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Classical: inductor current, 5 V with 2 H

Short answer

Initially iL=0.5 A. After closing, the Thevenin equivalent seen by the inductor has Vth=4.286 V, Rth=7.143 Ω, giving iL(∞)=0.6 A, τ=0.28 s, so iL(t)=0.6−0.1e^{-3.571t} A for t>0.

In the circuit of the figure below, the switch K is open and the circuit reaches a steady state. At t = 0, K is closed. Find the current in the inductor for t > 0. Use classical method. [Figure: 5 V source. A 10 Ω resistor runs from the source positive terminal to a middle node, then a 20 Ω resistor to the top-right node; a 2 H inductor is connected in parallel across the series 10 Ω + 20 Ω (from the source positive node to the top-right node). From the middle node, switch K in series with a 10 Ω resistor connects to the bottom wire. A 10 Ω resistor connects the top-right node to the bottom wire.]

Assumptions from the figure: 5 V source (node A at its + terminal). 10 Ω from A to node M, 20 Ω from M to node R; the 2 H inductor is connected from A to R (across the 10 Ω + 20 Ω series pair). From M, switch K and a 10 Ω resistor go to ground; a 10 Ω resistor connects R to ground. K is open until t = 0.

Before t = 0 (K open, dc steady state)

The inductor is a short, so V_R = 5 V and there is no voltage across the 10 Ω + 20 Ω path, hence no current in it. The inductor current flows only to the 10 Ω load at R:

i_L(0⁻) = 5/10 = 0.5 A = i_L(0⁺)

After t = 0 (K closed): Thevenin equivalent seen by the inductor

Open-circuit voltage (inductor removed), unknowns V_M, V_R with A = 5 V:

M:  (5 − V_M)/10 = V_M/10 + (V_M − V_R)/20
R:  (V_M − V_R)/20 = V_R/10   →  V_M = 3V_R
→  2(5 − 3V_R) = 2(3V_R) + 2V_R  →  10 = 14 V_R,  V_R = 0.714 V
V_oc = 5 − 0.714 = 4.286 V

Thevenin resistance (source shorted, so A = ground): looking between A and R,

R_th = 10 ∥ [20 + (10 ∥ 10)] = 10 ∥ 25 = 7.143 Ω

Final current and time constant:

i_L(∞) = V_oc/R_th = 4.286/7.143 = 0.6 A
τ = L/R_th = 2/7.143 = 0.28 s

Solution

i_L(t) = 0.6 + (0.5 − 0.6) e^(−t/0.28)
i_L(t) = 0.6 − 0.1 e^(−3.571t) A,   t > 0

Check: at t = 0⁺ the node equations with i_L = 0.5 A give a voltage 0.714 V across L, and L di/dt = 2 × 0.1 × 3.571 = 0.714 V ✓ (verified numerically). At t → ∞ the voltage across L is zero ✓.

Answer: i_L(t) = 0.6 − 0.1e^(−3.571t) A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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