An ideal 9.00 V battery, 15.0 Ω resistor and 0.300 H inductor are connected in series at t = 0. The inductor initially carries no current. Determine the time constant, initial current gradient, current and inductor voltage at t = 0.0400 s, and magnetic energy then. Explain the inductor voltage immediately after closing and long afterward, and describe one safe way to observe the transient.

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τ = 0.0200 s; (di/dt)_0 = 30.0 A s⁻¹; i(0.0400 s) = 0.519 A; V_L = 1.22 V; U_L = 0.0404 J. The LR time constant is τ = L/R = 0.300/15.0 = 0.0200 s. The final current is ε/R = 0.600 A. At t = 0 the current is zero, so the resistor drop is zero and L(di/dt) = ε. Therefore the initial gradient is 9.00/0.300 = 30.0 A s⁻¹. Current growth follows i = (ε/R)(1−e^(−t/τ)). At t = 0.0400 s = 2τ, i = 0.600(1−e⁻²) = 0.519 A. The inductor voltage is εe^(−t/τ) = 9.00e⁻² = 1.22 V. Stored magnetic energy is ½Li² = 0.0404 J. Initially self-induction takes the full battery voltage and opposes rapid current change; at steady state di/dt = 0 and its voltage tends to zero. Observe voltage with a properly rated data logger and include a flyback path before opening the circuit to suppress a damaging voltage spike.

Current growth in an LR circuit

Choose your AP course · School year 2026–27 · May 2027 exam

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