An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about _____ W.

We have two bulbs connected in series: Bulb 1 rated at 220 V, 100 W and Bulb 2 rated at 220 V, 60 W. The total voltage across the combination is 220 V.

The resistance of each bulb is found from its rating. For Bulb 1, $$R_1 = \frac{V^2}{P_1} = \frac{220^2}{100} = \frac{48400}{100} = 484 \; \Omega$$. For Bulb 2, $$R_2 = \frac{V^2}{P_2} = \frac{220^2}{60} = \frac{48400}{60} = \frac{2420}{3} \; \Omega$$.

Since the bulbs are in series, the total resistance is $$R = R_1 + R_2 = 484 + \frac{2420}{3} = \frac{1452 + 2420}{3} = \frac{3872}{3} \; \Omega$$.

The current through the series combination is $$I = \frac{V}{R} = \frac{220}{\frac{3872}{3}} = \frac{660}{3872} \text{ A}$$.

The power consumed by Bulb 1 (the 100 W bulb) is $$P = I^2 R_1 = \left(\frac{660}{3872}\right)^2 \times 484$$.

Now, $$\frac{660}{3872} = \frac{165}{968}$$, so $$I^2 = \frac{165^2}{968^2} = \frac{27225}{937024}$$.

Therefore, $$P = \frac{27225 \times 484}{937024} = \frac{13176900}{937024} \approx 14.06 \text{ W}$$.

Hence, the power consumed by the 100 W bulb is about $$\textbf{14}$$ W.