The power factor of an L-R series circuit is | vedclass

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(D) For the $L-R$ circuit: $\cos \phi_1 = \frac{R_1}{Z_1} = 0.6 = \frac{3}{5}$. Thus,$\sin \phi_1 = \frac{X_L}{Z_1} = 0.8 = \frac{4}{5}$. So,$X_L = \f

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$\frac{3\sqrt{3}}{4}$

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(D) For the $L-R$ circuit: $\cos \phi_1 = \frac{R_1}{Z_1} = 0.6 = \frac{3}{5}$. Thus,$\sin \phi_1 = \frac{X_L}{Z_1} = 0.8 = \frac{4}{5}$. So,$X_L = \frac{4}{3} R_1$.For the $C-R$ circuit: $\cos \phi_2 = \frac{R_2}{Z_2} = 0.5 = \frac{1}{2}$. Thus,$\sin \phi_2 = \frac{X_C}{Z_2} = \frac{\sqrt{3}}{2}$. So,$X_C = \sqrt{3} R_2$.When joined in series,the circuit becomes an $L-C-R$ circuit. For the power factor to be $1$,the circuit must be in resonance,meaning $X_L = X_C$.Equating the reactances: $\frac{4}{3} R_1 = \sqrt{3} R_2$.Therefore,the ratio $\frac{R_1}{R_2} = \frac{3\sqrt{3}}{4}$.

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