Discussion :: Passive Filters - General Questions (Q.No.4)
The maximum output voltage of a certain low-pass filter is 15 V. The output voltage at the critical frequency is
Answer: Option C
No answer description available for this question.
|Haseeb Abbas said: (Mar 5, 2012)|
|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.|
|Selam said: (Nov 19, 2022)|
|Vc = Vs * 0.707,
Vc =15v * 0.707,
Vc = 10.62.
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