I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

(Amp 1)
Lets assume we have a 200W @ 4-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 4), V = 28.2 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 28.2v, I = 7 amps


(Amp 2)
Lets assume we have a 200W @ 8-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 8), V = 40 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 40v, I = 5 amps


So in conclusion:
(Amp 1) V = IR, 28.2v = 7a * 4ohms
(Amp 1) P = IE, P = 7a * 28.2v = 197 Watts

(Amp 2) V = IR, 40v = 5a * 8ohms
(Amp 2) P = IE, P = 5a * 40v = 200 Watts


Is this correct?

Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

The short answers:

Yes, your calculations are correct.

No, the two amps will not draw the same current. Use your ohms' law he
Volts * Amps = Watts. If both devices draw 200 watts, the one being fed with
14.4 Volts will draw (200/14.4) or 13.9 amps. The one fed from 120 V house
current will draw (200/120) = 1.7 amps.

That's the easy part. However...

You have to check the wattage ratings carefully on the power amps.
Manufacturers (the lower-end ones, anyway) are incredibly sleazy in their
power ratings. Often they'll rate the amps at PEAK power, which is at least
twice what the amp can do on average. Futhermore, check the distortion %
(anything over 1% is not very pleasant). I remember reading a magazine
article 30 years ago saying you should take their published rating and
divide by 10 to get a realistic comparison to a good quality amp.

OK, that's for the output power. The input power could easily be twice that.
Standard Class B amplifiers are good to get 50% efficiency ratings. Meaning
of all the power you put in, 50% goes to your speakers. To put it another
way, take your (realistic) output rating and double it to get the input
requirement. Rather than guess, though, just check the nameplate - it should
tell you the maximum input current to the unit, or at least tell you a fuse
size (which will be about 25% over the maximum input current expected.)

Not sure why you want to know the voltage at the speaker terminals. It's not
going to be a simple calculation. The ohms' law formulae are for DC
circuits, though you can get away with using it for what we call sinusoidal
steady state (basically AC circuits where frequency, voltage and impedance
are constant - like a toaster or light bulb). It doesn't work well at all
for the complex signals fed to a speaker. The speaker coil also reacts quite
different than a basic resistor in that its own impedance changes radically
with changing frequencies. It's even effected by the type of enclosure it's
in. Add to that the very non-linear behavior of the crossover network
(assuming you have a midrange and tweeter) and you've got a pretty complex
system to try and characterize. So, back to my original question - why do
you want to know the speaker voltage? Perhaps you don't, though you solved
for it in your original post. Just choose speakers that have a sufficiently
high wattage rating to handle your power amp. To answer the obvious
question, then why do manufacturers show "impedance" on their speakers? That
number is what's called a "characteristic" or "nominal" impedance. Sort of
an average. It's meant to guide you for matching speaker combinations with
amp ratings. Yes, two 8 ohm speakers in parallel will show an impedance of 4
ohms to the amp. But if you put your ohmmeter across the speaker terminals
you'll get something around the 1-2 ohm range (or less for a high power
speaker) again, because the impedance changes with audio (AC) signal
applied.

Finally, you mentioned two different impedance ratings on the amplifiers. A
lot of amps can handle different speaker impedances (8, 4, and sometimes
even 2 ohms) and the power output is different for each one. Going from 8 to
4 ohm speakers usually brings a big increase in output power, almost double.
Again, check your amp's specs to see what it is - and even if it can handle
the lower impedance.

Hope this clarifies more than it confuses...

Mike Metzger

The short answers:

Yes, your calculations are correct.

No, the two amps will not draw the same current. Use your ohms' law he
Volts * Amps = Watts. If both devices draw 200 watts, the one being fed with
14.4 Volts will draw (200/14.4) or 13.9 amps. The one fed from 120 V house
current will draw (200/120) = 1.7 amps.

That's the easy part. However...

You have to check the wattage ratings carefully on the power amps.
Manufacturers (the lower-end ones, anyway) are incredibly sleazy in their
power ratings. Often they'll rate the amps at PEAK power, which is at least
twice what the amp can do on average. Futhermore, check the distortion %
(anything over 1% is not very pleasant). I remember reading a magazine
article 30 years ago saying you should take their published rating and
divide by 10 to get a realistic comparison to a good quality amp.

