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

Answer: Option C

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.

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