An ac circuit as shown in the figure has an inductor of inductance L and a resistor of resistance R connected in series. Using the phasor diagram, explain why the voltage in the circuit will lead the current in phase. The potential difference across the resistor is 160V and that across the inductor is 120V. Find the effective value of the applied voltage. If the effective current in the circuit be 1.0A. calculate the total impedance of the circuit. What will be the potential difference in the circuit when direct current is passed through the circuit?

a) ε is φ ahead of i because i and R are in phase and i.e., −Z and ε are in phase as ε = ε₀sin wt from diagram i = ε₀ −Zsin(ωt − φ) where, φ = tan⁻¹ XLR
b) i = 1.0A (given)
VR = 160V = iR ⇒ R = 160/1 = 160
VL = 120V = iXL ⇒ XL = 120/1 = 120
Vnet = √(VR² + VL²) = √(160)² + (120)² = √25600 + 14400 = 200: effective value of applied voltage
Z = √(R² + XL²) = √(160)² + (120)² = 200: Impedance
C) When direct current is passed through the circuit then XL will be zero (at ω = 0, inductor will act as a plane wire) and potential drop will only occur at R (i.e., i = ε/R)
For intermediate time E = iR + L(di/dt) where i will vary accordingly