OK, that's for the output power. The input power could easily be twice that.
Standard Class B amplifiers are good to get 50% efficiency ratings. Meaning
of all the power you put in, 50% goes to your speakers. To put it another
way, take your (realistic) output rating and double it to get the input
requirement. Rather than guess, though, just check the nameplate - it should
tell you the maximum input current to the unit, or at least tell you a fuse
size (which will be about 25% over the maximum input current expected.)

Not sure why you want to know the voltage at the speaker terminals. It's not
going to be a simple calculation. The ohms' law formulae are for DC
circuits, though you can get away with using it for what we call sinusoidal
steady state (basically AC circuits where frequency, voltage and impedance
are constant - like a toaster or light bulb). It doesn't work well at all
for the complex signals fed to a speaker. The speaker coil also reacts quite
different than a basic resistor in that its own impedance changes radically
with changing frequencies. It's even effected by the type of enclosure it's
in. Add to that the very non-linear behavior of the crossover network
(assuming you have a midrange and tweeter) and you've got a pretty complex
system to try and characterize. So, back to my original question - why do
you want to know the speaker voltage? Perhaps you don't, though you solved
for it in your original post. Just choose speakers that have a sufficiently
high wattage rating to handle your power amp. To answer the obvious
question, then why do manufacturers show "impedance" on their speakers? That
number is what's called a "characteristic" or "nominal" impedance. Sort of
an average. It's meant to guide you for matching speaker combinations with
amp ratings. Yes, two 8 ohm speakers in parallel will show an impedance of 4
ohms to the amp. But if you put your ohmmeter across the speaker terminals
you'll get something around the 1-2 ohm range (or less for a high power
speaker) again, because the impedance changes with audio (AC) signal
applied.

Finally, you mentioned two different impedance ratings on the amplifiers. A
lot of amps can handle different speaker impedances (8, 4, and sometimes
even 2 ohms) and the power output is different for each one. Going from 8 to
4 ohm speakers usually brings a big increase in output power, almost double.
Again, check your amp's specs to see what it is - and even if it can handle
the lower impedance.

Hope this clarifies more than it confuses...

Mike Metzger

The short answers:

Yes, your calculations are correct.

No, the two amps will not draw the same current. Use your ohms' law he
Volts * Amps = Watts. If both devices draw 200 watts, the one being fed with
14.4 Volts will draw (200/14.4) or 13.9 amps. The one fed from 120 V house
current will draw (200/120) = 1.7 amps.

That's the easy part. However...

You have to check the wattage ratings carefully on the power amps.
Manufacturers (the lower-end ones, anyway) are incredibly sleazy in their
power ratings. Often they'll rate the amps at PEAK power, which is at least
twice what the amp can do on average. Futhermore, check the distortion %
(anything over 1% is not very pleasant). I remember reading a magazine
article 30 years ago saying you should take their published rating and
divide by 10 to get a realistic comparison to a good quality amp.

OK, that's for the output power. The input power could easily be twice that.
Standard Class B amplifiers are good to get 50% efficiency ratings. Meaning
of all the power you put in, 50% goes to your speakers. To put it another
way, take your (realistic) output rating and double it to get the input
requirement. Rather than guess, though, just check the nameplate - it should
tell you the maximum input current to the unit, or at least tell you a fuse
size (which will be about 25% over the maximum input current expected.)

Not sure why you want to know the voltage at the speaker terminals. It's not
going to be a simple calculation. The ohms' law formulae are for DC
circuits, though you can get away with using it for what we call sinusoidal
steady state (basically AC circuits where frequency, voltage and impedance
are constant - like a toaster or light bulb). It doesn't work well at all
for the complex signals fed to a speaker. The speaker coil also reacts quite
different than a basic resistor in that its own impedance changes radically
with changing frequencies. It's even effected by the type of enclosure it's
in. Add to that the very non-linear behavior of the crossover network
(assuming you have a midrange and tweeter) and you've got a pretty complex
system to try and characterize. So, back to my original question - why do
you want to know the speaker voltage? Perhaps you don't, though you solved
for it in your original post. Just choose speakers that have a sufficiently
high wattage rating to handle your power amp. To answer the obvious
question, then why do manufacturers show "impedance" on their speakers? That
number is what's called a "characteristic" or "nominal" impedance. Sort of
an average. It's meant to guide you for matching speaker combinations with
amp ratings. Yes, two 8 ohm speakers in parallel will show an impedance of 4
ohms to the amp. But if you put your ohmmeter across the speaker terminals
you'll get something around the 1-2 ohm range (or less for a high power
speaker) again, because the impedance changes with audio (AC) signal
applied.

