An ideal inductor L = 0.38 H, resistor R = 28 Ω and ideal 15 V battery form a series circuit. At t = 0 the switch closes and the initial current is zero. Derive i(t) from L di/dt + Ri = ε. Find the time constant, current and inductor energy at t = τ, and the initial rate di/dt. Explain the inductor's behaviour immediately after switching and long afterward.

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i(t) = (ε/R)[1 − exp(−tR/L)]; τ = 0.01357 s; i(τ) = 0.3386 A; U_L(τ) = 0.02179 J; initial di/dt = 39.47 A s⁻¹. The differential equation contains the resistor's voltage drop and the self-induced emf opposing current change. Its steady current is ε/R and its homogeneous response decays as exp(−Rt/L). Applying i(0) = 0 gives i(t) = (ε/R)[1 − exp(−t/τ)], with τ = L/R = 0.01357 s. The final current is 0.5357 A. At one time constant i = (ε/R)(1 − e⁻¹) = 0.3386 A. The stored magnetic energy is U_L = ½Li² = 0.02179 J. At t = 0, Ri = 0, so the circuit equation gives di/dt = ε/L = 39.47 A s⁻¹. The current starts continuously from zero rather than jumping. Immediately after closing, the inductor supports the full battery voltage. Long afterward di/dt approaches zero, so an ideal inductor has no steady voltage drop and the resistor limits the current. The derived solution satisfies both limits.

Current growth and energy in an LR circuit · set 9

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