The gain and cut-off frequency (in radian/ sec) for the circuit shown below are :

Concept:

The given circuit is an active low-pass filter using an op-amp. The gain and cut-off frequency of this filter are derived based on its configuration.

Here, R₁ is the input resistor, and R₂ is in the feedback path along with capacitor C.

The standard form of the transfer function for such a filter is:

\(H(s) = \frac{V_o(s)}{V_i(s)} = \frac{-R_2/R_1}{1 + sCR_2}\)

From this expression, we can identify:

• Gain = \( -\frac{R_2}{R_1} \)
• Cut-off frequency (in radians/sec) = \( \omega_c = \frac{1}{CR_2} \)

Calculation:

Given:

Input resistor: R₁, Feedback resistor: R₂, Capacitor: C

Transfer function: \( H(s) = \frac{-R_2/R_1}{1 + sCR_2} \)

At low frequencies (s → 0), gain approaches \( -R_2/R_1 \)

Cut-off frequency occurs at the point where the magnitude of denominator becomes √2 times:

\( \omega_c = \frac{1}{CR_2} \)

Correct Option:

✅ The correct answer is: 1) –R₂/R₁, 1/CR₂