InterviewSolution
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InterviewSolution
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An inductor 200 mH, capacitor `500 mu F`, resistor 10 ohm are connected in series with a 100 V, variable frequency a.c. source. Calculate (i) frequency at which power factor of the circuit is unity (ii) current amplitude at this frequency (iii) Q factor. |
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Answer» Here `L=200mH = 200xx10^(-3)=0.2H` `C=500 mu F = 500 xx 10^(-6) = 5xx10^(-4)F` `R=10Omega and F_(V)=100V` (i) Power factor, `cos phi =1 ` (given) `(R)/(Z)=1 " " Z=R` `sqrt(R^(2)+(X_(L)-K_(C))^(2))=R` `R^(2)+(X_(L)-X_(C))^(2)=R^(2) rArr X_(L)-X_(C)=0` `rArr X_(L)=X_(C) rArr 2 pi v L = (1)/(2pi v C)` `rArr 4pi^(2)v^(2)LC=1` `:. v=(1)/(2pi sqrt(LC))=(1)/(2xx3.14sqrt(0.2xx5xx10^(-4)))=(1)/(2xx3.14xx10^(-2))=(100)/(6.28)=15.92 ~~ 16Hz`. (ii) Current amplitude at resonance `I_(0)=(E_(0))/(Z)=(sqrt(2 in_(rms)))/(R)=(1.414xx100)/(10)=14.14A` (iii) Q-factor `=(1)/(R) sqrt((L)/(C))=(1)/(10)sqrt((0.2)/(5xx10^(-4)))` `=(1)/(10)sqrt((2)/(50xx10^(-4)))=(1)/(10) sqrt((1)/(25xx10^(-4)))=(1)/(10xx5xx10^(-2))=(10^(2))/(50)=(100)/(50)=2`. |
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