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A series LCR circuit with a resistance of `100 sqrt(%) Omega` is connected to an ac source of 200 V. When the capacitor is removed from the circuit, current lags behind emf by `45^(@)`. When the inductor is removed from the circuit keeping the capacitor and resistor in the circuit, current leads by an angle of `tan^(-1)((1)/(2))`. Calculate the current and power dissipated in LCR circuit. |
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Answer» When capacitor is removed, `tan phi = (X_(L))/(R) ` (It is L - R circuit) `tan phi = tan 45^(@) = (X_(L))/(R) rArr X_(L)= R` When inductor is removed, `tan phi = (X_(C))/(R) = (1)/(2) = (X_(C))/(R) rArr X_(C)=(R)/(2)` Impendances `Z=sqrt(R^(2)+(X_(L)-X_(C))^(2))=sqrt(R^(2)+(R-(R)/(2))^(2))=(Sqrt(5))/(2) R` Now, `l_("rms")=(E_("rms"))/(Z) = (200)/(250) = 0.8A, cos phi = (R)/(Z) = (2) /(sqrt(5))` `P_(av)=V_("rms")l_("rms")cos phi = 64 sqrt(5) W` |
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