I have a mini stereo that comes with speakers rated at 4 ohms. I want to hook up the stereo to the in-wall speakers in the bathroom, which are rated at 8 ohms. Is this doable as-is? If not, are there any (reasonable) means of making it so? Thanks for any advice!

No prpblem. Just won't be as loud.

Issue of impedance matching.. in your case would not allow for maximum power transfer.. as KeepIt said.. would not be as loud.

Actually audio amps are not transmission lines so impedance matching is irrelevant, In fact there is a term that is related to audio amplifier output impedance and that is "damping factor".

If an amp has a damping factor of 100 and a rated load impedance of 8 ohms. The actual output impedance of the amp is 8/100 or 0.08 ohms. This actually describes an amp I built from scratch.

An amp is constrained by V or I (current). An amp designed for a 4 ohm load places a higher I constraint and a lower V constraint than an amplifier designed for 8 ohm speakers. P is roughly (I^2)*R If R is taken as Z and a constant, and since the available I is lower then the maximum power to the speaker will be lower.

You learn a lot when you build one from scratch and you fixed them for a while.

If you wanted your amp to See 4 ohms, just hook 2 8 ohm speakers in parallel(+ to +) and (- to -).

I'll test this by hooking the stereo up to a pair of old bookshelf speakers I have lying around. I believe those are 8 ohm units. I don't mind the bathroom speakers not being "loud enough" as long as they work. Don't plan to "sit" through an iTunes playlist, you know. I appreciate all the advice! :)

The formula for resistances in parallel is:

1/Rt = 1/R1 + 1/R2+... 1/Rn

Two 8's in parallel is 4.

Qoute "The formula for resistances in parallel is:
1/Rt = 1/R1 + 1/R2+ 1/Rn"
Or, the Reciprocal of the sum of Reciprocals, it is good for Resistors in Parallel, or Capacitors in Series.