5v Electret microphone to PC mute switch pop help
I think you've misinterpreted the circuit shown on the web page. It applies to any electret. Basically it shorts out only the AC (signal) by using a large capacitor but avoids switching the DC bias voltage which is the source of the pop.

When the mic is plugged in C1 experiences a step voltage (from the 5V supply/R2) which brings both terminals of the capacitor to the DC bias level of the electret.
The negative terminal of the capacitor then discharges through R1 (100k) back to (nearly) 0 volts leaving no voltage across R1 (after more than 5 time constants). From the microphone's perspective it 'sees' a parallel load of 100k (the impedance of the large capacitor being negligible). This is so large it has no effect on the mic's output.
When the mute switch is operated (closed) there is no large DC step because the voltage across R1 is 0V and hence no large POP. The microphone now 'sees' only the 100uF capacitor as a load because the 100k has been shorted out. This effectively shorts out any AC (signal) and hence mutes the mic. The DC bias circuit (R2) is not shorted out because C1 is effectively an open circuit to the DC signal.