Finally, you mentioned two different impedance ratings on the amplifiers. A
lot of amps can handle different speaker impedances (8, 4, and sometimes
even 2 ohms) and the power output is different for each one. Going from 8 to
4 ohm speakers usually brings a big increase in output power, almost double.
Again, check your amp's specs to see what it is - and even if it can handle
the lower impedance.

Hope this clarifies more than it confuses...

Mike Metzger

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

wrote in message
...
Thanks for the reply.

My main question was whether the two 200W amps would draw the same current
from
the power source if driven by the same input voltage. For example lets
say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V.
The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp
at
their rated impedance load.

I think both amps will draw the same current from the car
alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more
voltage
to the speaker terminals, while the 4-ohm amp will have to send more
current to
the speaker terminals.

Am I correct?

Assuming they are both putting out the same power, are driven from the same
supply, *AND* assuming the conversion efficiency is the same in both cases,
then yes the supply current would be the same regardless of load impedance.
In the real world, things may be slightly different however.

TonyP.

wrote in message
...
Thanks for the reply.

My main question was whether the two 200W amps would draw the same current
from
the power source if driven by the same input voltage. For example lets
say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V.
The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp
at
their rated impedance load.

I think both amps will draw the same current from the car
alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more
voltage
to the speaker terminals, while the 4-ohm amp will have to send more
current to
the speaker terminals.

Am I correct?

Assuming they are both putting out the same power, are driven from the same
supply, *AND* assuming the conversion efficiency is the same in both cases,
then yes the supply current would be the same regardless of load impedance.
In the real world, things may be slightly different however.

TonyP.

wrote in message
...
Thanks for the reply.

My main question was whether the two 200W amps would draw the same current
from
the power source if driven by the same input voltage. For example lets
say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V.
The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp
at
their rated impedance load.

I think both amps will draw the same current from the car
alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more
voltage
to the speaker terminals, while the 4-ohm amp will have to send more
current to
the speaker terminals.

Am I correct?

Assuming they are both putting out the same power, are driven from the same
supply, *AND* assuming the conversion efficiency is the same in both cases,
then yes the supply current would be the same regardless of load impedance.
In the real world, things may be slightly different however.

TonyP.

(Computer Prog) writes:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

(Amp 1)
Lets assume we have a 200W @ 4-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 4), V = 28.2 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 28.2v, I = 7 amps


(Amp 2)
Lets assume we have a 200W @ 8-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 8), V = 40 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 40v, I = 5 amps


So in conclusion:
(Amp 1) V = IR, 28.2v = 7a * 4ohms
(Amp 1) P = IE, P = 7a * 28.2v = 197 Watts

(Amp 2) V = IR, 40v = 5a * 8ohms
(Amp 2) P = IE, P = 5a * 40v = 200 Watts


Is this correct?

Yes.

Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.
--
% Randy Yates % "I met someone who looks alot like you,
%% Fuquay-Varina, NC % she does the things you do,
%%% 919-577-9882 % but she is an IBM."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

(Computer Prog) writes:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

(Amp 1)
Lets assume we have a 200W @ 4-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 4), V = 28.2 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 28.2v, I = 7 amps


(Amp 2)
Lets assume we have a 200W @ 8-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 8), V = 40 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 40v, I = 5 amps


So in conclusion:
(Amp 1) V = IR, 28.2v = 7a * 4ohms
(Amp 1) P = IE, P = 7a * 28.2v = 197 Watts

(Amp 2) V = IR, 40v = 5a * 8ohms
(Amp 2) P = IE, P = 5a * 40v = 200 Watts


Is this correct?

Yes.

Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.
--
% Randy Yates % "I met someone who looks alot like you,
%% Fuquay-Varina, NC % she does the things you do,
%%% 919-577-9882 % but she is an IBM."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

(Computer Prog) writes:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

(Amp 1)
Lets assume we have a 200W @ 4-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 4), V = 28.2 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 28.2v, I = 7 amps


(Amp 2)
Lets assume we have a 200W @ 8-ohm @ 14.4v mono amp.
First I want to compute voltage at the speaker terminals so I use:
Voltage = SQR(Watts*Resistance)
V = SQR(200w * 8), V = 40 Volts

Next I want current so I use:
I = Power / Volts
I = 200w / 40v, I = 5 amps


So in conclusion:
(Amp 1) V = IR, 28.2v = 7a * 4ohms
(Amp 1) P = IE, P = 7a * 28.2v = 197 Watts

(Amp 2) V = IR, 40v = 5a * 8ohms
(Amp 2) P = IE, P = 5a * 40v = 200 Watts


Is this correct?

Yes.

Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.
--
% Randy Yates % "I met someone who looks alot like you,
%% Fuquay-Varina, NC % she does the things you do,
%%% 919-577-9882 % but she is an IBM."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

In , on 02/05/04
at 01:31 PM, (Computer Prog) said:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

[ ... ]

Power is power is power. Assuming both amplifiers have similar
efficiencies, the draw from similar power sources will be the similar.
(certainly within a few percent, notice that I said "similar", not
"exactly the same")

It's true that the load current drawn by an eight Ohm load will be less
than the current drawn by a four Ohm load, but power is power. There
may be other advantages of one speaker vs. the other. For example, the
eight Ohm speaker will draw less current and you could use a smaller
wire size. If you are building thousands of systems, the $avings in
copper could be important.

One point to remember: Power amplfiers are constant voltage devices,
not constant power devices. Until you bump into some physical limit, a
given input signal will produce the same output voltage on a given
amplifier, regardless of the four or eight Ohm load.

When grapling with the issue of power draw, there is a certain "idling"
power consumed under no signal conditions. I'm not quite as optimistic
as another poster about amplifier efficiency. While 50% may be possible
at maximum power, I usually use 30% for quick estimates. If your
amplifier is a "class-D" or "switching" design, it could be in the
80-90% efficiency range (at full power).

Note: "Maximum Power" is measured a specified distortion and power
source voltage. Home amplifiers are normally specified at some very low
distortion. Automotive amplifiers are often rated at somewhat higher
distortions (sometimes 1% or 10%). This could cause the same amplifier
to be rated for much higher power when measured under automotive rules.
For most amplifiers, maximum power declines for lower source voltages.
This means that, for an automotive application, you'll have more power
available when the engine is charging the battery than when the engine
is OFF.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

In , on 02/05/04
at 01:31 PM, (Computer Prog) said:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

[ ... ]

Power is power is power. Assuming both amplifiers have similar
efficiencies, the draw from similar power sources will be the similar.
(certainly within a few percent, notice that I said "similar", not
"exactly the same")

It's true that the load current drawn by an eight Ohm load will be less
than the current drawn by a four Ohm load, but power is power. There
may be other advantages of one speaker vs. the other. For example, the
eight Ohm speaker will draw less current and you could use a smaller
wire size. If you are building thousands of systems, the $avings in
copper could be important.

One point to remember: Power amplfiers are constant voltage devices,
not constant power devices. Until you bump into some physical limit, a
given input signal will produce the same output voltage on a given
amplifier, regardless of the four or eight Ohm load.

When grapling with the issue of power draw, there is a certain "idling"
power consumed under no signal conditions. I'm not quite as optimistic
as another poster about amplifier efficiency. While 50% may be possible
at maximum power, I usually use 30% for quick estimates. If your
amplifier is a "class-D" or "switching" design, it could be in the
80-90% efficiency range (at full power).

Note: "Maximum Power" is measured a specified distortion and power
source voltage. Home amplifiers are normally specified at some very low
distortion. Automotive amplifiers are often rated at somewhat higher
distortions (sometimes 1% or 10%). This could cause the same amplifier
to be rated for much higher power when measured under automotive rules.
For most amplifiers, maximum power declines for lower source voltages.
This means that, for an automotive application, you'll have more power
available when the engine is charging the battery than when the engine
is OFF.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

In , on 02/05/04
at 01:31 PM, (Computer Prog) said:

I am trying to see how ohms law affects 2 amps of the same power, but
rated at different impedances.

