Ohm's law question

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If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
and I wire them in parallel, then put 100 watts through them, how many
watts will go through each driver?
I assume it is not merely 33, 67, respectively.

--Bryan
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

anahata wrote:
> CLASSACT@BRICK.NET wrote:
> > If I have two drivers, one of which is 8 ohm,
> > the other being 4 ohm,
> > and I wire them in parallel,
> > then put 100 watts through them, how many
> > watts will go through each driver?
> > I assume it is not merely 33, 67, respectively.
>
> Why not? At an output level where the total is
> 100W the 4 ohm will get 67 and the 8 ohm will get 33.
> (roughly - the quoted impedances are only nominal)
>
> Beware that the amplifier is seeing a load of
> 2.7 ohms when you do that!

Thank you. The power amp is rated for 8, 4 or 2 ohm operation, but I
am aware that only 2.7 ohms means don't turn the volume knob up past
about 5 or 6.
>
> --
> Anahata

--Bryan
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

Yes, it would be 33 Watts for the 8 ohm driver and 67 Watts for the 4
ohm driver. Here's a proof (using resistances instead of impedances):

The 100 Watts is going to see the 8 ohm and 4 ohm resistor as one
resistive load. The equation for total resistance for two resistors in
parallel is 1/Rt=1/Ra + 1/Rb. Therefore the total resistance of these
two resistors is 8/3 ohms (or 2.67 ohms). Notice it's less than either
resistor separately.

So through P=IV and ohms law of V=IR, you can get P=(Isquared)R.
Therefore, you can find the total current. 100 Watts = (Isquared)(2.67)
I come up with current total of 6.12 amps. Here I'm treating the two
loads as one.

We can then go back and find the total voltage across each resistor with
P=IV. 100 watts = (6.12 amps) V. Total volatge therefore equals 16.34
volts. And of course the resistors have the same voltage across them
since they are in parallel.

Since P=IV, you also can state with ohms law that P=(Vsquared)/R.
Therefore the power across the 8 ohms resistor is P=(16.34squared)/8
which equals 33. And the power across the 4 ohm resistor is
P=(16.34squared)/4 which is 67.

You could also go back and prove the same relationship by finding the
current through each resistor individually.

This proof is overly long, but I wanted to walk you through it. There
are other ways of arriving at the same answer, of course. Drivers are
of course theoretically inductive loads instead of resistive loads, but
in this theorectical case the calculations are the same.

In the real world, of course, speakers have both a resistive and
reactive component which gets you into power factors at specific
frequencies, but I don't think it's worth getting into that here. Just
know that each speaker will affect the frequency response of the other
because a complex impedance varies throughout the frequency range. I'm
guessing this is a trivial difference, but someone can correct me here
if I'm wrong. Anyway, people mostly match their speakers for this very
reason.

WARNING: The real issue here (as you may see) is that you are putting
an effective 2.67 load across your amp. Most amps are only rated for a
4 ohm load or higher. So while your drivers may be fine and be high
enough wattage, your amp may melt down becuase you are drawing too much
current from it.

Cheers,
Trevor de Clercq

CLASSACT@BRICK.NET wrote:
> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
> and I wire them in parallel, then put 100 watts through them, how many
> watts will go through each driver?
> I assume it is not merely 33, 67, respectively.
>
> --Bryan
>
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

Thanks. I think I'll play it safe and just go with the smaller 8 ohm
mains. Much more idiot proof.

--Bryan
 
Archived from groups: rec.audio.pro (More info?)

In article <1108909902.384045.259100@f14g2000cwb.googlegroups.com> CLASSACT@BRICK.NET writes:

> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
> and I wire them in parallel, then put 100 watts through them, how many
> watts will go through each driver?
> I assume it is not merely 33, 67, respectively.

Sure it is. That's how Ohm's Law works.

But you may not actually be able to put 100 watts into them, if you're
basing that on the fact that you have a "100 W" amplifier. The
parallel combination of the two speakers is 2.7 ohms. Many amplifiers
can't put out their full rated power into an impedance that low. Some
can. Some can put out more than their nominal 4-8 ohm rated power.



--
I'm really Mike Rivers (mrivers@d-and-d.com)
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me here: double-m-eleven-double-zero at yahoo
 
Archived from groups: rec.audio.pro (More info?)

> But you may not actually be able to put 100 watts into them, if you're
> basing that on the fact that you have a "100 W" amplifier.The
> parallel combination of the two speakers is 2.7 ohms. Many amplifiers
> can't put out their full rated power into an impedance that low. Some
> can. Some can put out more than their nominal 4-8 ohm rated power.

Krells do. The bigger ones keep doubling their output with each halving of the
load impedance, down to 1 ohm. (At least, that's what the specs state.)
 
