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Classical: inductor current, 10 V with 2 H

Short answer

Initial inductor current (switch open) is 1.333 A, final current (switch closed) is 1.2 A. The Thevenin resistance seen by the inductor is 7.143 Ohms, time constant 0.28 s. i_L(t) = 1.2 + 0.1333 e^(-3.571 t) A.

For the circuit shown in the following figure, use classical method to find the current in the inductor for t > 0. [Figure: 10 V source. A 10 Ω resistor runs along the top wire from the source positive node to the top-right node (bridging). A second row has a 10 Ω resistor from the source positive node to a middle node, then a 20 Ω resistor from the middle node to the right node. A 2 H inductor connects the right node to the bottom wire. From the middle node, a switch (t = 0) in series with a 10 Ω resistor connects to the bottom wire.]

Assumption (figure not clear): the switch is open for t < 0 and closes at t = 0. (The result for the opposite case, switch opening, is given at the end.)

Initial condition (t = 0⁻, switch open)

The inductor is a short circuit in steady state, so the 10 Ω (top) is in parallel with 10 Ω + 20 Ω = 30 Ω across the 10 V source.

i_L(0⁻) = 10/10 + 10/30 = 1.333 A = i_L(0⁺)

Final value (t → ∞, switch closed, L shorted)

Let V_m be the middle-node voltage; the right node is at 0 V.

(10 − Vm)/10 = Vm/10 + Vm/20   →  Vm = 4 V
i_L(∞) = 10/10 + 4/20 = 1 + 0.2 = 1.2 A

Time constant

Remove the source and look from the inductor terminals: R_th = 10 ∥ (20 + 10 ∥ 10) = 10 ∥ 25 = 7.143 Ω.

τ = L/R_th = 2/7.143 = 0.28 s

Classical solution

The differential equation is 2 di/dt + 7.143 i = 8.571 (Thevenin equivalent V_th = 8.571 V). Complete solution = forced + natural:

i_L(t) = 1.2 + A e^(−t/0.28)

i_L(0) = 1.333 → A = 0.1333

Answer: i_L(t) = 1.2 + 0.1333 e^(−3.571t) A, t > 0 (check: i_L(0) = 1.333 A, i_L(∞) = 1.2 A)

If instead the switch is closed for t < 0 and opens at t = 0: i_L(0) = 1.2 A, i_L(∞) = 10/10 + 10/30 = 1.333 A, R_th = 10 ∥ 30 = 7.5 Ω, τ = 0.2667 s, so i_L(t) = 1.333 − 0.1333 e^(−3.75t) A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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