Q. A coil has resistance 30 ohm and inductive reactance 20 ohm at 50 Hz frequency. If an ac source of 200 volt, 100 Hz is connected across the coil, the current in the coil will be
A) 20/√13 A
B) 2.0 A
C) 4.0 A
D) 8.0 A
Explanation
[Solution from source] Inductive reactance doubles with frequency: at 100 Hz XL = 40 Ω. Z = √(30² + 40²) = 50 Ω, so I = 200/50 = 4 A. Keeping XL = 20 Ω gives 20/√13 A, which is the trap. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.