JEE Main 2025 Apr 7 Shift 1, Physics Q22: LR, CR, LCR Circuits

Phase reference for each mid-node. In an RC voltage divider driven by $V_{\text{in}}$, the mid-node voltage has a phase relative to $V_{\text{in}}$ set by $\tan \theta =X_C/R$ (magnitude).

Branch 1 (R on top, C on bottom, $V_A$ across C): $V_A$ lags $V_{\text{in}}$ by an angle $\theta$ with $\tan \theta =X_C/R$.

Branch 2 (C on top, R on bottom, $V_B$ across R): $V_B$ leads $V_{\text{in}}$ by the complementary angle ${\theta }^{\prime }$ with $\tan {\theta }^{\prime }=R/X_C$.

By phasor construction the angle between $V_A$ and $V_B$ (i.e., the phase of $V_B-V_A$ relative to $V_{\text{in}}$) is $\theta +{\theta }^{\prime }$. The condition that $V_{\text{in}}$ and $(V_B-V_A)$ are $90^{\circ }$ out of phase requires

$\theta +{\theta }^{\prime }=90^{\circ }\Rightarrow \theta ={\theta }^{\prime }=45^{\circ }$

Setting $\theta =45^{\circ }$: $\tan 45^{\circ }=1\Rightarrow X_C=R\Rightarrow \frac{1}{\omega C}=R$.

Solve for $\omega$:

$\omega =\frac{1}{RC}=\frac{1}{(100\times 10^3)(100\times 10^{-12})}=\frac{1}{10^{-5}}=10^5\text{ rad/s}$

So $\omega =10^x$ gives $x=5$.

$\therefore$ Answer = 5.