To determine the appropriate resistance of the potentiometer that should be placed between the 2-ohm output source and the 10k-ohm input power amplifier, we need to consider the voltage divider formula:
Vout = Vin x (Rpot / (Rpot + Rin))
Where:
- Vout is the output voltage of the potentiometer
- Vin is the input voltage to the potentiometer
- Rpot is the resistance of the potentiometer
- Rin is the resistance of the load (in this case, the input impedance of the power amplifier)
We want to find a potentiometer value that will give us a suitable output voltage for the power amplifier, while also ensuring that the potentiometer and amplifier are properly matched in terms of impedance.
Assuming a standard line level signal with a peak voltage of 2V, the output impedance of the 2-ohm source is negligible compared to the load impedance of the power amplifier, so we can ignore it for this calculation.
If we want to maximize the signal level while minimizing the load on the source, we would want to choose a potentiometer resistance that is close to the load impedance of the power amplifier (10k-ohm). Therefore, a potentiometer value of 10k-ohm would be appropriate in this case.
However, if we want to minimize the noise and distortion in the system, we may want to choose a higher value potentiometer to increase the impedance seen by the source, which can reduce the loading effect and help maintain a cleaner signal. In this case, a 50k-ohm or 100k-ohm potentiometer may be more suitable, as long as the total resistance (potentiometer plus load) does not become too high and cause excessive signal attenuation.
Overall, the choice of potentiometer resistance will depend on your specific requirements for signal level, impedance matching, and noise/distortion performance.