Q.A series LCR circuit has $L=1\ \text{H}$, $C=1\ \mu\text{F}$ and $R=10\ \Omega$. Calculate

Given: $L=1\ \text{H}$, $C=1\ \mu\text{F}=1\times10^{-6}\ \text{F}$, $R=10\ \Omega$.

1. Resonant frequency.

   $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(1)(10^{-6})}} = \frac{1}{10^{-3}} = 1000\ \text{rad/s}$

   $f_0 = \frac{\omega_0}{2\pi} = \frac{1000}{6.283} \approx 159.2\ \text{Hz}$

2. Quality factor.

   $Q = \frac{\omega_0 L}{R} = \frac{1000\times1}{10} = 100$

3. Bandwidth. …

Compute $\omega_0=1/\sqrt{LC}$ and convert to $f_0$, compute $Q=\omega_0L/R$, then compute the bandwidth $\Delta\omega=R/L$ and convert t …

• Reporting $\omega_0$ (rad/s) as though it were $f_0$ (Hz) without dividing by $2\pi$. …