Electrical Engineering - Passive Filters - Discussion
Discussion Forum : 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?
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?Discussion:
7 comments Page 1 of 1.
Bharath said:
8 years ago
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.
Vo= 1/(R * (XL-Xc)/(Xl*Xc) + 1).
So max. voltage is occured, If XL=XC.
Hence, not exactly parallel.
Nabeel ALRAWI said:
9 years ago
It is not a parallel resonant, it is a series resonant, so the calculation done by Priyadarshini is correct.
Sanju said:
9 years ago
It's wrong. For parallel resonant circuit, fr = sqrt ( (1/LC) - (R^2/L^2) )/(2 * pi).
Sewak said:
9 years ago
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.
The resistor value is required while calculating the critical frequency. R = Xc, R = Xl.
Ahmed said:
9 years ago
Where is the resistor value?
Sarath said:
1 decade ago
How to work out without calculator?
PRIYADARSHINI said:
1 decade ago
Fc = 1/(2*pi*sqrt(L*C)) = 1/(2*pi*sqrt(8u*120p) = 5.136Mhz app = to 5.14 Mhz.
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