jun 2009 11:49:40 -0700 (PDT), pcardinale@volcanomail.com
wrote:

On Jun 3, 7:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net
wrote:
On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:





Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

It doesn't if you measuring what's called apparent power.
But, with AC lines people usually think in terms of RMS power.

Actually, it's 'average', not 'RMS'. RMS power is a useless quantity.
RMS is applicable to voltage & current, but not power.

Paul Cardinale

Not to disagree with the point you're trying to make, but as an aside,
RMS power can be used to describe temperature rise situations
such as motor duty cycle.

dave y.

You are right, but this is a specialized application involved in application

of motors to cyclic loads and the definition is clear to those dealing with
this situation or similar situations where temperature is the critical
limiting factor.

However, somehow, the error that has crept in is that many refer to V*I
(*power factor) where V and I are rms quantities,as rms power which is
nonsense (i.e. Zzbun's statement) Unfortunately this error is quite common
and reflects a lack of basic understanding . - it is this that Paul is
correctly addressing.


--
Don Kelly
dhky@shawcross.ca
remove the x to reply

 

[zzbunker@netscape.net]

On Jun 7, 10:17 pm, " Don Kelly" <d...@shaw.ca> wrote:

"dave y." <nos...@myhouse.com> wrote in message

news:nl8i25hf5efir9l1eqdh8ajt6km6ni8r5h@4ax.com...



On Thu, 4 Jun 2009 11:49:40 -0700 (PDT), pcardin...@volcanomail.com
wrote:

On Jun 3, 7:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net
wrote:
On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700 nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

It doesn't if you measuring what's called apparent power.
But, with AC lines people usually think in terms of RMS power.

Actually, it's 'average', not 'RMS'.  RMS power is a useless quantity..
RMS is applicable to voltage & current, but not power.

Paul Cardinale

Not to disagree with the point you're trying to make, but as an aside,
RMS power can be used to describe temperature rise situations
such as motor duty cycle.

dave y.

You are right, but this is a specialized application involved in application
of motors to cyclic loads and the definition is clear to those dealing with
this situation or similar situations where temperature is  the critical
limiting factor.

 However, somehow, the error that has crept in is that many refer to V*I
(*power factor) where V and I are rms quantities,as rms power which is
nonsense (i.e. Zzbun's statement) Unfortunately this error is quite  common
and reflects a lack of basic understanding . - it is this that Paul is
correctly addressing.

But, people like me have actually worked on high voltage, rather
than
with GE imbeciles, usually tell physicist to fuck off anyway so it
doesn't matter.

--
Don Kelly
d...@shawcross.ca
remove the x to reply- Hide quoted text -

- Show quoted text -

 

[Don Kelly]

<zzbunker@netscape.net> wrote in message
news:e5b7dbdc-fa3f-4638-bdf0-cea5bfbbc5b2@i6g2000yqj.googlegroups.com...
On Jun 7, 10:17 pm, " Don Kelly" <d...@shaw.ca> wrote:

"dave y." <nos...@myhouse.com> wrote in message

news:nl8i25hf5efir9l1eqdh8ajt6km6ni8r5h@4ax.com...



On Thu, 4 Jun 2009 11:49:40 -0700 (PDT), pcardin...@volcanomail.com
wrote:

On Jun 3, 7:23 pm, "zzbun...@netscape.net" <zzbun...@netscape.net
wrote:
On Jun 3, 9:29 pm, GreenXenon <glucege...@gmail.com> wrote:

Hi:

Let's say there are two AC [Alternating Current] signals of the same
voltage and amperage. However, one of them is 50 Hz and the other is
60 Hz.

Will the 60 Hz signal contain more watts than the 50 Hz signal?

I ask because...

... let's say there are two equally-large laser beams which contain
the same light intensity [i.e. number of photon-per-second-per-
square-meter]. If one beam is of 400 nm wavelength and the other is
700 nm, the 400 nm light beam will contain more watts than the 700
nm
one because a shorter-wavelength [i.e. higher-frequency] photon
possesses more energy in it than a longer-wavelength [or lower-
frequency] photon.

I am wondering if the same analogy applies to electric current.

It doesn't if you measuring what's called apparent power.
But, with AC lines people usually think in terms of RMS power.

Actually, it's 'average', not 'RMS'. RMS power is a useless quantity.
RMS is applicable to voltage & current, but not power.

Paul Cardinale

Not to disagree with the point you're trying to make, but as an aside,
RMS power can be used to describe temperature rise situations
such as motor duty cycle.

dave y.

You are right, but this is a specialized application involved in
application
of motors to cyclic loads and the definition is clear to those dealing
with
this situation or similar situations where temperature is the critical
limiting factor.

However, somehow, the error that has crept in is that many refer to V*I
(*power factor) where V and I are rms quantities,as rms power which is
nonsense (i.e. Zzbun's statement) Unfortunately this error is quite common
and reflects a lack of basic understanding . - it is this that Paul is
correctly addressing.

But, people like me have actually worked on high voltage, rather
than
with GE imbeciles, usually tell physicist to fuck off anyway so it
doesn't matter.

And who is the physicist? Not me,and the highest voltage that I have played
with was only in the order of 300KV
If youir statement was RMS power as a result of RMS voltage and current then
it is fundamentally wrong. Common usage maybe -for audio salesmen, not for
technicians and engineers in the power industry who should know better.

If I have attributed this to you in error, then I apologise.

--
Don Kelly
dhky@shawcross.ca
remove the x to reply