JEE MAIN - Physics (2025 - 7th April Morning Shift - No. 23)
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 $\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)$ is $90^{\circ}$. The input signal frequency is $10^x \mathrm{rad} / \mathrm{sec}$, where ' $x$ ' is __________ .
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Responder
5
Explicação
Input voltage
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$$\begin{aligned} & \theta+\theta=90^{\circ} ; \theta=45^{\circ} \\ & \tan \theta=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}} \\ & \mathrm{X}_{\mathrm{C}}=\mathrm{R} \Rightarrow \frac{1}{\mathrm{~W}_{\mathrm{C}}}=\mathrm{R} \\ & \mathrm{~W}=\frac{1}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{100 \times 10^3 \times 100 \times 10^{-12}} \\ & =\frac{10^{12}}{10^7}=10^5 \end{aligned}$$