An alternating e.m.f. of 0 2 ~V is applied across an LCR series circuit having R =4 , C =80 F and L =200 mH . At resonance the voltage drop across the inductor is

Step-by-step solution

At resonance X _ L = X _ C Current = 0.2 R = 0.2 4 = 0.1 2 =0.05 ~A Now, X _ L = L =2 f L aligned f &= 1 2 LC X _ L &= 2 L 2 LC = L C X _ L &= 200 10⁻³ 80 10⁻⁶ = 10² 2 =50 V _ L &= i L =0.05 50=2.5 ~V aligned