Resistor of 2 , inductor of 100 H and capacitor of 400 pF are connected in series across an a.c. source of e_ r m s =0.1 volt. At resonance, voltage drop across inductor is

Step-by-step solution

We know, at resonance X_C=X_L L= 1 C = ( 1 L C ) The potential drop across inductor is V_L=i X_L=i L C LCR circuit in series and at resonance i= V R = 0.1 2 ~A =0.05 ~A Now, V_L=0.05 ( 100 10⁻⁶ 400 10⁻¹² ) V =2.5 10⁻² 10^3 ~V =25 ~V