A series LCR circuit has `R=5 Omega, L=40 mH` and `C=1mu F`, the bandwidth of the circuit is

Options

A. 10 Hz
B. 20 Hz
C. 30 Hz
D. 40 Hz

Explanation

To find the bandwidth of a series LCR circuit with given values of resistance (R), inductance (L), and capacitance (C), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Resistance, \( R = 5 \, \Omega \)
- Inductance, \( L = 40 \, \text{mH} = 40 \times 10^{-3} \, \text{H} \)
- Capacitance, \( C = 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F} \)

2. **Calculate the resonant frequency \( f_0 \)**:
The resonant frequency \( f_0 \) for an LCR circuit is given by the formula:
\[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \]
Substituting the values of \( L \) and \( C \):
\[ f_0 = \frac{1}{2 \pi \sqrt{(40 \times 10^{-3}) \times (1 \times 10^{-6})}} \]

3. **Calculate \( LC \)**:
\[ LC = 40 \times 10^{-3} \times 1 \times 10^{-6} = 40 \times 10^{-9} \, \text{H} \cdot \text{F} \]

4. **Calculate \( \sqrt{LC} \)**:
\[ \sqrt{LC} = \sqrt{40 \times 10^{-9}} = \sqrt{40} \times 10^{-4.5} \approx 6.32 \times 10^{-5} \, \text{s} \]

5. **Calculate \( f_0 \)**:
\[ f_0 = \frac{1}{2 \pi (6.32 \times 10^{-5})} \approx \frac{1}{3.98 \times 10^{-4}} \approx 2510 \, \text{Hz} \]

6. **Calculate the bandwidth \( \Delta f \)**:
The bandwidth \( \Delta f \) for a series LCR circuit is given by:
\[ \Delta f = \frac{R}{2 \pi L} \]
Substituting the values:
\[ \Delta f = \frac{5}{2 \pi (40 \times 10^{-3})} \]

7. **Calculate \( 2 \pi L \)**:
\[ 2 \pi L = 2 \pi (40 \times 10^{-3}) \approx 0.2513 \, \text{s} \]

8. **Calculate \( \Delta f \)**:
\[ \Delta f = \frac{5}{0.2513} \approx 19.88 \, \text{Hz} \approx 20 \, \text{Hz} \]

9. **Final Result**:
The bandwidth of the circuit is approximately \( 20 \, \text{Hz} \).

### Summary:
The bandwidth of the series LCR circuit is \( 20 \, \text{Hz} \).

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