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InterviewSolution
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An inductor 200 mH, capacitor `500mu F` and resistor `10 Omega` are connected in series with a 100 V variable frequency ac source. What is the frequency at which the power factor of the circuit is unity?A. `10.22 Hz`B. `12.4 Hz`C. `19.2 Hz`D. `15.9 Hz` |
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Answer» Correct Answer - D Here `L = 200 mH = 200xx10^(-3)H=0.2H` `C=500 mu F=500xx10^(-6)=5xx10^(-4)F` `R=10 Omega` and `V_("rms")=100 V` Power factor `=cos phi =(R )/(Z)=1` (Given) `therefore Z=R rArr sqrt(R^(2)+(X_(L)-X_(C ))^(2))=R` `R^(2)+(X_(L)-X_(C ))^(2)=R^(2) rArr X_(L)-X_(C )=0` or `X_(L)=X_(C )` (This is resonance condition) `2pi upsilon L=(1)/(2pi upsilon C) rArr 4pi^(2)upsilon^(2)LC=1` `therefore upsilon=(1)/(2pi sqrt(LC))=(1)/(2xx3.14sqrt(0.2xx5xx10^(-4))` `=(1)/(2xx3.14xx10^(-2))=(100)/(6.28)=15.9 Hz` |
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