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.

Solution

Physics sir · 3 steps

The rating gives you each bulb's resistance; once in series, the bigger resistance takes more of the power.

1. Resistances from ratings. A bulb rated V, P has R = V²/P. So R₁ = 220²/100 = 484 Ω for the 100 W bulb and R₂ = 220²/60 ≈ 806.7 Ω for the 60 W bulb.
2. Share of voltage. In series the voltage divides in the ratio of resistances. Since R ∝ 1/P, the 100 W bulb gets R₁/(R₁+R₂) = (1/100)/(1/100 + 1/60) = 60/160 = 3/8 of 220 V, that is 82.5 V.

   P₁ = V₁²/R₁ = (82.5)²/484 = 100 × (3/8)² ≈ 14.06 W

3. So. The 100 W bulb consumes about 14.06 W, which is 14 W. It glows dimmer than the 60 W bulb, which takes about 23.4 W.

The common slip: 23. Mixing up the two resistances, or assuming the 100 W bulb takes the larger share, gives 100 × (5/8)² ≈ 23 W, which is the 60 W bulb's power.

The idea to keep. A higher wattage bulb has lower resistance. In series the current is the same, so P = I²R and the lower wattage bulb actually glows brighter.

Answer from the NTA final answer key. Solution by PrepWiser.