In an $$LCR$$ series circuit, an inductor 30 mH and a resistor 1 $$\Omega$$ are connected to an AC source of angular frequency 300 rad s$$^{-1}$$. The value of capacitance for which the current leads the voltage by 45$$^\circ$$ is $$\frac{1}{x} \times 10^{-3}$$ F. Then the value of $$x$$ is ___.

In an LCR series circuit with $$L = 30$$ mH $$= 30 \times 10^{-3}$$ H, $$R = 1\ \Omega$$, and angular frequency $$\omega = 300$$ rad s$$^{-1}$$, we need to find the capacitance for which the current leads the voltage by $$45^\circ$$.

The inductive reactance is $$X_L = \omega L = 300 \times 30 \times 10^{-3} = 9\ \Omega$$.

For the current to lead the voltage, the circuit must be capacitively dominant, meaning $$X_C > X_L$$. The phase angle $$\phi$$ satisfies:

$$\tan\phi = \frac{X_C - X_L}{R}$$

With $$\phi = 45^\circ$$, $$\tan 45^\circ = 1$$, so:

$$X_C - X_L = R \implies X_C - 9 = 1 \implies X_C = 10\ \Omega$$

Since $$X_C = \frac{1}{\omega C}$$:

$$\frac{1}{\omega C} = 10 \implies C = \frac{1}{300 \times 10} = \frac{1}{3000} = \frac{1}{3} \times 10^{-3}\ \text{F}$$

Comparing with $$C = \frac{1}{x} \times 10^{-3}$$ F, we get $$x = 3$$.

Therefore, the value of $$x$$ is $$\boxed{3}$$.