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An AC circuit containing 800mH inductor and a `60 muF` capacitor is in series with `15 Omega` resistance. They are connected to 230 V ,50 Hz AC supply. Obtain average power transferred to each element and total power absorbed. |
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Answer» Reactance of the inductor `X_(L) = omegaL = 2pivL= 2xx 3.14 xx 50 xx 800 xx 10^(-3)=251.3Omega` Capacitive reactance, `X_(C) =1/(omegaC)=1/(2pivC) = 1/(2 xx 3.14 xx 50 xx 60 xx 10^(-3)) = 251.3 Omega` `therefore` Impedance, `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt((15)^(2) + (251.3-53.05)^(2))` `rArr Z=198.8Omega` `therefore I_(rms) = V_(rms)/Z = 230/198.8 = 1.157`A Note that the inductance and capacitor do not consume power over a cycle. Now, average power dissipated in the resistor. `I_(rms)^(2).R=(1.157)^(2) xx 15` =20.07 W Total power consumed = `V_(rms).I_(rms).cosphi` `=V_(rms).I_(rms).R/Z` `=(230 xx 1.157 xx 15)/(198.8) = 20.07`W Note that the total power consumed always is same as that consumed in resistor. |
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