In the circuit shown in the figure,if the val | vedclass

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(A) Given: $V_{rms} = 220 	ext{ V}$,$I_{rms} = 2.2 	ext{ A}$,$L = \frac{1}{\pi} 	ext{ H}$,$\omega = 100\pi 	ext{ rad/s}$.First,calculate the total

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$\frac{1}{\sqrt{2}}$

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(A) Given: $V_{rms} = 220 	ext{ V}$,$I_{rms} = 2.2 	ext{ A}$,$L = \frac{1}{\pi} 	ext{ H}$,$\omega = 100\pi 	ext{ rad/s}$.First,calculate the total impedance of the circuit: $Z = \frac{V_{rms}}{I_{rms}} = \frac{220}{2.2} = 100 \ \Omega$.The circuit consists of an inductor $L$ in series with a box containing a resistor $R = 100 \ \Omega$ and a capacitor $C$. The total impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Calculate inductive reactance: $X_L = \omega L = (100\pi) 	imes (\frac{1}{\pi}) = 100 \ \Omega$.Since $Z = 100 \ \Omega$ and $R = 100 \ \Omega$,we have $Z = R$,which implies $X_L - X_C = 0$,so $X_C = X_L = 100 \ \Omega$.The box contains a resistor $R = 100 \ \Omega$ and a capacitor $X_C = 100 \ \Omega$. The impedance of the box is $Z_{box} = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = 100\sqrt{2} \ \Omega$.The power factor of the box is $\cos \phi = \frac{R}{Z_{box}} = \frac{100}{100\sqrt{2}} = \frac{1}{\sqrt{2}}$.

In the circuit shown in the figure,if the value of $R.M.S$ current is $2.2 \text{ A}$,the power factor of the box is:

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