A series LCR circuit of $$R = 5$$ $$\Omega$$, $$L = 20$$ mH and $$C = 0.5$$ $$\mu$$F is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is ___ $$\times 10^2$$ W.

At resonance in a series LCR circuit, the impedance is purely resistive: $$Z = R$$. Therefore, the current at resonance is maximum and equals:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{R} = \frac{250}{5} = 50 \text{ A}$$

The power dissipated at resonance is:

$$P = I_{\text{rms}}^2 \times R = (50)^2 \times 5 = 2500 \times 5 = 12500 \text{ W}$$

Expressing in units of $$10^2$$ W: $$P = 125 \times 10^2$$ W.

Therefore the answer is $$125$$.