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L = 8.1 mH, C = 12.5 muF 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. |
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Answer» Solution :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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