A series R-C combination is connected to an A | vedclass

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(C) Given: $\omega = 500 \ rad/s$.The impedance $Z$ of a series $R-C$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.Gi

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(C) Given: $\omega = 500 \ rad/s$.The impedance $Z$ of a series $R-C$ circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{\omega C}$.Given $Z = R\sqrt{1.25}$,we have $Z^2 = 1.25R^2$.Substituting into the impedance formula: $R^2 + X_C^2 = 1.25R^2$.$X_C^2 = 0.25R^2$.$X_C = 0.5R$.Since $X_C = \frac{1}{\omega C}$,we have $\frac{1}{\omega C} = 0.5R$.Rearranging for the time constant $	au = RC$: $RC = \frac{1}{0.5\omega}$.Substituting $\omega = 500 \ rad/s$: $	au = \frac{1}{0.5 	imes 500} = \frac{1}{250} = 0.004 \ s$.Converting to milliseconds: $	au = 0.004 	imes 1000 \ ms = 4 \ ms$.

$A$ series $R-C$ combination is connected to an $AC$ voltage of angular frequency $\omega = 500 \ rad/s$. If the impedance of the $R-C$ circuit is $R\sqrt{1.25}$,the time constant (in $ms$) of the circuit is:

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