Archived from groups: rec.audio.pro (More info?)

Mike Rivers wrote:

>
> Sure it is. That's how Ohm's Law works.

Since we are talking about AC here, doesn't the applied frequency also
effect these numbers some, and the balance between differently rated
speakers?
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

CLASSACT@BRICK.NET wrote:
> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
> and I wire them in parallel, then put 100 watts through them, how many
> watts will go through each driver?
> I assume it is not merely 33, 67, respectively.

Why not? At an output level where the total is 100W the 4 ohm will get
67 and the 8 ohm will get 33. (roughly - the quoted impedances are only
nominal)

Beware that the amplifier is seeing a load of 2.7 ohms when you do that!

--
Anahata
anahata@treewind.co.uk -+- http://www.treewind.co.uk
Home: 01638 720444 Mob: 07976 263827
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

CLASSACT@BRICK.NET wrote:

> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
> and I wire them in parallel, then put 100 watts through them, how many
> watts will go through each driver?
> I assume it is not merely 33, 67, respectively.

Um, oh gosh! Let's see, the total R (Rt) presented to the amp is easy,
since:

1/Rt = 1/R1 + 1/R2, or Rt = R1R2/(R1 + R2), or Rt = 8*4/(8+4) = 2.67 Ohms

But the 100W rating of the amp is into a specified load, maybe 8 Ohms,
maybe 4 Ohms, maybe less, dunno w/o looking at it's specs! You can *get*
more P out of any amp, believe me I know! The power rating on amplifiers
is into a specific load at a specific THD rating. Overdrive the amp and
it'll do it, but it will produce a signal of a lot more THD! Also driving
a lower impedance load will cause the amp to put out more than the rated
power, always.

Are you trying not to overdrive the speaker drivers, to avoid blowing the
cones? Contact me if it's a real concern. After all Bryan, I *am* a EE by
degree! ;-)

John_Kuthe...
 
Archived from groups: rec.audio.pro (More info?)

John Kuthe wrote:
> CLASSACT@BRICK.NET wrote:
>
>
>>If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
>>and I wire them in parallel, then put 100 watts through them, how many
>>watts will go through each driver?
>>I assume it is not merely 33, 67, respectively.
>
>
> Um, oh gosh! Let's see, the total R (Rt) presented to the amp is easy,
> since:
>
> 1/Rt = 1/R1 + 1/R2, or Rt = R1R2/(R1 + R2), or Rt = 8*4/(8+4) = 2.67 Ohms
>
> But the 100W rating of the amp is into a specified load, maybe 8 Ohms,
> maybe 4 Ohms, maybe less, dunno w/o looking at it's specs! You can *get*
> more P out of any amp, believe me I know! The power rating on amplifiers
> is into a specific load at a specific THD rating. Overdrive the amp and
> it'll do it, but it will produce a signal of a lot more THD! Also driving
> a lower impedance load will cause the amp to put out more than the rated
> power, always.


One might add that the amplifier may not be designed to work into the
2.7-ohm load, so using it that way for very long (more power = much less
time) could overheat it and cause it to fail.
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

Good assumption, however.............
The total load for an 8 and 4 ohm driver in parrallel is actually 2.67
Ohms...............
So, Your amp may be putting out more than 100watts with that load.

Greg Dwinell
Orange Sound



<CLASSACT@BRICK.NET> wrote in message
news:1108909902.384045.259100@f14g2000cwb.googlegroups.com...
> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
> and I wire them in parallel, then put 100 watts through them, how many
> watts will go through each driver?
> I assume it is not merely 33, 67, respectively.
>
> --Bryan
>
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

Pan paniscus wrote:
> anahata wrote:
>
>>CLASSACT@BRICK.NET wrote:
>>
>>>If I have two drivers, one of which is 8 ohm,
>>>the other being 4 ohm,
>>>and I wire them in parallel,
>>>then put 100 watts through them, how many
>>>watts will go through each driver?
>>>I assume it is not merely 33, 67, respectively.
>>
>>Why not? At an output level where the total is
>>100W the 4 ohm will get 67 and the 8 ohm will get 33.
>>(roughly - the quoted impedances are only nominal)
>>
>>Beware that the amplifier is seeing a load of
>>2.7 ohms when you do that!
>
>
> Thank you. The power amp is rated for 8, 4 or 2 ohm operation, but I
> am aware that only 2.7 ohms means don't turn the volume knob up past
> about 5 or 6.
>
>
the "volume" knob only affects input sensitivity
it has no effect on the amp output or load
the output section of your amp has no idea what position the "volume "
control is in
You can still drive the amp deep into overload even witht he "volume"
knob at 1, with a sufficently strong input
 
Archived from groups: rec.audio.pro (More info?)

