A light bulb rated as 60 W at 220 V has a potential difference of 110 V across its ends. The power dissipated in this light bulb is:

The correct answer is 15 W.

Key Points

• The power dissipated by a resistive device can be calculated using the formula P = V²/R, where P is power, V is voltage, and R is resistance.
• The resistance of the bulb is derived from its rated values using R = V²/P. For a 60 W bulb at 220 V, R = (220)² / 60 = 806.67 Ω.
• When the potential difference is reduced to 110 V, the power dissipated is recalculated as P = (110)² / 806.67 ≈ 15 W.
• This reduction in power occurs because the power dissipation is proportional to the square of the voltage applied across the bulb.
• Thus, the power dissipated by the bulb at 110 V is approximately 15 W.

Additional Information

• Ohm's Law:
  • Ohm's Law states that the current (I) through a conductor is directly proportional to the voltage (V) across it, and inversely proportional to its resistance (R): I = V/R.
  • This principle helps in understanding the relationship between power, voltage, and resistance in electrical circuits.
• Power Formulae in Circuits:
  • Power (P) can also be expressed as P = IV or P = I²R, depending on the known variables.
  • These formulas are vital in analyzing energy consumption and dissipation in electrical devices.
• Resistive Load Behavior:
  • For resistive loads like incandescent light bulbs, power dissipation is strongly dependent on the voltage applied across the load.
  • Halving the voltage results in quartering the power dissipation, as power is proportional to the square of the voltage.
• Practical Implications:
  • Operating a bulb at a lower voltage than its rated value reduces brightness and power output.
  • Such reduced voltage operation can potentially extend the lifespan of the bulb but at the cost of reduced efficiency.
• Electrical Safety:
  • Understanding power dissipation is essential for ensuring the safe operation of electrical devices and preventing overheating or fire hazards.