JEE Advanced · Physics · 22. AC Circuits
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
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(D) 4
Step-by-step Solution
Detailed explanation
\(Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}\)
\(\therefore \quad R^2+X_C^2=1.25 R^2\)
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.
\(\therefore \quad R^2+X_C^2=1.25 R^2\)
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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