Classical: loop currents with 1 H and 1 μF

Assumptions: the network is initially relaxed. Left mesh (i₁): 10 V, switch and 2 Ω. Shared branch: 3 Ω in series with 1 H. Right mesh (i₂): 1 Ω and C = 1 μF.

Mesh equations (t > 0)

Mesh 1:  10 = 2 i₁ + 3(i₁ − i₂) + d(i₁ − i₂)/dt
Mesh 2:  0 = i₂ + 10⁶∫i₂ dt + 3(i₂ − i₁) + d(i₂ − i₁)/dt

Using the operator s = d/dt (and 1/s for ∫):

(5 + s) i₁ − (3 + s) i₂ = 10
−(3 + s) i₁ + (4 + s + 10⁶/s) i₂ = 0

Characteristic equation

Δ = (5 + s)(4 + s + 10⁶/s) − (3 + s)² = 0
  → 3s² + (10⁶ + 11)s + 5×10⁶ = 0
  → s² + 333 337 s + 1.667×10⁶ = 0
s₁ = −5.0,   s₂ = −3.333×10⁵ (1/s)

Both roots are real, so the response is overdamped.

Steady state (t → ∞)

The capacitor is open (i₂ = 0) and the inductor is short:

i₁(∞) = 10/(2 + 3) = 2 A,   i₂(∞) = 0

Initial values at t = 0⁺

The inductor is open and the capacitor is a short, so the 2 Ω and 1 Ω are in series with the 10 V source: i₁(0⁺) = i₂(0⁺) = 10/3 = 3.333 A.

Constants

Writing i₁ = 2 + A₁e^(s₁t) + B₁e^(s₂t) and i₂ = A₂e^(s₁t) + B₂e^(s₂t), the first mesh equation gives A₂/A₁ = (5 + s)/(3 + s) ≈ 0.00001 (for s₁) and 0.999994 (for s₂). With i₁(0) = i₂(0) = 3.333:

A₁ = −2.0000,  B₁ = 3.3334
A₂ = −2×10⁻⁵,  B₂ = 3.3334

Result

i₁(t) = 2 − 2 e^(−5t) + 3.333 e^(−3.33×10⁵ t) A
i₂(t) ≈ 3.333 e^(−3.33×10⁵ t) A   (plus a negligible −2×10⁻⁵ e^(−5t))

Because τ₂ is only 3 μs, i₂ is a very short pulse that charges the capacitor. i₁ jumps to 3.33 A at t = 0⁺, drops to almost zero within a few microseconds (the inductor current i₁ − i₂ starts from zero), and then rises to its final value of 2 A with the slow time constant 1/5 = 0.2 s.

Answer: i₁(t) = 2 − 2e^(−5t) + 3.33e^(−3.33×10⁵t) A, i₂(t) ≈ 3.33e^(−3.33×10⁵t) A.