MHT CET202312 May 2023Evening ShiftPhysicsAlternating CurrentActual
Resistor of 2 , inductor of 100 H and capacitor of 400 pF are connected in series across a source of e _ rms =0.1 Volt. At resonance, voltage drop across inductor is
Options
- A25 ~V
- B2.5 ~V
- C250 ~V
- D20 ~V
Correct answer
A. 25 ~V
Step-by-step solution
At resonance condition, X _ C = X _ L The impedance is given as: aligned & Z = R ^2+ ( X _ L - X _ C )^2 & Z = R =2 & I _ rms = e _ mms R & I _ rms = 0.1 2 =0.05 ~A & = 1 LC = 1 10⁻⁴ 4 10⁻¹⁰ & =5 10^6 aligned The voltage-drop across the inductor is, aligned & V = I _ rms X _ L = I _ rms L & V =0.05 10⁻⁴ 5 10^6 & ~V =25 ~V aligned