Two resistors of 200 Ω and 400 Ω are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the 400 Ω resistance. The potential drop across the bulb is V.
- 25
- 50
- 66.6
- 100
Answer
(2) 50
Solution
Bulb resistance: Rb = V2P = (200)2100 = 400 Ω
Bulb ∥ 400 Ω: Rp = 400 400400 + 400 = 200 Ω
Circuit: 200 Ω in series with Rp = 200 Ω across 100 V
I = 100200 + 200 = 0.25 A
Potential drop across bulb = IRp = 0.25 200 = 50 V
Answer: (2)