For ac circuit shown in figure, R = 100 k$$\Omega$$ and C = 100 pF and the phase difference between $$V_{in}$$ and $$(V_B - V_A)$$ is $$90^{\circ}$$. The input signal frequency is $$10^x$$ rad/sec, where 'x' is ______.

Using AC voltage divider relation with complex impedance $$Z_C = \frac{1}{j\omega C}$$:

$$V_A = V_{in} \left( \frac{Z_C}{R + Z_C} \right) = V_{in} \left( \frac{1}{1 + j\omega RC} \right)$$

$$V_B = V_{in} \left( \frac{R}{R + Z_C} \right) = V_{in} \left( \frac{j\omega RC}{1 + j\omega RC} \right)$$

Evaluating differential potential difference phasor ($$V_B - V_A$$): $$V_B - V_A = V_{in} \left( \frac{j\omega RC - 1}{1 + j\omega RC} \right)$$

$$\angle(V_B - V_A) - \angle V_{in} = \angle(j\omega RC - 1) - \angle(1 + j\omega RC)$$

$$\Delta \phi = (\pi - \tan^{-1}(\omega RC)) - \tan^{-1}(\omega RC) = \pi - 2\tan^{-1}(\omega RC)$$

$$\frac{\pi}{2} = \pi - 2\tan^{-1}(\omega RC) \implies 2\tan^{-1}(\omega RC) = \frac{\pi}{2} \implies \tan^{-1}(\omega RC) = \frac{\pi}{4}$$

$$\omega RC = 1 \implies \omega = \frac{1}{RC}$$

$$\omega = \frac{1}{10^5 \times 10^{-10}} = \frac{1}{10^{-5}} = 10^5\text{ rad/sec}$$

$$10^5 = 10^x \implies x = 5$$