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:

Correct — C, 15 W. What a bulb's rating actually fixes is its resistance, not its power: '60 W at 220 V' means that when 220 volts is applied across it, it dissipates 60 watts, from which R equals V squared divided by P, that is 220 squared divided by 60, about 806.7 ohms. Take that resistance to a 110 volt supply and the power becomes V squared divided by R, that is 110 squared divided by 806.7, which is 15 watts. The proportional route is quicker and safer under time pressure. For a fixed resistance, power varies as the square of the applied voltage. The voltage here is halved, so the power falls by a factor of two squared, that is four — and a quarter of 60 watts is 15 watts. This is why a bulb run on a sagging supply glows so much more dimly than the voltage drop alone suggests.

• (a)30 W — The commonest error — treating power as proportional to voltage and simply halving 60. Power goes as the square of the voltage at fixed resistance, so halving the voltage quarters the power.
• (b)45 W — Corresponds to three-quarters of the rated power, which no combination of the given figures produces. At 110 V across a fixed 806.7 ohm filament the dissipation is 15 W.
• (d)2 W — Far too small. Dividing by four, not by thirty, is what the halved voltage requires; 2 W would correspond to a voltage of about 40 V across the same filament.

For a resistor obeying Ohm's law the electrical power dissipated can be written three ways — P equals VI, P equals I squared R, and P equals V squared over R — and which one is convenient depends on what is held fixed. When the resistance is the fixed property of a device and the applied voltage changes, the third form is the one to use, and it shows the power varying as the square of the voltage. A bulb's marked rating is a pair of numbers, a power and the voltage at which that power is delivered; together they encode the resistance, which is the property the bulb actually carries with it.

The single most useful habit for this class of question is to convert a rating into a resistance before doing anything else. Once R is in hand, every later question about the same device — at a different voltage, in series with another bulb, on a fluctuating supply — becomes routine. It is also worth being honest about the idealisation: school problems treat the filament resistance as constant, but tungsten's resistance rises steeply with temperature, so a bulb running at half voltage is much cooler and its resistance is appreciably lower than the rated figure, which lifts the real power somewhat above 15 watts. The exam expects the constant-resistance answer.

• The rating '60 W at 220 V' fixes the bulb's resistance: R = V squared / P = 220 squared / 60, about 806.7 ohms.
• At 110 V the same resistance dissipates P = V squared / R = 110 squared / 806.7 = 15 W.
• At constant resistance power varies as the square of the applied voltage, so halving the voltage reduces the power to one quarter.
• The three standard forms of electrical power are P = VI, P = I squared R and P = V squared / R.
• In reality a tungsten filament's resistance falls as it cools, so a bulb run at half voltage draws slightly more than the ideal 15 W.

• Scaling power linearly with voltage instead of with the square of the voltage.
• Assuming the bulb still delivers its rated power on any supply; the rating is only valid at the rated voltage.
• Forgetting that the constant-resistance assumption is an idealisation for a filament lamp.

As a bulb-rating numerical at a changed voltage, as a series-and-parallel comparison of two differently rated bulbs, or as an energy-consumption calculation in kilowatt hours.

• practice — not a real PYQ

  An electric heater rated 1000 W at 220 V is operated at 110 V. Assuming the resistance stays constant, the power consumed is:

  • (a)1000 W
  • (b)500 W
  • (c)250 W
  • (d)100 W

  Answer(c) 250 W — at fixed resistance power varies as the square of the voltage, so halving the voltage gives a quarter of the power.

• practice — not a real PYQ

  The resistance of a bulb rated 100 W at 200 V is:

  • (a)2 ohms
  • (b)20 ohms
  • (c)200 ohms
  • (d)400 ohms

  Answer(d) 400 ohms — R = V²/P = 200 × 200 / 100 = 400 ohms.