Question

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:

Hint:

In a series RC circuit with a 90° phase difference between the voltage and current, the frequency is determined by the time constant \( \tau = RC \) and the phase condition \( \omega RC = 1 \).

Answer 1

Correct Answer: 5

We are given that: \( R = 100 \, \text{k}\Omega \) \( C = 100 \, \text{pF} \) The phase difference between \( V_{\text{in}} \) and \( (V_B - V_A) \) is 90°. The phase difference \( \phi \) in a series RC circuit is given by: \[ \tan \phi = \frac{1}{\omega RC} \] Given that \( \phi = 90^\circ \), we have: \[ \tan 90^\circ = \infty \quad \Rightarrow \quad \frac{1}{\omega RC} = \infty \] This implies: \[ \omega RC = 1 \] Where: \( \omega \) is the angular frequency (in rad/sec) \( R = 100 \times 10^3 \, \Omega \) \( C = 100 \times 10^{-12} \, \text{F} \) Substituting these values into the equation: \[ \omega \times (100 \times 10^3) \times (100 \times 10^{-12}) = 1 \] Simplifying: \[ \omega \times 10^{-6} = 1 \] \[ \omega = 10^6 \, \text{rad/sec} \] The angular frequency \( \omega \) is related to the frequency \( f \) by: \[ \omega = 2 \pi f \] Thus: \[ 10^6 = 2 \pi f \] Solving for \( f \): \[ f = \frac{10^6}{2 \pi} \approx 1.59 \times 10^5 \, \text{Hz} \] We are asked to express the frequency in the form \( 10^x \): \[ f \approx 10^5 \, \text{Hz} \] Therefore, \( x = 5 \). %
Final Answer x = 5