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In a series \(L-C\) - \(R\) circuit, the inductive reactance is twice the resistance and the capacitive reactance is \(1 / 3\) rd of the inductive reactance. The power factor of the circuit is
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Verified Answer
The correct answer is:
0.6
In \(L-C-R\) circuit,
\(X_L=2 R \text { and } X_C=\frac{X_L}{3}\)
Impedance of \(L-C-R\) circuit is given as
\(\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(X_L-\frac{X_L}{3}\right)^2} \\
& =\sqrt{R^2+\left(\frac{2 X_L}{3}\right)^2}=\sqrt{R^2+\frac{4 X_L^2}{9}}
\end{aligned}\)
\(\begin{aligned} & =\sqrt{R^2+\frac{4}{9}(2 R)^2}=\sqrt{R^2+\frac{16}{9} R^2} \\ & =\sqrt{\frac{25 R^2}{9}} \Rightarrow Z=\frac{5}{3} R \\ & \text { Power factor }=\cos \phi \\ & =\frac{R}{Z}=\frac{R}{5 R / 3}=\frac{3}{5}=0.6 \\ & \end{aligned}\)
\(X_L=2 R \text { and } X_C=\frac{X_L}{3}\)
Impedance of \(L-C-R\) circuit is given as
\(\begin{aligned}
Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\
& =\sqrt{R^2+\left(X_L-\frac{X_L}{3}\right)^2} \\
& =\sqrt{R^2+\left(\frac{2 X_L}{3}\right)^2}=\sqrt{R^2+\frac{4 X_L^2}{9}}
\end{aligned}\)
\(\begin{aligned} & =\sqrt{R^2+\frac{4}{9}(2 R)^2}=\sqrt{R^2+\frac{16}{9} R^2} \\ & =\sqrt{\frac{25 R^2}{9}} \Rightarrow Z=\frac{5}{3} R \\ & \text { Power factor }=\cos \phi \\ & =\frac{R}{Z}=\frac{R}{5 R / 3}=\frac{3}{5}=0.6 \\ & \end{aligned}\)
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