Inductor voltage and time for i = i0/2

Part 1: V_L = ε e^(−Rt/L)

Circuit: battery ε, resistance R, inductance L in series, switched on at t = 0.

1. Current growth: i = (ε/R)(1 − e^(−Rt/L)), obtained from ε = iR + L di/dt.
2. Voltage across the resistor: V_R = iR = ε(1 − e^(−Rt/L))
3. Loop rule: V_L = ε − V_R = ε − ε(1 − e^(−Rt/L))

V_L = ε e^(−Rt/L)

(Equivalently V_L = L di/dt = L · (ε/R)(R/L) e^(−Rt/L) = ε e^(−Rt/L).) At t = 0, V_L = ε (all of the emf), and it decays to zero as the current settles at ε/R.

Part 2: when V_L = V_R

1. Set V_L = V_R: since V_L + V_R = ε, each equals ε/2.
2. ε e^(−Rt/L) = ε/2 → e^(−Rt/L) = ½
3. Rt/L = ln 2

t = (L/R) ln 2 = 0.693 τ, where τ = L/R.

Current at that time

i = (ε/R)(1 − ½) = ½ (ε/R) = i₀/2, where i₀ = ε/R is the final current. So at t = 0.693 L/R the voltage across L equals that across R, and the current is half its final value.