An inductor of inductance $$2$$ $$\mu$$H is connected in series with a resistance, a variable capacitor and an AC source of frequency $$7$$ kHz. The value of capacitance for which maximum current is drawn into the circuit is $$\frac{1}{x}$$ F, where the value of $$x$$ is ______. (Take $$\pi = \frac{22}{7}$$)

We need to find the value of $$x$$ where the capacitance for maximum current is $$\dfrac{1}{x}$$ F.

Maximum current in a series LCR circuit occurs at resonance, where $$\omega^2 L C = 1 \implies C = \dfrac{1}{\omega^2 L}$$.

Given $$f = 7$$ kHz $$= 7000$$ Hz, $$L = 2\,\mu$$H $$= 2 \times 10^{-6}$$ H, and $$\pi = \dfrac{22}{7}$$, we have $$\omega = 2\pi f = 2 \times \dfrac{22}{7} \times 7000 = 2 \times 22000 = 44000 \text{ rad/s}$$.

Then $$\omega^2 = (44000)^2 = 1936 \times 10^6$$ and $$C = \dfrac{1}{\omega^2 L} = \dfrac{1}{1936 \times 10^6 \times 2 \times 10^{-6}} = \dfrac{1}{1936 \times 2} = \dfrac{1}{3872} \text{ F}$$.

Since $$C = \dfrac{1}{x}$$ F, it follows that $$x = 3872$$.

The answer is $$\boxed{3872}$$.