[ ... ]

Power is power is power. Assuming both amplifiers have similar
efficiencies, the draw from similar power sources will be the similar.
(certainly within a few percent, notice that I said "similar", not
"exactly the same")

It's true that the load current drawn by an eight Ohm load will be less
than the current drawn by a four Ohm load, but power is power. There
may be other advantages of one speaker vs. the other. For example, the
eight Ohm speaker will draw less current and you could use a smaller
wire size. If you are building thousands of systems, the $avings in
copper could be important.

One point to remember: Power amplfiers are constant voltage devices,
not constant power devices. Until you bump into some physical limit, a
given input signal will produce the same output voltage on a given
amplifier, regardless of the four or eight Ohm load.

When grapling with the issue of power draw, there is a certain "idling"
power consumed under no signal conditions. I'm not quite as optimistic
as another poster about amplifier efficiency. While 50% may be possible
at maximum power, I usually use 30% for quick estimates. If your
amplifier is a "class-D" or "switching" design, it could be in the
80-90% efficiency range (at full power).

Note: "Maximum Power" is measured a specified distortion and power
source voltage. Home amplifiers are normally specified at some very low
distortion. Automotive amplifiers are often rated at somewhat higher
distortions (sometimes 1% or 10%). This could cause the same amplifier
to be rated for much higher power when measured under automotive rules.
For most amplifiers, maximum power declines for lower source voltages.
This means that, for an automotive application, you'll have more power
available when the engine is charging the battery than when the engine
is OFF.

-----------------------------------------------------------
spam:
wordgame:123(abc):14 9 20 5 2 9 18 4 at 22 15 9 3 5 14 5 20 dot 3 15
13 (Barry Mann)
[sorry about the puzzle, spammers are ruining my mailbox]
-----------------------------------------------------------

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

---
Frayed knot...

Look at it this way:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

+-------+ +------+
120V-----| MAGIC |-----| 200W |
| BOX | | SPKR |--- MUCHO SPL
120V-----| |-----| |
+-------+ +------+


Since the 120V mains is supplying _all_ the power into the system,

and since

P = IE

where P is power in watts,
I is current in amperes, and
E is voltage in Volts

we can rearrange

P = IE into

I = P/E

in order to solve for the current the mains has to supply in order for
the load to have 200 watts to make noise with. Don't forget that the
magic box is taking 200 watts out of the wall socket and doing whatever
it has to do to transform that power into 200 watts the speaker can use.

Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:


I = 200W/120V = 1.666...A ~ 1.7A


OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

--
John Fields

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

---
Frayed knot...

Look at it this way:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

+-------+ +------+
120V-----| MAGIC |-----| 200W |
| BOX | | SPKR |--- MUCHO SPL
120V-----| |-----| |
+-------+ +------+


Since the 120V mains is supplying _all_ the power into the system,

and since

P = IE

where P is power in watts,
I is current in amperes, and
E is voltage in Volts

we can rearrange

P = IE into

I = P/E

in order to solve for the current the mains has to supply in order for
the load to have 200 watts to make noise with. Don't forget that the
magic box is taking 200 watts out of the wall socket and doing whatever
it has to do to transform that power into 200 watts the speaker can use.

Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:


I = 200W/120V = 1.666...A ~ 1.7A


OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

--
John Fields

On Fri, 06 Feb 2004 05:45:23 GMT, "
wrote:

Thanks for the reply.

My main question was whether the two 200W amps would draw the same current from
the power source if driven by the same input voltage. For example lets say they
are both car audio amps. The first amp is rated at 200W @ 4-ohms @ 14.4V. The
second is rated 200W @ 8-ohms @ 14.4V. In this example I do run each amp at
their rated impedance load.

I think both amps will draw the same current from the car alternator/batter when
running at 200W? The difference is the 8-ohm amp will have to send more voltage
to the speaker terminals, while the 4-ohm amp will have to send more current to
the speaker terminals.

Am I correct?

---
Frayed knot...

