The AC coming from a 6-0-6 center tap transformer is rectified using two diodes. This rectified pulsating output is smothened to true DC by a 100µF. This true DC output is then scaled with a multi turn pot.

Hallo Prahlad,

don't forget, that this 100µF capacitor must be able to be discharged also. Otherwise, you will not detect a dip of mains voltage!

How to choose the right time constant?
From C = dQ / dU = I x dt / dU follows:

I = C x dU / dt

So, assume your transformer delivers 6Veff, then this makes 6Veff x 1.414 - 1V = 7.5V at the capacitor, assuming, that transformer output is loaded by rated load.
If you know tolerate a ripple caused by discharge during 100Hz halfwave of 1% (peak to peak), then you can tolerate a dU = 0.075V. So, this yields:

I = 100µF x 0.075V / 10msec = 750µA

This corresponds to a discharging resistor of R = U / I = 7.5V / 750µA = 10kOhm. And the time constant is tau = R x C = 10kOhm x 100µF = 1sec.

Mmh, this is a bit long. Keeping in mind, that a peak to peak ripple of 1% yields an error of 0.5%, you could even decrease R. This would worsen the ripple a bit, but at the same time fast up your mains voltage detetctor:

Assume you can tolerate a ripple of 4% (peak to peak), then this would cause an error of 2%. This would correspond to dU = 0.3V and I = 100µF x 0.3V / 10msec = 3mA. This would mean to use a discharging resistor of R = 2.5kOhm in combination with storage capacitor of 100µF. Time constant then is tau = R x C = 2.5kOhm x 100µF = 250msec, which is much better, I think.

Kai