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This question is like this one. I have two 1 ohm resistors, six 2 ohm resistors, and want 2 ohms total. All resistors must be powered (since they are actually speakers.) How do I do this? Ideally, a systematic way, or a computer algorithm should be presented to do this, but I am mostly concerned with the above problem.

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    \$\begingroup\$ Are you sure your speakers have impedances of 1 and 2 Ohms? That is possible, but would be unusual. The vast majority of speakers are either 4 or 8 Ohms. \$\endgroup\$ May 28, 2014 at 14:07
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    \$\begingroup\$ This looks like a homework problem, with the "speakers" used as a justification for them all being "powered". I can see at least one way to do it (with many trivial re-arrangements), given no constraint for them to be equally powered. Just play around with series-parallel resistances until you find a way. Hint: The one I came up with is 4||4. \$\endgroup\$ May 28, 2014 at 14:12
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    \$\begingroup\$ How do you want the input power distributed among the speakers? Should they all get the same voltage, the same current, the same power, or some other distribution altogether? \$\endgroup\$
    – Dave Tweed
    May 28, 2014 at 14:15
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    \$\begingroup\$ Wait, should I mark it as homework? It is an intermediate part of a homework problem. (Design a car audio system.) \$\endgroup\$
    – PyRulez
    May 28, 2014 at 16:15
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    \$\begingroup\$ Possible duplicate of Tool, algorithm or method to know which resistors to use for an equivalent resistance \$R_T\$? \$\endgroup\$ Oct 5, 2017 at 8:45

1 Answer 1

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I'd use a transformer like this: -

schematic

simulate this circuit – Schematic created using CircuitLab

The transformer turns ratio would convert the 3 ohms of the inter-wired 6x 2 ohms speakers to 1.5 ohm seen on the primary. This, being in series with 2x parallel 1ohm speakers makes a total impedance of 2 ohms seen by the voltage source.

The turns ratio is 1:\$\sqrt2\$ by the way

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  • \$\begingroup\$ @SpehroPefhany nearly... if 1 v rms applied, the two 1 ohm speakers would share 62.5 mW and the six 2 ohm speakers would share 562.5 mW. Of course they could all have been wired in series and use a transformer to get 2 ohms!! \$\endgroup\$
    – Andy aka
    May 28, 2014 at 14:37
  • \$\begingroup\$ If 1A RMS is applied, the 1 ohm speakers share 500mW so 250mW each. The other speakers share 1.5W six ways, so 250mW each. Or did I make some stupid mistake? \$\endgroup\$ May 28, 2014 at 14:44
  • \$\begingroup\$ @SpehroPefhany my brain is starting to ache LOL. Your explanation sounds perfect so WTF is wrong with mine.... thinking if there is 1V applied, there will be 0.25 volts across the || 1ohm speakers and 0.75 volts across the network of 2 ohm speakers. Square the ratio of voltages to give power and this comes to 9:1?!?!?! It's been a long day!! \$\endgroup\$
    – Andy aka
    May 28, 2014 at 14:52
  • \$\begingroup\$ @Andyaka Okay, 0.25V is 62.5mW each. 0.75V at the primary of the transformer, which is 1.5 ohms looking in, so 375mW divide by six is 62.5mW each. Some guys (like you) can come up with a perfect design intuitively! \$\endgroup\$ May 28, 2014 at 14:57
  • \$\begingroup\$ @spehro. I did design it to be equal power but got into a muddle analysing it and didn't bother mentioning it. Then you said it's equal and that's when my brain went wrong proper lol. Back home and one long steady drive later and it all makes sense! \$\endgroup\$
    – Andy aka
    May 28, 2014 at 16:49

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