Look at it this way:

Here's a wall socket supplying 200 watts into a magic box that's
converting that into what a speaker needs to make 200 watts worth of
noise:

+-------+ +------+
120V-----| MAGIC |-----| 200W |
| BOX | | SPKR |--- MUCHO SPL
120V-----| |-----| |
+-------+ +------+


Since the 120V mains is supplying _all_ the power into the system,

and since

P = IE

where P is power in watts,
I is current in amperes, and
E is voltage in Volts

we can rearrange

P = IE into

I = P/E

in order to solve for the current the mains has to supply in order for
the load to have 200 watts to make noise with. Don't forget that the
magic box is taking 200 watts out of the wall socket and doing whatever
it has to do to transform that power into 200 watts the speaker can use.

Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:


I = 200W/120V = 1.666...A ~ 1.7A


OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.

--
John Fields

On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

John Fields writes:

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?

John,

I took the poster's question to be "Given a *specific* power source,
pwill both 200W amps draw the same amount of current?" He also clarified
that this is what he meant in an adjacent post.

Yes, obviously, as you pointed out from the equation I = P/E, that if
E changes, then I will also change for a given P.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

John Fields writes:

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?

John,

I took the poster's question to be "Given a *specific* power source,
pwill both 200W amps draw the same amount of current?" He also clarified
that this is what he meant in an adjacent post.

Yes, obviously, as you pointed out from the equation I = P/E, that if
E changes, then I will also change for a given P.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

John Fields writes:

On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?

John,

I took the poster's question to be "Given a *specific* power source,
pwill both 200W amps draw the same amount of current?" He also clarified
that this is what he meant in an adjacent post.

Yes, obviously, as you pointed out from the equation I = P/E, that if
E changes, then I will also change for a given P.
--
% Randy Yates % "Maybe one day I'll feel her cold embrace,
%% Fuquay-Varina, NC % and kiss her interface,
%%% 919-577-9882 % til then, I'll leave her alone."
%%%% % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket
in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in
order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the
combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

- FLINT

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket
in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in
order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the
combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

- FLINT

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket
in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in
order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the
combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

- FLINT

"John Fields" wrote in message
news
On Fri, 06 Feb 2004 23:52:46 GMT, "flint"
wrote:


"John Fields" wrote in message
news
On Fri, 06 Feb 2004 13:37:43 GMT, Randy Yates wrote:


Will both 200W amps draw the same amount of current from the power
source (car battery/alternator in a car application or wall socket
in
a home application)?

Yes.

Really? That's amazing!

Considering that

P = IE

we can, without a lot of trouble, rearrange it into

I = P/E

and solve for the current required from various voltage sources in
order
to dissipate a given amount of power in a load.

For a 14V source supplying 200 watts, we can solve for the current by
writing

I = 200W/14V ~14.3A

while for 120V sourced from the mains:

I = 200W/120V ~ 1.67A

Is 14.3A equal to 1.67A?


--
John Fields

Current is not a value that can stand alone. Any current value must be
accompanied by the voltage to mean anything because power is the
combination
of voltage and current.

---
Other than the fact that current _can_ stand alone inasmuch as it's
defined as a given number of charges moving past a fixed point in a
given amount of time, you seem to have a remarkable grasp of the
obvious. Be that as it may, I'm not sure what point you're trying to
make.

Are you disputing my argument?

--
John Fields

My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

- FLINT

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:


My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

---
In many cases the voltage, or power, is immaterial. A fuse, for
example, is rated in amperes, with the voltage rating being secondary in
importance and relating only to the fuse's ability to interrupt the
plasma which will occur once the arc has been struck. Granted that the
fundamental mechanism which will cause the fusible element to melt is
the I²R loss and the subsequent heating of the link, to failure, that's
usually of little or no importance when specifying a fuse. Another
example is common hookup wire, with a size chosen to safely pass a given
current. Here, however, when selecting a wire size, some thought is
usually given to the voltage drop across a given length of wire with a
particular current flowing through it, and sometimes to the temperature
rise of the conductor.

