For the AC circuit shown in the figure, $ R = 100 \, \text{k}\Omega $ and $ C = 100 \, \text{pF} $, and the phase difference between $ V_{\text{in}} $ and $ (V_B - V_A) $ is 90°. The input signal frequency is $ 10^x $ rad/sec, where $ x $ is:
Since both branches are identical, the phase difference between \( V_A \) and \( V_B \) and \( V_{in} \) are the same, but in opposite directions. Thus, the phase difference between \( V_{in} \) and \( V_A \) must be \( 45^\circ \), as \( V_{in} \) and \( |V_A - V_D| \) have a difference of \( 90^\circ \). Now, clearly: \[ |R| = (xc) \] Given: \[ 100 \times 10^3 = \frac{10^{12}}{w \times 100} \] Solving for \(w\): \[ w = 10^5 \, \text{rad/s} \] \[ \boxed{w = 10^5 \, \text{rad/s}} \]
