A series LCR circuit with $$L = \frac{100}{\pi}$$ mH, $$C = \frac{10^{-3}}{\pi}$$ F and $$R = 10$$ $$\Omega$$, is connected across an AC source of 220 V, 50 Hz supply. The power factor of the circuit would be _____.

Given: $$L = \frac{100}{\pi}$$ mH $$= \frac{0.1}{\pi}$$ H, $$C = \frac{10^{-3}}{\pi}$$ F, $$R = 10 \; \Omega$$, $$f = 50$$ Hz.

$$\omega = 2\pi f = 100\pi$$ rad/s.

$$X_L = \omega L = 100\pi \times \frac{0.1}{\pi} = 10 \; \Omega$$

$$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times \frac{10^{-3}}{\pi}} = \frac{1}{100 \times 10^{-3}} = \frac{1}{0.1} = 10 \; \Omega$$

Since $$X_L = X_C$$, the circuit is at resonance.

Impedance: $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100 + 0} = 10 \; \Omega$$

Power factor: $$\cos\phi = \frac{R}{Z} = \frac{10}{10} = 1$$

The answer is $$\boxed{1}$$.