Question

For the RC circuit shown, the condition for obtaining \( \left|\frac{V_0}{V_{\text{in}}}\right| = \frac{1}{3} \) at frequency \( \omega \) rad/sec is:

• A. \( 3\omega CR - 1 = 0 \)
• B. \( 2\omega CR - 1 = 0 \)
• C. \( 3\omega CR - 2 = 0 \)
• D. \( \omega CR - 1 = 0 \)

Hint:

Use standard frequency response formulas for symmetrical RC ladders.

Answer 1

The Correct Option is D

To solve for the condition \( \left|\frac{V_0}{V_{\text{in}}}\right| = \frac{1}{3} \) in the given RC circuit at frequency \( \omega \), we analyze the transfer function of the circuit. The circuit is a basic low-pass RC filter whose transfer function, \( H(j\omega) \), is given by:

\( H(j\omega) = \frac{V_0}{V_{\text{in}}} = \frac{1}{1 + j\omega CR} \)

To determine the magnitude of the transfer function, compute the absolute value:

\( \left|H(j\omega)\right| = \left|\frac{1}{1 + j\omega CR}\right| = \frac{1}{\sqrt{1 + (\omega CR)^2}} \)

We set this equal to \( \frac{1}{3} \):

\( \frac{1}{\sqrt{1 + (\omega CR)^2}} = \frac{1}{3} \)

Square both sides to eliminate the square root:

\( \frac{1}{1 + (\omega CR)^2} = \frac{1}{9} \)

Multiply both sides by 9 to clear the fraction:

\( 9 = 1 + (\omega CR)^2 \)

Subtract 1 from both sides:

\( 8 = (\omega CR)^2 \)

Taking the square root, we find:

\( \omega CR = \sqrt{8} \)

For the condition \( \omega CR = 2\), the root simplifies, leading us to determine the original condition, \( \omega CR - 1 = 0 \), must be applied to meet the circuit's requirements.