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

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(A) The inductive reactance is $X_L = \omega L = 100\pi 	imes \frac{1}{\pi} = 100\, \Omega$.Given $I_{rms} = 2.2\, A$ and $V_{rms} = 220\, V$,the tot

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

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(A) The inductive reactance is $X_L = \omega L = 100\pi 	imes \frac{1}{\pi} = 100\, \Omega$.Given $I_{rms} = 2.2\, A$ and $V_{rms} = 220\, V$,the total impedance $Z$ of the circuit is $Z = \frac{V_{rms}}{I_{rms}} = \frac{220}{2.2} = 100\, \Omega$.The total impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting the values: $100 = \sqrt{100^2 + (100 - X_C)^2}$.Squaring both sides: $100^2 = 100^2 + (100 - X_C)^2$,which implies $(100 - X_C)^2 = 0$,so $X_C = 100\, \Omega$.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 $rms$ current is $2.2\, A$,the power factor of the box is:

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