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In a series L-R circuit, X_{L} = R and power factor of the circuit is P_{1}. When a capacitor with capacitance C such that X_{L} = X_{C} is put in series, the power factor becomes P2. Find P_{1}/ P_{2} .

In a LC-R Circuit, Impedence , Z = ( R^{2} + (X_{L}_{}- Xc)^{2})^{1/2} and Power Factor , P = R/Z

Case 1:-When X_{L} = R, Z = (2R^{2})^{1/2 }= (2)^{1/}^{2}R and so P_{1}_{}= R/ (2)^{1/}^{2}R => P_{1}_{}= 1/ (2)^{1/2}

Case 2 :- WHen X_{L} = X_{C}, Z = R AND SO Power Factor P_{2} = R/R = 1

So Ratio is P_{1} : P_{2} = 1/(2)^{1/2 }: 1

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