Choosing a resistance for volume (input side) potentiometer
I have an amp that delivers 30W to each of the two (2) channels. Each channel contains a crossover composed of a 6 ohm and a 4 ohm speaker in parallel with each other.


That's all irrelevant other than you have a stereo power-amplifer. The potentiometer will be going to the input. The output power doesn't matter.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Your setup.


  To add a simple volume knob between the source (a phone or iPod) and the input jack of the amp, what is the logical process for determining a resistance value, and type (Lin/Log)?



The phone and iPod are designed to drive earbuds in the range 8 to 32 Ω. Using the 10:1 rule any pot over 80 Ω would not load the output significantly.
The amplifier input is typically >= 100 kΩ (but check the specifications). To prevent it loading the potentiometer too much we'll drive it with a source impedance of &lt;= 10 kΩ.
Due to the logarithmic response of the human ear to sound pressure levels a logarithmic potentiometer is required to get a perceived even increase in volume with rotation.


A 1 kΩ to 10 kΩ pot should be fine.

If you use too high a value for the potentiometer the amplifier input impedance will load it excessively and distort the volume curve. See Electrically loaded resistive Potentiometer! for more on the subject.