Exam

JEE Main

Year

2025

Date

7 April

Shift

Morning Shift

Subject

Physics

Chapter

Alternating Current

JEE Main 2025 · 7 April · Morning Shift · Physics · Alternating Current

For ac circuit shown in figure, R=100 k and C=100 pF and the phase...

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 ______

Step-by-step solution

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