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JEE Main 2026 Physics question

PhysicsCurrent ElectricityJEE Main 2026

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.

Answer: (B)

Solution

Bulb resistance = 200²/100 = 400 Ω, in parallel with the 400 Ω resistor giving 200 Ω. Total resistance = 200 +

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Question ID VK-015147