NTA Abhyas JEE Main2020PhysicsElectromagnetic InductionPractice
The current through an inductor of impedance 10 Ω lags behind the voltage by a phase of 60 ° when just the inductor is connected to the ac source. Now the inductor is connected to a 5 Ω resistance in series, then the net impedance of the circuit is
Options
- A15 Ω
- B12 Ω
- C13.2 Ω
- D18 Ω
Correct answer
C. 13.2 Ω
Step-by-step solution
10 = r 2 + X L 2 and X L r = tan ⁡ 60 ° 10 = r 2 + r 3 2 or r = 5   Ω ,   X L = 5 3   Ω Z = 5 + 5 2 + 5 3 2 = 175 = 13.2   Ω