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

Correct answer: 4 Z= R^2+X_C^2 =R 1.25 R^2+X_C^2=1.25 R^2 or X_C=(R)/(2) or (1)/(ω C)=(R)/(2) aligned Time constant & =C R=(2)/(ω) & =(2)/(500) s=4 ms aligned 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.

JEE Advanced 2011Paper 2

Exam: JEE Advanced
Year: 2011
Paper: Paper 2
Subject: Physics
Chapter: Alternating Current

JEE Advanced 2011 · Paper 2 · Physics · Alternating Current

A series R-C combination is connected to an AC voltage of angular...

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

Correct answer

Step-by-step 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.

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