See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

The impedance of the circuit is given by Z = 2 2 2 2 2 (R L ) [R (2 fL) ]     = 2 2 [(12) {2 3.14 50 (0.05 / 3.14)} ]     = (144 25)  = 13 ohm Current in the circuit i = E/Z = 130 13 = 10 amp. Inductive reactance of coil XL = L = 2 fL  XL = 2 × 50 × 0.05        = 5 ohm. Potential difference across inductance VL = i × XL = 10 × 5 = 50 volt.