# Electrical Engineering - Passive Filters - Discussion

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

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

Explanation:

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.