# Electrical Engineering - Passive Filters - Discussion

### Discussion :: Passive Filters - General Questions (Q.No.4)

4.

The maximum output voltage of a certain low-pass filter is 15 V. The output voltage at the critical frequency is

 [A]. 0 V [B]. 15 V [C]. 10.60 V [D]. 21.21 V

Explanation:

No answer description available for this question.

 Haseeb Abbas said: (Mar 5, 2012) fc-70.7% Vc=(15*70.7)/100 Vc=10.60volts

 Arundeep said: (May 17, 2012) Critical frequency is less then 70. 7% by, O/P voltage.

 Swe said: (Oct 28, 2012) How did 70. 7 arise, can you please explain me ?

 Ganga said: (Jan 18, 2014) Critical frequency < 1/sqrt(2) of o/p voltage. => critical freq = 15/sqrt(2) = 10.605 V.

 A_Jediknight_Ib said: (Mar 6, 2014) I think it is more intuitive to think of it in terms of dB. Max voltage is 15V, convert this to dB. 15V = 20*log(15) = 23.52dB. Voltage at the critical freq will be 3dB less than this max. 23.52dB - 3dB = 20.52dB. Then convert back to voltage 20.52dB = 10.62V.

 Nyce said: (Mar 27, 2015) Critical frequency is the 70.7% of the original value or -3 db in terms of db.

 Mitrabhanu said: (Aug 25, 2016) How to convert dB into the voltage? Please answer this.

 Shailendra said: (Sep 15, 2016) Db to Voltage: 20log(V) = 20.52, log(V) = 20.52/20 = 1.026, V = 10^(1.026), V = 10.62.

 Udaypratapsingh said: (Sep 17, 2019) As we know, that voltage output is 70.7% of input voltage at critical frequency.

 Vin said: (Dec 14, 2019) So, which one is the correct formula of all the formulae that has been presented here. I am confused.