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I have my old Technics SU V560 amp that I used to drive Wharfedale E20 speakers in the living room.

The amp has two stereo outputs, selectable from buttons on the front.

Recently I was given a set of Kef Model 104 AB speakers which sound nicer than the Wharfedales so I use the Kefs in the living room and the Wharfedales in the dining room.

Both sets of speakers are 8 Ω impedance but the Wharfedales are much, much louder than the Kefs. So much so that if I set the volume for the Wharfedales then the Kefs are very quiet.

There is only one volume control, so I can't change the volume of each speaker set individually.
Both sets of speakers have exactly the same type of speaker cable, all cable are the same length.

• How do I reduce the volume of the Wharfedales?

I've tried switching the outputs around but this hasn't made any difference. I also bought several 2 Ω 50 W resistors in the hope that putting them in series with the Wharfedales would quieten them down but they didn't - even linking them in series to a total of 10 Ω seemed to have no effect.

Any ideas out there?

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  • \$\begingroup\$ the "E" in E20 stands for efficiency; they are made to be loud with modest amplification. don't the wharfedales have a volume knob on each speaker, under the front grille? Do they support bi-wiring? You can also change the placement to soften the sound: fire them to the ceiling or into the corner; you want ambient sound in the the dining room, not critical listening perfect imaging. \$\endgroup\$
    – dandavis
    Commented Jun 18, 2020 at 16:23

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From a quick search, the Wharfedale E20 have 95 dB at 1W, 1m sensitivity. That's quite high. The Kef on the other hand have lower sensitivity, 85dB at 1W, 1m. So the latter will produce 10x less acoustic power with the same input power (watts). So there is no actual problem, just different speaker efficiencies.

If you add a resistor in series with the more efficient loudspeaker, you're making a voltage divider with R2 being the loudspeaker and R1 being the resistor.

enter image description here

Thus, voltage on the loudspeaker is given by the usual voltage divider formula:

\$ V_{out} = V_{in} \frac{R2}{R1 R2} \$

Problem: here's an impedance graph for a random "4 ohm" speaker from the internet...

enter image description here

As you can see, speaker impedance varies substantially with frequency, and thus the divider ratio will do the same. So the tone balance of the loudspeaker will be affected, and it will sound different. You can still try it though, for example a 68 ohm resistor of suitable wattage, like 5-10 watts.

Another option would be to get another amplifier and use an amp per pair of loudspeakers, so you get two independent volume knobs.

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    \$\begingroup\$ Thank you, I'll try the resistor solution - buying another amp sounds a bit of an expensive option, plus I don't really have the space. \$\endgroup\$
    – doug
    Commented Jun 19, 2020 at 11:02
  • \$\begingroup\$ So if there were 10 2 Ω power resistors to begin with, make R1 6 Ω and R2 (parallel to one Wharfedale) 4 Ω for maximum effect without further investment. \$\endgroup\$
    – greybeard
    Commented Jun 16 at 6:42

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