Classical: inductor current and voltage at 10 ms

Assumptions from the figure: the 20 V and 10 V sources are in series aiding (total 30 V, with the 10 V negative terminal at the reference). The + terminal of the 20 V source feeds a 30 Ω resistor to node N. From N, a 20 Ω resistor and the 0.5 H inductor go to the reference. The switch connects the junction of the two sources (10 V above the reference) to N. The switch is open for a long time, and closes at t = 0.

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

i(0⁻) = 30/(30 + 20) = 0.6 A = i(0⁺)

After t = 0 (switch closed)

The switch connects N directly to the 10 V point, so the 20 Ω + L branch is across 10 V (the 30 Ω and 20 V source no longer matter to it).

L di/dt + 20 i = 10
0.5 di/dt + 20 i = 10  →  di/dt + 40 i = 20

• Particular (final) value: i(∞) = 10/20 = 0.5 A.
• Complementary: s = −40, τ = L/R = 0.5/20 = 25 ms.

i(t) = 0.5 + K e^(−40t),   i(0⁺) = 0.6  →  K = 0.1
i(t) = 0.5 + 0.1 e^(−40t) A,   t > 0

Voltage across the inductor at t = 10 ms

v_L = L di/dt = 0.5 × (−40)(0.1) e^(−40t) = −2 e^(−40t) V
v_L(10 ms) = −2 e^(−0.4) = −2 × 0.6703 = −1.341 V

Answer: i(t) = 0.5 + 0.1e^(−40t) A; v_L(10 ms) = −1.34 V.