JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2025
- Hard
- Numerical
For ac circuit shown in figure, \(\mathrm{R}=100 \mathrm{k} \Omega\) and \(\mathrm{C}=100 \mathrm{pF}\) and the phase difference between \(\mathrm{V}_{\text {in }}\) and \(\left(V_B-V_A\right)\) is \(90^{\circ}\). The input signal frequency is \(10^x \mathrm{rad} / \mathrm{sec}\), where ' x ' is ______.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation

\(\theta+\theta=90^{\circ} ; \theta=45^{\circ}\)
\(\tan \theta=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}\)
\(\mathrm{X}_{\mathrm{C}}=\mathrm{R} \Rightarrow \frac{1}{\mathrm{~W}_{\mathrm{C}}}=\mathrm{R}\)
\(\mathrm{W}=\frac{1}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{100 \times 10^3 \times 100 \times 10^{-12}}\)
\(=\frac{10^{12}}{10^7}=10^5\)
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