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Initial values with charged capacitor, two meshes

Short answer

At t = 0+, i1 = 1.5 mA, i2 = 0.5 mA, di2/dt = −8.75×10⁻⁵ A/s, d²i2/dt² = 1.406×10⁻⁵ A/s². Initial capacitor voltages are V0 on C1 and zero on C2; derivatives obtained by differentiating mesh equations.

In the given network, the capacitor C1 is charged to voltage V0 and switch K is closed at t = 0. When R1 = 2 MΩ, V0 = 1000 V, R2 = 1 MΩ, C1 = 10 μF and C2 = 20 μF, solve for i1, i2, di2/dt, d²i2/dt² at t = 0⁺. [Figure: capacitor C1 (charged to V0) at the left; switch K in the top wire of the left mesh (closes at t = 0); resistor R2 in the middle vertical branch; resistor R1 in the top wire of the right mesh; capacitor C2 on the right; loop current i1 in the left mesh and i2 in the right mesh]

Assumptions: C₂ is initially uncharged; C₁ is charged to V₀ (voltage polarity such that it drives i₁ in the direction shown); i₁ and i₂ are the mesh currents of the left and right loops, R₂ is the common branch.

Mesh equations for t > 0

Mesh 1:  V₀ − (1/C₁)∫i₁ dt = R₂(i₁ − i₂)
Mesh 2:  R₂(i₂ − i₁) + R₁ i₂ + (1/C₂)∫i₂ dt = 0

At t = 0⁺ (v_C2 = 0, so ∫i₂ = 0; C₁ still has V₀)

V₀ = R₂(i₁ − i₂)          (R₁ + R₂) i₂ = R₂ i₁
V₀ = R₂ i₁ · R₁/(R₁ + R₂)
i₁(0⁺) = V₀(R₁ + R₂)/(R₁R₂) = 1000 × 3×10⁶ /(2×10¹²) = 1.5 mA
i₂(0⁺) = R₂ i₁/(R₁ + R₂) = 1.5/3 = 0.5 mA

First derivatives

Differentiate Mesh 1: −i₁/C₁ = R₂(i₁′ − i₂′)

i₁′ − i₂′ = −i₁/(R₂C₁) = −1.5×10⁻³/(10⁶ × 10⁻⁵) = −1.5×10⁻⁴ A/s

Differentiate Mesh 2: R₂(i₂′ − i₁′) + R₁ i₂′ + i₂/C₂ = 0

R₁ i₂′ = R₂(i₁′ − i₂′) − i₂/C₂ = 10⁶(−1.5×10⁻⁴) − 0.5×10⁻³/(2×10⁻⁵)
       = −150 − 25 = −175
di₂/dt (0⁺) = −175/(2×10⁶) = −8.75×10⁻⁵ A/s
di₁/dt (0⁺) = −8.75×10⁻⁵ − 1.5×10⁻⁴ = −2.375×10⁻⁴ A/s

Second derivatives

Differentiate again:

i₁″ − i₂″ = −i₁′/(R₂C₁) = 2.375×10⁻⁵ A/s²
R₁ i₂″ = R₂(i₁″ − i₂″) − i₂′/C₂
       = 10⁶(2.375×10⁻⁵) + 8.75×10⁻⁵/(2×10⁻⁵) = 23.75 + 4.375 = 28.125
d²i₂/dt² (0⁺) = 28.125/(2×10⁶) = 1.406×10⁻⁵ A/s²
d²i₁/dt² (0⁺) = 3.781×10⁻⁵ A/s²

Answer: i₁(0⁺) = 1.5 mA, i₂(0⁺) = 0.5 mA, di₂/dt = −8.75×10⁻⁵ A/s, d²i₂/dt² = 1.406×10⁻⁵ A/s².

✓ Answer checked by Claude Opus 5.5 (high effort)

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