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A circuit draws a power of 550 watt from a source of 220 volt, `50 Hz`. The power factor of the circuit is 0.8 and the current lags in phase behind the potential difference. To make the power factor of the circuit as 1.0, The capacitance should be connected in series with it isA. `(1)/(42 pi)xx10^(-2)F`B. `(1)/(41pi)xx10^(-2)F`C. `(1)/(5pi)xx10^(-2)F`D. `(1)/(84pi)xx10^(-2)F` |
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Answer» Correct Answer - A As the current lags behind the potential difference, the circuit contains resistance and inductance. Power, `P=V_("rms")xxi_("rms")xxcos phi` Here, `I_("rms")=(V_("rms"))/(Z)`, where `Z=sqrt([(R )^(2)+(omega L)^(2)])` `therefore P = (V_("rms")^(2))/(Z)xxcos phi` or `Z=(V_("rms")^(2))/(P)xx cos phi` So, `Z=((220)^(2)xx0.8)/(550)=70.4 Omega` Now, Power factor `cos phi =(R )/(Z)` or `R = Z cos phi` `therefore R = 70.4xx0.8=56.32 Omega` Further, `Z^(2)=R^(2)+(omega L)^(2)` or `omega L = sqrt((Z^(2)-R^(2)))` or `omega L = sqrt((70.4)^(2)-(56.32)^(2))=42.2 Omega` When the capacitor is connected in the circuit, `therefore Z = sqrt([R^(2)+(omega L-(1)/(omega C))^(2)])` According to the question, Power factor `cos phi = 1` Hence, `R = Z cos phi = Zxx1, R=Z` `therefore Z=sqrt(Z^(2)+(omega L-(1)/(omega C))^(2)) , (omega L-(1)/(omega C))^(2)=0` `therefore C = (1)/(omega(omega L))=(1)/(2pi upsilon(42.2))=(1)/(2pixx50xx42.2)~~(1)/(42pi)xx10^(-2)F` |
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