NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
A coil of inductive reactance 3 1 Ω has a resistance of 8 Ω . It is placed in series with a condenser of capacitive reactance 2 5 Ω . The combination is connected to an ac source of 110 V . The power factor of the circuit is
Options
- A0 .33
- B0 .56
- C0 .64
- D0 .80
Correct answer
D. 0 .80
Step-by-step solution
X L = 3 1 Ω , X C = 2 5 Ω , R = 8 Ω Impedance of series L C R is Z = R 2 + X L - X C 2 = 8 2 + 3 1 - 2 5 2 = 6 4 + 3 6 = 1 0 Ω Power factor, cos ϕ = R Z = 8 1 0 = 0.8