An ideal OPAMP circuit with a sinusoidal input is shown in the figure. The 3 dB frequency is the frequency at which the magnitude of the voltage gain decreases by 3 dB from the maximum value. Which of the options is/are correct?
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A capacitor in series with the input of an op-amp amplifier always indicates a high-pass filter. The cutoff is found from the input RC network, not from the feedback network.
The 3 dB frequency is $\dfrac{1000}{3}\ \text{rad/s}$.
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The Correct Option isB, C
Solution and Explanation
The circuit consists of a capacitor in series with the input, followed by a resistor before reaching the non-inverting terminal of the op-amp. A series capacitor at the input produces a high-pass filter behavior, because low-frequency signals (including DC) are blocked while higher frequencies pass through.
The op-amp configuration given is a non-inverting amplifier with feedback resistors \(R_f = 2\,\text{k}\Omega\) and \(R = 1\,\text{k}\Omega\). Thus, the mid-band (maximum) gain is:
\[
A_0 = 1 + \frac{R_f}{R} = 1 + \frac{2000}{1000} = 3.
\]
Step 1: Determine the high-pass cutoff frequency.
The cutoff frequency \( \omega_c \) is determined by the series capacitor–resistor combination at the input:
\[
C = 1\,\mu\text{F}, \qquad R = 1\,\text{k}\Omega.
\]
The high-pass cutoff angular frequency is:
\[
\omega_c = \frac{1}{RC} = \frac{1}{(1000)(1\times 10^{-6})} = 1000\ \text{rad/s}.
\]
Step 2: Verify 3 dB point.
For a high-pass filter, the magnitude at the cutoff frequency satisfies:
\[
|A(j\omega_c)| = \frac{A_0}{\sqrt{2}}
\]
which matches the 3 dB drop definition.
Thus the calculated cutoff frequency is indeed the 3 dB frequency.
Therefore:
- The circuit is a high-pass filter.
- The 3 dB frequency is 1000 rad/s. Final Answer: (B) and (C)
