Electrical Engineering - Passive Filters - Discussion

Discussion :: Passive Filters - General Questions (Q.No.2)

2. 

An RL low-pass filter consists of a 5.6 mH coil and a 3.3 k resistor. The output voltage is taken across the resistor. The circuit's critical frequency is

[A]. 93.8 kHz
[B]. 93.8 Hz
[C]. 861 Hz
[D]. 86.12 kHz

Answer: Option A

Explanation:

No answer description available for this question.

Shrutali said: (Sep 1, 2011)  
How to find critical frequency in this problem ?

Asha said: (Oct 15, 2011)  
Critical frequency=R/2*PIE*L

Shankarling.Patil said: (Apr 27, 2012)  
What is critical frequency. Please tell me friends.

Naga Mallesh said: (Jul 27, 2012)  
When R = XL its called critical frequency.

Shahul said: (Apr 5, 2013)  
Critical frequency/Bandwidth = R/ (2*3. 14*L).

Jon said: (Sep 22, 2013)  
Using R = XL.

Then R = 2*pie*freq*L.

Freq. = R/(2*pie*L).

= 3300/(2*3.1416*.0056).

Freq. = 93,787 Hz or 93.8 KHz.

Pesce said: (Apr 26, 2015)  
Critical frequency is a frequency at which I is maximum in RLC circuit.

i.e, when R=sqrt (XL*XL-Xc*Xc).

Here, Xc = 0 (since RC ckt).

So.

R = XL.

But XL = 2*pie*f*L.

Hence R = 2*pie*f*L.

F = R/2*pie*L.

Mshal said: (Feb 15, 2016)  
At Resonant state of series RLC circuit, Z = R = (XL-XC).

Since R & L are given, R = XL but XL = 2*π*F*L.

Therefore 3.3k ohms = 2*3.1416*F*5.6 millihenry;

F = 93.78 HZ.

Sana said: (Jul 8, 2016)  
Is critical and cut off frequency are same?

Sonu Kumar said: (Apr 1, 2018)  
F =R/2πL,
= 3300/2*3.14*.0056,
= 3300/6.28*.0056,
= 33*10^8/628*56,
= 33*10^8/35168,
= 93835hz or,
= 93.8khz.

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