A series resonant LCR circuit has a quality factor (Q-factor)=0.4. If `R=2k Omega, C=0.1 mu F` then the value of inductance is

Options

A. `0.1 H`
B. `0.064 H`
C. 2 H
D. 5 H

Explanation

To find the value of inductance \( L \) in a series resonant LCR circuit with a given quality factor \( Q \), resistance \( R \), and capacitance \( C \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Quality factor \( Q = 0.4 \)
- Resistance \( R = 2 \, k\Omega = 2000 \, \Omega \)
- Capacitance \( C = 0.1 \, \mu F = 0.1 \times 10^{-6} \, F \)

2. **Use the formula for the quality factor \( Q \):**
The quality factor for a series resonant circuit is given by the formula:
\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} \]

3. **Rearrange the formula to solve for \( L \):**
First, we can rearrange the equation:
\[ Q \cdot R = \sqrt{\frac{L}{C}} \]
Now, squaring both sides gives:
\[ (Q \cdot R)^2 = \frac{L}{C} \]
Therefore, we can express \( L \) as:
\[ L = (Q \cdot R)^2 \cdot C \]

4. **Substitute the known values into the equation:**
\[ L = (0.4 \cdot 2000)^2 \cdot (0.1 \times 10^{-6}) \]

5. **Calculate \( Q \cdot R \):**
\[ Q \cdot R = 0.4 \cdot 2000 = 800 \]

6. **Square the result:**
\[ (Q \cdot R)^2 = 800^2 = 640000 \]

7. **Multiply by \( C \):**
\[ L = 640000 \cdot (0.1 \times 10^{-6}) = 640000 \cdot 0.0000001 = 0.064 \, H \]

8. **Convert to milliHenries:**
\[ L = 0.064 \, H = 64 \, mH = 6.4 \, mH \]

### Final Answer:
The value of inductance \( L \) is \( 0.064 \, H \) or \( 6.4 \, mH \).