99 Percentile Qs Bank for JEE MainPhysicsAlternating Current
A coil has a resistance of 30 and an inductive reactance of 20 at 50 ~Hz frequency. If an AC source of 200 ~V , 100 ~Hz is connected across the coil, the current in the coil is
Options
- A2 ~A
- B20 13 ~A
- C4 ~A
- D8 ~A
Correct answer
C. 4 ~A
Step-by-step solution
Inductive reactance, X _ L = L 20=2 (50) L...(1) When AC source of 200 ~V , 100 ~Hz is connected across the coil then aligned & X _ L =2 ( f ) L =2 (100) L =2 (50) L 2 & X _ L ^ =20 2=40 aligned Impedance, Z = R ^2+ ( X _ L )^2 = (30)^2+(40)^2 =50 Current in the coil, I= V Z = 200 50 =4 A