Even if you go ahead, ignore all threaded advice, and parallel the two
different ohms speakers, and the 100W amplifier survives, I reckon the
four-ohm one will drown out the eight-ohm one by its very loudness!
Are you by any chance hiding from us curious that you have 2 pairs of each
type (stereo)?
Jim
"Joe Sensor" <crabcakes@emagic.net> wrote in message
news:37s88lF5f0snuU1@individual.net...
> Mike Rivers wrote:
>
>>
>> Sure it is. That's how Ohm's Law works.
>
> Since we are talking about AC here, doesn't the applied frequency also
> effect these numbers some, and the balance between differently rated
> speakers?
 
Archived from groups: rec.audio.pro (More info?)

Jim Gregory wrote:

> Even if you go ahead, ignore all threaded advice, and parallel the two
> different ohms speakers, and the 100W amplifier survives, I reckon the
> four-ohm one will drown out the eight-ohm one by its very loudness!

Assuming equal efficiency, the difference between the two is 3 dB.
 
Archived from groups: rec.audio.pro (More info?)

In article <37s88lF5f0snuU1@individual.net> crabcakes@emagic.net writes:

> Since we are talking about AC here, doesn't the applied frequency also
> effect these numbers some, and the balance between differently rated
> speakers?

Yes, but you have to work with nominal numbers if you want to ever get
the job done. No 8 ohm speaker is really 8 ohms except maybe at a
couple of frequencies, and it can change with the cabinet design (how
that affects the resonance and damping of the speaker).

--
I'm really Mike Rivers (mrivers@d-and-d.com)
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me here: double-m-eleven-double-zero at yahoo
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

On Sun, 20 Feb 2005 11:18:31 -0500, Trevor de Clercq
<declerct@REMOVETHISnewschool.edu> wrote:

>Yes, it would be 33 Watts for the 8 ohm driver and 67 Watts for the 4
>ohm driver. Here's a proof (using resistances instead of impedances):

>...

>This proof is overly long, but I wanted to walk you through it. There
>are other ways of arriving at the same answer, of course. Drivers are
>of course theoretically inductive loads instead of resistive loads, but
>in this theorectical case the calculations are the same.
>
>In the real world, of course, speakers have both a resistive and
>reactive component which gets you into power factors at specific
>frequencies, but I don't think it's worth getting into that here. Just
>know that
> each speaker will affect the frequency response of the other
^^^^ ^^^^^^ ^^^ ^^^^^^^^^ ^^^^^^^^ ^^ ^^^ ^^^^^
>because a complex impedance varies throughout the frequency range. I'm
>guessing this is a trivial difference, but someone can correct me here
>if I'm wrong. Anyway, people mostly match their speakers for this very
>reason.

Just to point out a mistake in an otherwise excellent post (someone
will come along and point out that correcting others' posts is the
purpose of Usenet...), two important assumptions are that 1. the
speakers are in parallel, as you said earlier, and 2. the amplifier
has such a high damping factor (or equivalently, low output impedance)
that it doesn't matter what the load is on it (within its rating,
which the OP later said goes down to 2 ohms), so that whatever
changing impedance one speaker has in relation to frequency, it won't
affect the voltage to the other speaker, thus its acoustic output
won't be affected by the other speaker.
The rest of this post shows you know what you're talking about, so
I can only guess you either forgot the word "not" in the above, or you
somehow got mentally turned around and were thinking the speakers were
wired in series.


>WARNING: The real issue here (as you may see) is that you are putting
>an effective 2.67 load across your amp. Most amps are only rated for a
>4 ohm load or higher. So while your drivers may be fine and be high
>enough wattage, your amp may melt down becuase you are drawing too much
>current from it.
>
>Cheers,
>Trevor de Clercq
>
>CLASSACT@BRICK.NET wrote:
>> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
>> and I wire them in parallel, then put 100 watts through them, how many
>> watts will go through each driver?
>> I assume it is not merely 33, 67, respectively.
>>
>> --Bryan

-----
http://mindspring.com/~benbradley
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

> <CLASSACT@BRICK.NET> wrote in message
> news:1108909902.384045.259100@f14g2000cwb.googlegroups.com...
>> If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
>> and I wire them in parallel, then put 100 watts through them, how many
>> watts will go through each driver?
>> I assume it is not merely 33, 67, respectively.

Never assume. It is indeed "merely 33, 67 respectively." Since the
voltage across both dirvers is the same, the power dissipated will be
inversely proportional to the resistance. Since the total power is 100
watts, 1/3 will go to the 8 ohm driver and 2/3 to the 4 ohm driver.

