1. **State the problem:** We have an electric bulb marked 220V and 100W. We need to find: - The resistance of the filament. - The energy consumed in kilowatt-hours if used 5 hours daily for one month. 2. **Formulas and rules:** - Power formula: $$P = \frac{V^2}{R}$$ where $P$ is power, $V$ is voltage, and $R$ is resistance. - Energy consumed: $$E = P \times t$$ where $E$ is energy in watt-hours, $P$ is power in watts, and $t$ is time in hours. - Convert energy to kilowatt-hours by dividing by 1000. 3. **Calculate resistance:** Given $P=100$ W and $V=220$ V, $$R = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484\ \Omega$$ 4. **Calculate energy consumed in one month:** - Daily usage time $t = 5$ hours - Number of days in a month (assume 30 days) - Total time $T = 5 \times 30 = 150$ hours - Energy in watt-hours: $$E = 100 \times 150 = 15000\ \text{Wh}$$ - Convert to kilowatt-hours: $$E_{kWh} = \frac{15000}{1000} = 15\ \text{kWh}$$ **Final answers:** - Resistance of filament: $484\ \Omega$ - Energy consumed in one month: $15\ \text{kWh}$