Electrical Engineering - Passive Filters - Discussion

Discussion :: Passive Filters - General Questions (Q.No.7)


A parallel resonant band-pass filter consists of a 6.8 resistor in series with a parallel network made up of an 8 H coil and a 120 pF capacitor. The output is taken across the capacitor/coil. Assume RW = 0 . What is the center frequency of the filter?

[A]. 5.14 MHz
[B]. 514 kHz
[C]. 5.03 MHz
[D]. 503 kHz

Answer: Option A


No answer description available for this question.

Priyadarshini said: (Jun 5, 2015)  
Fc = 1/(2*pi*sqrt(L*C)) = 1/(2*pi*sqrt(8u*120p) = 5.136Mhz app = to 5.14 Mhz.

Sarath said: (Jul 29, 2015)  
How to work out without calculator?

Ahmed said: (Aug 6, 2016)  
Where is the resistor value?

Sewak said: (Oct 28, 2016)  
Resistor value not required for calculation of centre (resonance) frequency. Xl = Xc.

The resistor value is required while calculating the critical frequency. R = Xc, R = Xl.

Sanju said: (Nov 11, 2016)  
It's wrong. For parallel resonant circuit, fr = sqrt ( (1/LC) - (R^2/L^2) )/(2 * pi).

Nabeel Alrawi said: (Dec 3, 2016)  
It is not a parallel resonant, it is a series resonant, so the calculation done by Priyadarshini is correct.

Bharath said: (Mar 19, 2017)  
Find voltage expression, we will be having (Xl-Xc) expression in the denominator.
Vo= 1/(R * (XL-Xc)/(Xl*Xc) + 1).
So max. voltage is occured, If XL=XC.
Hence, not exactly parallel.

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