JEE Main 2024 Physics question
When a dc voltage of 100V is applied to an inductor, a dc current of 5A flows through it. When an ac voltage of 200V peak value is connected to inductor, its inductive reactance is found to be 20√3 Ω. The power dissipated in the circuit is W.
Answer: 250
Solution
R = 100/5 = 20 Ω. Z = √(20² + 1200) = 40 Ω; = (200/√2)/40 and P = ²R = (25/2)(20) = 250 W.