A series LCR circuit is connected to an a.c. | vedclass

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(B) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$,where $R$ is the resistance and $Z$ is the impedance of the circuit.Giv

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$0.8$

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(B) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$,where $R$ is the resistance and $Z$ is the impedance of the circuit.Given:Resistance $R = 80 \ \Omega$Inductive reactance $X_L = 70 \ \Omega$Capacitive reactance $X_C = 130 \ \Omega$The impedance $Z$ of a series $LCR$ circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values:$Z = \sqrt{80^2 + (70 - 130)^2}$$Z = \sqrt{80^2 + (-60)^2}$$Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \ \Omega$.Now,the power factor $x = \cos \phi = \frac{R}{Z} = \frac{80}{100} = 0.8$.

$A$ series $LCR$ circuit is connected to an a.c. source of $230 \ V, 50 \ Hz$. The circuit contains a resistance of $80 \ \Omega$,an inductor having inductive reactance $70 \ \Omega$,and a capacitor of capacitive reactance $130 \ \Omega$. The power factor of the circuit is $x$. The value of $x$ is

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