SolveItJEE Main
Physics · 2025

JEE Main · 7 April 2025, Shift 1 · Q49

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…

For ac circuit shown in figure, $\displaystyle \mathrm{R}=100 \mathrm{k} \Omega$ and $\displaystyle \mathrm{C}=100 \mathrm{pF}$ and the phase difference between $\displaystyle V_{\text {in }}$ and $\displaystyle \left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)$ is $\displaystyle 90^{\circ}$. The input signal frequency is $\displaystyle 10^x \mathrm{rad} / \mathrm{sec}$, where ' $\displaystyle x$ ' is $\displaystyle \_\_\_\_$. Figure: JEE Main Physics 2025, Alternating Current
ShareWhatsAppTelegram

More from Alternating Current

JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.