--
John Fields

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:


My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

---
In many cases the voltage, or power, is immaterial. A fuse, for
example, is rated in amperes, with the voltage rating being secondary in
importance and relating only to the fuse's ability to interrupt the
plasma which will occur once the arc has been struck. Granted that the
fundamental mechanism which will cause the fusible element to melt is
the I²R loss and the subsequent heating of the link, to failure, that's
usually of little or no importance when specifying a fuse. Another
example is common hookup wire, with a size chosen to safely pass a given
current. Here, however, when selecting a wire size, some thought is
usually given to the voltage drop across a given length of wire with a
particular current flowing through it, and sometimes to the temperature
rise of the conductor.

--
John Fields

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:


My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

---
In many cases the voltage, or power, is immaterial. A fuse, for
example, is rated in amperes, with the voltage rating being secondary in
importance and relating only to the fuse's ability to interrupt the
plasma which will occur once the arc has been struck. Granted that the
fundamental mechanism which will cause the fusible element to melt is
the I²R loss and the subsequent heating of the link, to failure, that's
usually of little or no importance when specifying a fuse. Another
example is common hookup wire, with a size chosen to safely pass a given
current. Here, however, when selecting a wire size, some thought is
usually given to the voltage drop across a given length of wire with a
particular current flowing through it, and sometimes to the temperature
rise of the conductor.

--
John Fields

On Sat, 07 Feb 2004 02:34:40 GMT, "flint"
wrote:


My point was only in regards to power ratings. A person cannot only consider
the current draw and assume anything without also considering voltage or
resistance or Power. Since they are all inter-related, simply stating one
thing is 1 amp and another is 14 amps doesn't relate to anything. I deal
with this all the time with large systems where the tech will ask what the
current rating is and I always reply with a question asking what the supply
voltage is. They get confused because they do not understand how power, and
thus Ohm's law, works.

---
In many cases the voltage, or power, is immaterial. A fuse, for
example, is rated in amperes, with the voltage rating being secondary in
importance and relating only to the fuse's ability to interrupt the
plasma which will occur once the arc has been struck. Granted that the
fundamental mechanism which will cause the fusible element to melt is
the I²R loss and the subsequent heating of the link, to failure, that's
usually of little or no importance when specifying a fuse. Another
example is common hookup wire, with a size chosen to safely pass a given
current. Here, however, when selecting a wire size, some thought is
usually given to the voltage drop across a given length of wire with a
particular current flowing through it, and sometimes to the temperature
rise of the conductor.

--
John Fields

Hello,

Thanks for the reply, but I think you misinterpreted my question. In my example I
wanted both amps to run off the same input voltage, like a car audio application.

Amp1 - 200W @ 14.4V @ 8-ohms
Amp1 - 200W @ 14.4V @ 4-ohms

Assume that both amps are connect to their proper impedance load. Will both amps
draw the same current from the car alternator when they are producing the 200W off
the 14V input voltage?

My reason for asking this question in the first place was because my brother just
purchased a "High Output" stereo upgrade for his new Harley Davidson Electra Glide
Ultra Classic motorcycle. This setup comes with an external amp and 4 coaxial
speakers. The Harley dealer is telling him the external amp is rated 50x4 @ 8-ohms.
This sounds wrong to me since I cannot see how a motorcycle amplifier would be rated
@ 8-ohms since that would require more voltage at each speaker and we are only
starting with ~ 13V at the input power input. I wanted to know if a 200W 8-ohm 14V
(input voltage) amp would draw less current than a 200 Watt 4-ohm 14V (input voltage)
amp. I think the answer is no.

Here is a link to the Harley part:

http://www.harley-davidson.com/gma/g...bmLocale=en_US

Thanks.



Knowing that the magic box is only going to take 200 watts out of the
wall because that's all the speaker wants, we can plug in the power and
the voltage the magic box needs to do that (into the formula) and solve
for the current:

I = 200W/120V = 1.666...A ~ 1.7A

OK, so now what happens if we plug the magic box into a 14V supply and
the speaker still wants 200 watts? Well, the magic box has no choice
but to get whatever current it needs from the 14V supply to make that
happen, so if we use our I = P/E formula again it comes out:

I = P/E = 200W/14V = 14.286A ~ 14.3A

So, if you've got a 200 watt load, it doesn't make any difference what
its impedance is, the primary power source has to supply that power, so
the current it'll have to supply will depend on its voltage.