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QPages 330-331: Q18: HC Verma: Concepts of Physics - Volume 2 Alternating Current Pages 330-331: Q18 Solution: Following figure shows a typical circuit for a low-pass filter. …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the frequency $f$ directly into the low‑pass formula $V_0=V_i/\sqrt{1+(\omega RC)^2}$, forgetting that $\omega=2\pi f$. This missing factor of $2\pi$ (or using the wrong component values) leads to an output voltage off by roughly an order of magnitude. Always convert $f$ to angular frequency before evaluating $\omega RC$.

Following figure shows a typical circuit for a low-pass filter. An AC input Vi=10 mVV_i = 10 \text{ mV} is applied at the left end and the output V0V_0 is received at the right end. Find the output voltage for ν=10 kHz\nu = 10 \text{ kHz}, 1.0 MHz1.0 \text{ MHz} and 10.0 MHz10.0 \text{ MHz}. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.

Step-by-Step Verified Solution

**Answer:** (a) V0≈1.6 mVV_0 \approx 1.6 \, \text{mV} (b) ≈0.16 mV\approx 0.16 \, \text{mV} (c) ≈0.016 mV\approx 0.016 \, \text{mV}. **Step-by-Step Explanation:** This is a low-pass RC filter. V0=Vi1+(ωRC)2V_0 = \frac{V_i}{\sqrt{1 + (\omega RC)^2}}. Given R=1 kΩR = 1 \, \text{k}\Omega, C=10 nFC = 10 \, \text{nF}. For f=10 kHzf = 10 \, \text{kHz}, 1 MHz1 \, \text{MHz}, 10 MHz10 \, \text{MHz}, calculate accordingly [citation:4].

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