An LCR series circuit of capacitance `62.5 nF` and resistance of `50Omega`, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is _____ mH. (Take `pi^2= 10`)

Answer

To solve the problem, we need to find the inductance \( L \) in an LCR series circuit for maximum current. The condition for maximum current in an LCR circuit is that the circuit is at resonance, which occurs when the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance \( C = 62.5 \, \text{nF} = 62.5 \times 10^{-9} \, \text{F} \)
- Resistance \( R = 50 \, \Omega \)
- Frequency \( f = 2.0 \, \text{kHz} = 2000 \, \text{Hz} \)

2. **Calculate the angular frequency \( \omega \):**
\[ \omega = 2 \pi f = 2 \pi \times 2000 = 4000 \pi \, \text{rad/s} \]

3. **Set up the resonance condition:**
At resonance, \( X_L = X_C \).
- The inductive reactance \( X_L = \omega L \)
- The capacitive reactance \( X_C = \frac{1}{\omega C} \)

Therefore, we have:
\[ \omega L = \frac{1}{\omega C} \]

4. **Rearranging the equation to find \( L \):**
\[ L = \frac{1}{\omega^2 C} \]

5. **Substituting the values:**
\[ L = \frac{1}{(4000 \pi)^2 \times (62.5 \times 10^{-9})} \]

6. **Calculating \( (4000 \pi)^2 \):**
\[ (4000 \pi)^2 = 16000000 \pi^2 \]
Given that \( \pi^2 = 10 \), we have:
\[ (4000 \pi)^2 = 16000000 \times 10 = 160000000 \]

7. **Substituting back into the equation for \( L \):**
\[ L = \frac{1}{160000000 \times 62.5 \times 10^{-9}} \]

8. **Calculating the denominator:**
\[ 160000000 \times 62.5 \times 10^{-9} = 10 \times 10^{-9} = 10^{-8} \]

9. **Final calculation for \( L \):**
\[ L = \frac{1}{10^{-8}} = 0.1 \, \text{H} = 100 \, \text{mH} \]

### Final Answer:
The value of inductance \( L \) is **100 mH**.