Norm Strong
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

<normanstr...@comcast.net> wrote:
> > <CLASSACT@BRICK.NET> wrote in message
> > news:1108909902.384045.259100@f14g2000cwb.googlegroups.com...
> >> If I have two drivers, one of which is 8 ohm, the other being 4
ohm,
> >> and I wire them in parallel, then put 100 watts through them, how
many
> >> watts will go through each driver?
> >> I assume it is not merely 33, 67, respectively.
>
> Never assume. It is indeed "merely 33, 67 respectively." Since the

> voltage across both dirvers is the same, the power dissipated will be

> inversely proportional to the resistance. Since the total power is
100
> watts, 1/3 will go to the 8 ohm driver and 2/3 to the 4 ohm driver.

Thanks for your help, but I got lucky today and now own four B-52
cabinets with an 18 and horn, so it'll be four nice safe ohms per side.
Super deal. At Guitar Center's Presidents Day sale I got all 4 for
under $900! Together with the QSC RMX-2450 we should be nice and loud
>
> Norm Strong

--Bryan
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

On Sun, 20 Feb 2005 11:52:14 -0800, Pan paniscus wrote:

> Thanks. I think I'll play it safe and just go with the smaller 8 ohm
> mains. Much more idiot proof.

They'll always build the better idiot.

- dsb


--
"My doctor says that I have a malformed public-duty gland and a natural
deficiency in moral fibre, and that I an therefore excused from saving
Universes." -- Ford Prefect
 
Archived from groups: alt.punk,rec.audio.pro (More info?)

Ben,

I guess I was just not assuming that the amp was for all practical
purposes an ideal voltage source. My post was written before we knew
that the amp was rated for operation down to two ohms. As I said, it
would be a trivial difference, but perhaps it is so trivial I shouldn't
even have bothered mentioning it. Without knowing the output impedance
of the amp, though, it's tough to say 100%. Of course most modern amps
are designed such that their output impedance is low enough where it's
not a factor. But you are right, I should have left that part of my
post out. Why unnecessarily overcomplicate the issue? Sometimes I
overthink problems....

Cheers,
Trevor de Clercq

Ben Bradley wrote:
> On Sun, 20 Feb 2005 11:18:31 -0500, Trevor de Clercq
> <declerct@REMOVETHISnewschool.edu> wrote:
>
>
>>Yes, it would be 33 Watts for the 8 ohm driver and 67 Watts for the 4
>>ohm driver. Here's a proof (using resistances instead of impedances):
>
>
>>...
>
>
>>This proof is overly long, but I wanted to walk you through it. There
>>are other ways of arriving at the same answer, of course. Drivers are
>>of course theoretically inductive loads instead of resistive loads, but
>>in this theorectical case the calculations are the same.
>>
>>In the real world, of course, speakers have both a resistive and
>>reactive component which gets you into power factors at specific
>>frequencies, but I don't think it's worth getting into that here. Just
>>know that
>> each speaker will affect the frequency response of the other
>
> ^^^^ ^^^^^^ ^^^ ^^^^^^^^^ ^^^^^^^^ ^^ ^^^ ^^^^^
>
>>because a complex impedance varies throughout the frequency range. I'm
>>guessing this is a trivial difference, but someone can correct me here
>>if I'm wrong. Anyway, people mostly match their speakers for this very
>>reason.
>
>
> Just to point out a mistake in an otherwise excellent post (someone
> will come along and point out that correcting others' posts is the
> purpose of Usenet...), two important assumptions are that 1. the
> speakers are in parallel, as you said earlier, and 2. the amplifier
> has such a high damping factor (or equivalently, low output impedance)
> that it doesn't matter what the load is on it (within its rating,
> which the OP later said goes down to 2 ohms), so that whatever
> changing impedance one speaker has in relation to frequency, it won't
> affect the voltage to the other speaker, thus its acoustic output
> won't be affected by the other speaker.
> The rest of this post shows you know what you're talking about, so
> I can only guess you either forgot the word "not" in the above, or you
> somehow got mentally turned around and were thinking the speakers were
> wired in series.
>
>
>
>>WARNING: The real issue here (as you may see) is that you are putting
>>an effective 2.67 load across your amp. Most amps are only rated for a
>>4 ohm load or higher. So while your drivers may be fine and be high
>>enough wattage, your amp may melt down becuase you are drawing too much
>>current from it.
>>
>>Cheers,
>>Trevor de Clercq
>>
>>CLASSACT@BRICK.NET wrote:
>>
>>>If I have two drivers, one of which is 8 ohm, the other being 4 ohm,
>>>and I wire them in parallel, then put 100 watts through them, how many
>>>watts will go through each driver?
>>>I assume it is not merely 33, 67, respectively.
>>>
>>>--Bryan
>
>
> -----
> http://mindspring.com/~benbradley