For ac circuit shown in figure, R=100 k Ω and C=100 pF and the phase difference between V in and ( V B - V A ) is 90 ° . The input signal frequency is 10 × rad / sec , where ' x ' is

Physics · Alternating Current

JEE Main 2025 — 7 April, Morning Shift — Question 69

For ac circuit shown in figure, $R=100 \mathrm{k} \Omega$ and $C=100$ pF and the phase difference between

$\mathrm{V}_{\text {in }}$ and $\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)$ is $90^{\circ}$ .

The input signal frequency is $10^{\times} \mathrm{rad} / \mathrm{sec}$ , where ' $x$ ' is $\qquad$

Numerical answer — enter this value.

Since both branch are identical. So phase difference between

$V_{A}$ and $V_{\text {in }}$ and $V_{B}$

and $V_{\text {in }}$ are same but in opposite direction.

So, phase difference between

$\mathrm{V}_{\text {in }}$ and $\mathrm{V}_{\mathrm{A}}$ must be

$45^{\circ}$ as $V_{\text {in }}$

and $\left|V_{A}-V_{D}\right|$ has difference of $90^{\circ}$ . So, clearly $|R|=(x c)$

\begin{aligned} & \Rightarrow 100 \times 10^{3}=\frac{10^{12}}{w \times 100} \\ & \Rightarrow w=10^{5} \mathrm{rad} / \mathrm{s} \end{aligned}

Answer key and solution verified before publishing.

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