Bandwidth of a series RLC resonant circuit
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A series $RLC$ circuit has $R = 10\ \Omega$, $L = 1\ \text{mH}$ and $C = 1\ \mu\text{F}$.
Its $3\text{-dB}$ bandwidth, in Hz, is ________.
Show the step-by-step derivation
Answer
1591.5 Hz (accept 1580 to 1600) Hz (accept 1580 to 1600)
Step-by-step derivation
- Find the resonant frequency, where the inductive and capacitive reactances cancel: $$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(1\times 10^{-3})(1 \times 10^{-6})}} = \frac{1}{\sqrt{1 \times 10^{-9}}}.$$
- Evaluating the square root: $\sqrt{10^{-9}} = 3.1623 \times 10^{-5}$, so $$\omega_0 = 3.1623 \times 10^{4}\ \text{rad/s}, \qquad f_0 = \frac{\omega_0}{2\pi} = 5032.9\ \text{Hz}.$$
- Find the quality factor. For a series $RLC$ circuit, $$Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{1 \times 10^{-3}}{1 \times 10^{-6}}} = \frac{1}{10}\sqrt{1000}.$$
- With $\sqrt{1000} = 31.623$: $$Q = \frac{31.623}{10} = 3.1623.$$
- Bandwidth is the resonant frequency divided by $Q$: $$BW = \frac{f_0}{Q} = \frac{5032.9}{3.1623} = \mathbf{1591.5\ Hz}.$$
- Shortcut worth knowing. For a series $RLC$ the bandwidth in rad/s is simply $R/L$, independent of $C$: $$BW = \frac{R}{L} = \frac{10}{10^{-3}} = 10^{4}\ \text{rad/s} \;\Rightarrow\; \frac{10^4}{2\pi} = 1591.5\ \text{Hz}.$$ Same answer in one line. ✓
The trap this question is built around
Using the parallel RLC formula $Q = R\sqrt{C/L}$. The two are reciprocals of each other, so in a series circuit a larger $R$ lowers $Q$ and widens the bandwidth, while in a parallel circuit it does the opposite.