An inductor 200 mH, capactior `500 mu F`, resistor 10 ohm are connected in series with a 100 V, varialbe 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.

Explanation

Here `L=200 mH=200xx10^(-3)=0.2 H`
`C=500 muF=500xx10^(-6)=5xx10^(-4) F`
`R=10 Omega` and `E_(v)=100 V`
(i) Power factor ,`cos phi=1` (given)
`R/Z=1`
`Z=R`
`sqrt(R^(2)+(X_(L)-X_(C)))=R`
`R^(2)+(X_(L)-X_(C))^(2)=R^(2)rArrX_(L)-X_(C)=0 rArr X_(L)=X_(C)`
`2pivL=1/(2pivC) therefore v=1/(2pisqrt(L_(C))) =1/(2xx3.14sqrt(0.2xx5xx10-4))`
`=1/(2xx3.14xx10^(-2))=100/6.28=15.92~~16 Hz`
(ii)Current amplitude at resonance
`I=E_(v)/Z=E_(v)/R=100/10=10A`
(iiI)`Q`-factor `=1/R sqrt(L/C)=1/10sqrt(0.2/(5xx10^(-4)))`
`=1/10sqrt(2/(50xx10^(-4))) = 1/10sqrt(1/(25xx10^(-4)))=1/(10xx5xx10^(-2)) =1/(10xx5xx10^(-2))=10^(2)/50xx100/50=2`