Q. An ac voltage V = 220 sin(2 × 10³ t) Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is: (Given : L = 10 mH, C = 25 μF, R = 100 Ω)
A) 22.0 A
B) 2.2 A
C) 5.5 A
D) 11.0 A
Explanation
[Solution from source] ω = 2000 rad s⁻¹: $X_L=\omega L=20\ \Omega$ and $X_C=1/(\omega C)=20\ \Omega$. They cancel (resonance), so Z = R = 100 Ω and $I_0=220/100=2.2$ A. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.