An LCR series circuit of capacitance 62.5 nF and resistance of 50 $$\Omega$$, 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 $$\pi^2 = 10$$)

For maximum current amplitude in an LCR series circuit, the circuit must be at resonance, where $$X_L = X_C$$.

Given: $$C = 62.5$$ nF $$= 62.5 \times 10^{-9}$$ F, $$R = 50 \; \Omega$$, $$f = 2.0$$ kHz $$= 2000$$ Hz, and $$\pi^2 = 10$$.

At resonance:

$$\omega L = \frac{1}{\omega C}$$

$$L = \frac{1}{\omega^2 C}$$

The angular frequency:

$$\omega = 2\pi f = 2\pi \times 2000 = 4000\pi$$

Computing $$\omega^2$$:

$$\omega^2 = (4000\pi)^2 = 16 \times 10^6 \times \pi^2 = 16 \times 10^6 \times 10 = 1.6 \times 10^8$$

Therefore:

$$L = \frac{1}{1.6 \times 10^8 \times 62.5 \times 10^{-9}} = \frac{1}{1.6 \times 10^8 \times 6.25 \times 10^{-8}} = \frac{1}{10} = 0.1 \text{ H}$$

Converting to millihenry:

$$L = 0.1 \text{ H} = 100 \text{ mH}$$

Therefore, the answer is $$\mathbf{100}$$.