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

The provided values are \( R = 100 \, \text{k}\Omega \) and \( C = 100 \, \text{pF} \). The phase difference between \( V_{\text{in}} \) and \( (V_B - V_A) \) is 90°. For a series RC circuit, the phase difference \( \phi \) is defined by \( \tan \phi = \frac{1}{\omega RC} \). Given \( \phi = 90^\circ \), \( \tan 90^\circ = \infty \), which implies \( \frac{1}{\omega RC} = \infty \). This further simplifies to \( \omega RC = 1 \). Substituting the values \( \omega \) (angular frequency in rad/sec), \( R = 100 \times 10^3 \, \Omega \), and \( C = 100 \times 10^{-12} \, \text{F} \), we get \( \omega \times (100 \times 10^3) \times (100 \times 10^{-12}) = 1 \). This simplifies to \( \omega \times 10^{-6} = 1 \), yielding \( \omega = 10^6 \, \text{rad/sec} \). The relationship between angular frequency \( \omega \) and frequency \( f \) is \( \omega = 2 \pi f \). Therefore, \( 10^6 = 2 \pi f \). Solving for \( f \), we obtain \( f = \frac{10^6}{2 \pi} \approx 1.59 \times 10^5 \, \text{Hz} \). Expressing the frequency in the form \( 10^x \), we have \( f \approx 10^5 \, \text{Hz} \). Thus, \( x = 5 \).
Final Answerx = 5