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