Here is my theory as to why they make a difference, with analysis and measurements.
Amplifiers demand current from the power-line when the capacitors in their power-supplies become momentarily discharged due to high-current transients in the music signal. This discharge condition must be quickly recharged from the power-line, through the power-supply transformer, or a voltage sag will occur. Such voltage sags can cause audible distortion at the loudspeakers. If the power-line has significant series inductance in the path from the power panel to the amplifier, this can prevent the capacitor bank from recharging in time to prevent a voltage sag from occurring at the amplifier output transistors. With a low-inductance cable, the voltage drop across the cable will be insignificant during high-current transients, minimizing the voltage sag. This allows all of the current needed by the output transistors to be supplied when they need it, resulting in fast, dynamic response to transient signals.
A typical 6-foot 14 AWG rubber cord and 25 feet of ROMEX has inductance of 7.2 uH and resistance of 235 mohms, ignoring the plug resistance effect. Therefore, the voltage drop at 20kHz will be I*(wL+R)= I*(.905+.235) = I*(1.14). With a 6-foot low-inductance cord and 25 feet of ROMEX, the inductance is 5.9 uH and the total resistance is 147 mohms. This is an 18% reduction in inductance and a 37% reduction in resistance. The voltage drop for this combination will be I(wL+R) = I(.741+.147) = I(.888). So at a fixed dynamic current I, the voltage drop in the entire power feed at 20kHz is 22% smaller with the low-inductance power cord. I would consider 22% to be significant. The reality is even more compelling. When you add in lower plug and receptacle resistance and the fact that the di/dt on the power cord will have spectra well above 20kHz with some amplifiers, the low-inductance cord makes an even bigger difference.
And please don't give me a lot of flack about 60Hz current in power cords. The currents are very high in frequency - just measure them...
Amplifiers demand current from the power-line when the capacitors in their power-supplies become momentarily discharged due to high-current transients in the music signal. This discharge condition must be quickly recharged from the power-line, through the power-supply transformer, or a voltage sag will occur. Such voltage sags can cause audible distortion at the loudspeakers. If the power-line has significant series inductance in the path from the power panel to the amplifier, this can prevent the capacitor bank from recharging in time to prevent a voltage sag from occurring at the amplifier output transistors. With a low-inductance cable, the voltage drop across the cable will be insignificant during high-current transients, minimizing the voltage sag. This allows all of the current needed by the output transistors to be supplied when they need it, resulting in fast, dynamic response to transient signals.
A typical 6-foot 14 AWG rubber cord and 25 feet of ROMEX has inductance of 7.2 uH and resistance of 235 mohms, ignoring the plug resistance effect. Therefore, the voltage drop at 20kHz will be I*(wL+R)= I*(.905+.235) = I*(1.14). With a 6-foot low-inductance cord and 25 feet of ROMEX, the inductance is 5.9 uH and the total resistance is 147 mohms. This is an 18% reduction in inductance and a 37% reduction in resistance. The voltage drop for this combination will be I(wL+R) = I(.741+.147) = I(.888). So at a fixed dynamic current I, the voltage drop in the entire power feed at 20kHz is 22% smaller with the low-inductance power cord. I would consider 22% to be significant. The reality is even more compelling. When you add in lower plug and receptacle resistance and the fact that the di/dt on the power cord will have spectra well above 20kHz with some amplifiers, the low-inductance cord makes an even bigger difference.
And please don't give me a lot of flack about 60Hz current in power cords. The currents are very high in frequency - just measure them...