A series $R-C$ combination is connected to an $\mathrm{AC}$ voltage of angular frequency $\omega=500…

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

Solution

$Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}$ $ \therefore \quad R^2+X_C^2=1.25 R^2 $ or $\quad X_C=\frac{R}{2}$ or $\quad \frac{1}{\omega C}=\frac{R}{2}$ $\begin{aligned} \therefore \text { Time constant } & =C R=\frac{2}{\omega} \\ & =\frac{2}{500} \mathrm{~s}=4 \mathrm{~ms}\end{aligned}$ $\therefore$ Answer is 4 . Analysis of Question (i) Question is very simple. (ii) I think this is one of the simplest formula based question of this paper. `

Asked in: JEE Advanced 2011 (Paper 2)

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