If a light bulb operates at 220 V and resistance of the bulb is 484 Ω, then the value of powerdissipated will be
Correct answer: C. 100.0 W.
- A. 0.4 W.
- B. 2.2 W.
- C. 100.0 W.
- D. 1064.8 W.
Explanation
To find the power dissipated in the light bulb, use the formula P = V²/R, where V is the voltage (220 V) and R is the resistance (484 Ω). Substituting these values gives P = (220)² / 484 = 100 W. Therefore, the correct answer is 100 W. The other options are incorrect due to miscalculations or misunderstandings of the formula.
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Steady current is the continuous flow of charge at a constant rate through a conductor. The topic covers current, charge, potential difference, resistance, Ohm's law, resistivity and electrical power, along with series and parallel circuits, while distinguishing conventional current direction from electron motion.
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