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In an L-C-R A.C. series circuit L = 5H, omega = "100 rads"^(-1), R= 100 Omega and power factor is 0.5. Calculate the value of capacitance of the capacitor. |
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Answer» SOLUTION :FIRST Method : Power factor `cosdelta=R/sqrt(R^2+(omegaL-1/(omegaC))^2)` Square on both SIDES `cos^2delta=R^2/(R^2+(omegaL-1/(omegaC))^2)` but , cos`delta=0.5=1/2` `therefore 1/4=R^2/(R^2+(omegaL-1/(omegaC))^2)` `therefore R^2+(omegaL-1/(omegaC))^2` `therefore (omegaL-1/(omegaC))^2=3R^2` `therefore omegaL-1/(omegaC)=sqrt3R` `therefore omegaL-sqrt3R=1/(omegaC)` `therefore C=1/omega(1/(omegaL-sqrt3R))` `=1/100(1/(100xx5-sqrt3xx100))` `=10^(-2)/(500-173.2)=10^(-2)/326.8=306xx10^(-7)` `=30.6xx10^(-6)` F =30.6 `muF` Second Method: Power factor `cosdelta=0.5` `therefore delta=pi/3` rad Now tan`delta=(omegaL-1/(omegaC))/R` `therefore "tan"pi/3 xx R=omegaL-1/(omegaC)` `therefore sqrt3xx100=100xx5-1/(100xxC)` `therefore 1.732xx100=500-1/(100C)` `therefore 1/(100C) =500-173.2` `therefore 1/(100C) =326.8` `therefore C=1/(326.8xx100)`=0.00003059 `approx 30.6xx10^(-6)F=30.6 muF` |
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