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L = 8.1 mH, C = 12.5 muF and R = 100 Ome...

L = 8.1 mH, C = 12.5 `mu`F and R = 100 `Omega` are connected in series with A.C. source of 230 V and frequency 500 Hz. Calculate voltage across two ends of resistance.

Text Solution

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Impedance for a L-C-R series circuit,
`|Z|=sqrt(R^2+(X_L-X_C))^2`
where `X_L=omegaL=2pifL`
`=2xx3.14xx500xx8.1xx10^(-3)=25.4 Omega`
`X_C=1/(omegaC)=1/(2pifC)`
`=1/(2xx3.14xx500xx12.5xx10^(-6))=25.4Omega`
`therefore |Z|=sqrt((10)^2+(25.4-25.4)^2)=10Omega`
Now `I_"rms"=V_"rms"/"|Z|"=230/100`=2.3 A
`therefore` The voltage across the ends of resistance ,
`V_R=I_(rms)xxR`=2.3 x 100
`therefore V_R`= 230 V
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