Question

For the AC circuit shown in the figure, \(R = 100\, \text{k}\Omega\) and \(C = 10\, \text{pF}\), the phase difference between \(V_{in}\) and \(V_{out}\) is \(90^\circ\) at the input signal frequency of ............. kHz. (Round off to 2 decimal places) 

Hint:

For an RC circuit, a phase shift of \(90^\circ\) occurs when the capacitive reactance equals the resistance: \(X_C = R\).

Answer 1

Correct Answer: 159 - 159.3

Step 1: Condition for phase difference.
For an RC circuit, the phase difference between \(V_{in}\) and \(V_{out}\) is \(90^\circ\) when: \[ \omega R C = 1 \]

Step 2: Substitute given values.
\[ R = 100 \times 10^3\, \Omega, C = 10 \times 10^{-12}\, \text{F} \] \[ \omega = \frac{1}{R C} = \frac{1}{(100 \times 10^3)(10 \times 10^{-12})} = 10^6\, \text{rad/s} \]

Step 3: Find frequency.
\[ f = \frac{\omega}{2\pi} = \frac{10^6}{2\pi} = 1.59 \times 10^5\, \text{Hz} = 159.15\, \text{kHz} \]

Step 4: Conclusion.
Hence, the input frequency for a \(90^\circ\) phase difference is \(159.15\, \text{kHz}\).