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In an L-C-R A.C. series circuit L = 5H, ...

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.

Text Solution

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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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