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| 1. |
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. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> Method : <br/>Power factor `cosdelta=R/sqrt(R^2+(omegaL-1/(omegaC))^2)` <br/>Square on both <a href="https://interviewquestions.tuteehub.com/tag/sides-1207029" style="font-weight:bold;" target="_blank" title="Click to know more about SIDES">SIDES</a> `cos^2delta=R^2/(R^2+(omegaL-1/(omegaC))^2)` <br/> but , cos`delta=0.5=1/2` <br/> `therefore 1/4=R^2/(R^2+(omegaL-1/(omegaC))^2)` <br/>`therefore R^2+(omegaL-1/(omegaC))^2` <br/> `therefore (omegaL-1/(omegaC))^2=3R^2` <br/> `therefore omegaL-1/(omegaC)=sqrt3R` <br/> `therefore omegaL-sqrt3R=1/(omegaC)` <br/> `therefore C=1/omega(1/(omegaL-sqrt3R))` <br/>`=1/100(1/(100xx5-sqrt3xx100))` <br/> `=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-2)/(500-173.2)=10^(-2)/326.8=306xx10^(-7)` <br/> `=30.6xx10^(-6)` F <br/>=30.6 `muF` <br/>Second Method: <br/>Power factor `cosdelta=0.5` <br/>`therefore delta=pi/3` rad <br/> Now tan`delta=(omegaL-1/(omegaC))/R` <br/> `therefore "tan"pi/3 xx R=omegaL-1/(omegaC)` <br/> `therefore sqrt3xx100=100xx5-1/(100xxC)` <br/> `therefore 1.732xx100=500-1/(100C)` <br/>`therefore 1/(100C) =500-173.2` <br/> `therefore 1/(100C) =326.8` <br/>`therefore C=1/(326.8xx100)`=0.00003059 <br/> `approx 30.6xx10^(-6)F=30.6 muF`</body></html> | |