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Obtain the resonant frequency and Q - factor of a series LCR circuit with L = 3 H, C=27muF, R=7.4 Omega. It is desired to improve the sharpness of resonance of circuit by reducing its full width at half maximum by a factor of 2. Suggest a suitable way. |
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Answer» Solution :Here `L=3H, C=27muF=27xx10^(-6)F and R=7.4Omega` `therefore"Resonant frequency v, "=(1)/(2pisqrt(LC))=(1)/(2xx3.14sqrt(3xx27xx10^(-6)))=17.7Hz` `"andQ - factor "=(X_(L))/(R )=(L omega)/(R )=(Lxx2piv)/(R )=(3xx2xx3.14xx17.7)/(7.4)=45` IMPROVING the resonance of the circuit by reducing its full width at half of maximum power by a factor of 2 means that Q - factor is doubled. The simplest method for this is to reduce the value of resistance R so that `Q.=2Q rArr (X_(L))/(R.)=2(X_(L))/(R) rArr R.=(R )/(2)=(7.4)/(2)=3.7Omega` |
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