NEET2026PhysicsChapterActual
An ideal inductor of inductance L = 2 H is connected to an AC source as shown in the figure. If the RMS voltage of the source is 220 V and its frequency is 50 Hz , the peak current in the circuit is approximately: (Take 2 = 1.41 )
Options
- A1.10 A
- B0.78 A
- C1.55 A
- D3.11 A
Correct answer
C. 1.55 A
Step-by-step solution
The inductive reactance X_L of the circuit is calculated as: X_L = L = 2 f L Substituting the given values f = 50 Hz and L = 2 H : X_L = 2 50 2 = 200 The RMS current I_ rms in the circuit is: I_ rms = V_ rms X_L Substituting V_ rms = 220 V : I_ rms = 220 200 = 1.1 A The peak current I₀ is related to the RMS current by the relation: I₀ = 2 I_ rms Using the given value 2 = 1.41 : I₀ = 1.41 1.1 = 1.551 A The peak current is approximately 1.55 A . Answer: 1.55 A