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Calculate the resonent frequency of Q-factor (Quality factor) of a series L-C-R circuit containing a pure inductor of inductance 4H, capacitor of capacitance 27 muF and resister of resistance 8.4Omega. |
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Answer» Solution :`v_(0)=(1)/(2pisqrt(LC))` Finding `v_(0)=(1)/(2xx3.142sqrt(4xx27xx10^(-6)))=15.31Hz` `Q=(X_(L))/(R)=(omega_(0)L)/(R)=(2piv_(0)L)/(R)` ARRIVING at `Q=(2xx3.142xx15.31xx4)/(8.4)=48.8` Detailed Answer: Solution, Here, `L=4H,C=27xx10^(-6)F`, R=`8.4-2` Resonant frequency `f_(0)=(1)/(2pisqrt(LC))` `f_(0)=(1)/(3.142xxsqrt(4xx27xx10^(-6)))` `f_(0)=(1xx10^(3))/(2xx3.142xxsqrt(108))=(1xx10^(3))/(6.284xx10.4)` `f_(0)=(1xx10^(3))/(65.353)=15.30Hz` Q-factor `Q=(1)/(R)sqrt((L)/(C))=(1)/(8.4)xxsqrt((4)/(27xx10^(-6)))=(1)/(8.4)=(2xx10^(3))/(3sqrt(3))` `Q=(2000)/(43.65)=45.81` |
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