Overcoming resistance (as in "Ohm"s)

[ChrisGreaves]

Just checking up on my physics math here ...

Having won a Noresco 4-264 Stereo (Quadraphonic) tuner/amp out of the recycling room up the street, I'd be a fool not to go back and keep an eye out for speaker enclosures ditto, right?
Sure enough last night a Yamaha NS-AP100 is sitting there, 9" x 6" x 5".
Pretty. (But where is its mate?)

The speaker label reads "Impedance 6 ohms"
The amplifier reads "8 ohms minimum"

If I remember Mr Puzey's lectures correctly, the resistance in parallel is the reciprocal of the sum of the reciprocals.
So if I have my 6 ohm speaker and another 8 ohm speaker, I divide "one" by (1/6)+(1/8)
So I divide "one" by (8/48)+(6/48)
So I divide "one" by (14/48)
To arrive at about 3.5 ohm for the two speakers in parallel.

If I then couple this set up with another 8 ohm speaker in series I'll get a resistance of 8+3.5 = 11.5 ohms.
Which should satisfy one speaker outlet ("8 ohms minimum") on the amplifier.
Correct?

[Leif]

Yes.

But I don't think you can expect to get an equal volume out of each speaker...

[StuartR]

Approximately,

These numbers are IMPEDANCE, not resistance. That means that they vary with frequency and sum in a somewhat more complex way.

[ChrisGreaves]

Yes. But I don't think you can expect to get an equal volume out of each speaker...

Thanks, Leif for the confirmation.
FWIW I'm a real "low-volume" guy, the music is so soft that most visitors don't notice it for a while.
My plan is to maintain an amplifier in each room, so that each room will have a volume (control) between the signal and the speaker.

[ChrisGreaves]

Approximately, These numbers are IMPEDANCE, not resistance. That means that they vary with frequency and sum in a somewhat more complex way.

Thanks Stuart.
From which I infer that the current in the speaker circuits is AC, not DC, right?

http://en.wikipedia.org/wiki/Electrical_impedance
http://in.answers.yahoo.com/question/index?qid=20090819085746AAmQLnX
http://answers.yahoo.com/question/index?qid=20070726034233AAKUeho
http://www.physlink.com/education/askexperts/ae517.cfm

[StuartR]

...
From which I infer that the current in the speaker circuits is AC, not DC, right?
...

Absolutely spot on. And the frequency of the AC is the same as the frequency of the sound waves that will be produced by the speakers.

[ChrisGreaves]

And the frequency of the AC is the same as the frequency of the sound waves that will be produced by the speakers.

As in FRequency Modulation?

[StuartR]

...
As in FRequency Modulation?

NO. Frequency Modulation is used to superimpose an audio wave on a radio frequency signal, so it can be transmitted. It requires decoding to extract the audio signal.
The frequency in the speaker wires is JUST the sound frequency, it causes the speaker cone to oscillate at that frequency, directly generating the sound.

[HansV]

ResistanceIsFutile.gif

[ChrisGreaves]

The frequency in the speaker wires is JUST the sound frequency,

Thanks for your patience, Stuart.

So if you were trying to explain this to a stale mushroom, that had spent the last year in the bottom of the 'fridge, (grin), could you say, very guardedly, that in this case the frequency corresponds to the wiggle in the groove of an old gramophone record?

[ChrisGreaves]

ResistanceIsFutile.gif

Oh, I don't know ...

Spoiler

And as we all know, there's a lot that I don't know!

[StuartR]

...in this case the frequency corresponds to the wiggle in the groove of an old gramophone record?

Exactly. Yes.

[ChrisGreaves]

Exactly. Yes.

Thanks so much.
I think I'll quit dumpster-diving for a few days.
I think that'd make you happy too!

[HansV]

Is that you, Chris?

dumpster-dive.jpg

[ChrisGreaves]

Is that you, Chris?

dumpster-dive.jpg

Close. A true resemblance would have one of my few remaining friends standing nearby with a long-suffering look of patience ...

[stuck]

ResistanceIsFutile.gif

never had you down as Borg

[Leif]

never had you down as Borg

There's one Björn every minute.