Question

For the circuit given below, choose the angular frequency \(\omega_0\) (in rad/s) at which the voltage across the capacitor has maximum amplitude?

• A. 1000
• B. 100
• C. 1
• D. 0

Hint:

A series \(RC\) acts as a low-pass voltage divider for the capacitor: the capacitor voltage is largest at DC and decreases monotonically with frequency.

Answer 1

The Correct Option is D

Step 1: For a series \(RC\) circuit with source amplitude \(V_s\), the capacitor voltage amplitude is \[ |V_C| = V_s \frac{|Z_C|}{|Z_R+Z_C|} = V_s \frac{\tfrac{1}{\omega C}}{\sqrt{R^2+\left(\tfrac{1}{\omega C}\right)^2}}. \]

Step 2: Let \[ x = \frac{1}{\omega C}. \] Then \[ |V_C| = V_s \frac{x}{\sqrt{R^2+x^2}}. \] As \(\omega \to 0\) (i.e., \(x \to \infty\)): \[ |V_C| \to V_s \frac{x}{x\sqrt{1+\tfrac{R^2}{x^2}}} \to V_s. \] For any finite \(\omega > 0\), \(x\) is finite and \(|V_C| < V_s\).

Hence, the maximum amplitude occurs at \[ \omega_0 = 0 \ \text{rad/s (DC)}. \] Final Answer: \(\boxed{0}\).