A series RLC circuit has $R=30,\Omega$, $L=0.4,\text{H}$ and $C=25,\mu \text{F}$. It is connected to an AC source of $110,\text{V rms}$ at $f=50,\text{Hz}$. (a) Show that the circuit is at resonance. (b) Find the rms current at resonance. (c) Calculate the quality factor $Q$ of the circuit. (d) Determine the bandwidth $\Delta f$ (in Hz) of the circuit.

Step-by-step solution

1. Calculate angular frequency $\omega =2\pi f=2\pi \times 50=314.16,\text{rad/s}$.
2. Compute resonant frequency: $f_0=\frac{1}{2pisqrtLC}$. Here $LC=0.4\times 25\times 10^{-6}=1\times 10^{-5}$. Thus $f_0=1/(2\pi \times \sqrt{1\times 10^{-5}})\approx 50.3,\text{Hz}$.
3. Since $fapprox{f}_0$, the circuit is at resonance.
4. At resonance the impedance equals $R$, so $I_{\text{rms}}=\frac{V}{R}=\frac{110}{30}=3.667,\text{A}$.
5. Quality factor: $Q=\frac{\omega L}{R}=\frac{314.16\times 0.4}{30}\approx 4.19$.
6. Bandwidth: $\Delta f=\frac{f_0}{Q}=\frac{50.3}{4.19}\approx 12.0,\text{Hz}$.

Concept tested: RLC resonance and bandwidth

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked