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Derivatives at t=0+: 2 μF, 10 V, 1 H

Short answer

At 0⁺, capacitor short, inductor open: i1 = 10/10 = 1 A, i2 = 0. Derivatives: di1/dt = −49 990 A/s, di2/dt = 10 A/s; d²i1/dt² ≈ 2.499×10⁹ A/s², d²i2/dt² = −500 100 A/s².

In the circuit shown in figure below, switch K is closed at time t = 0. Find the values of i1, i2, di1/dt, di2/dt, d²i1/dt², d²i2/dt² at t = 0⁺. [Figure: 10 V source; switch K and a 2 μF capacitor in the top wire of the left mesh (current i1); a 10 Ω resistor in the middle vertical branch; a 10 Ω resistor in the top wire of the right mesh and a 1 H inductor on the right (current i2)]

Assumptions: the capacitor (C = 2 μF) and inductor (L = 1 H) are initially relaxed. Left mesh (current i₁): 10 V, switch K, C, common 10 Ω. Right mesh (current i₂): common 10 Ω, 10 Ω and L.

Mesh equations

Mesh 1:  10 = (1/C)∫i₁ dt + 10(i₁ − i₂)
Mesh 2:  10(i₂ − i₁) + 10 i₂ + L di₂/dt = 0

At t = 0⁺ (v_C = 0, i_L = 0)

i₂(0⁺) = 0
Mesh 1:  10 = 0 + 10(i₁ − 0)  →  i₁(0⁺) = 1 A

First derivatives

Mesh 2:  10(0 − 1) + 0 + 1·i₂′ = 0  →  di₂/dt(0⁺) = 10 A/s
Mesh 1 differentiated:  0 = i₁/C + 10(i₁′ − i₂′)
        10(i₁′ − 10) = −1/(2×10⁻⁶) = −5×10⁵
        di₁/dt(0⁺) = 10 − 5×10⁴ = −49 990 A/s

Second derivatives

Mesh 2 differentiated:  10(i₂′ − i₁′) + 10 i₂′ + i₂″ = 0
        i₂″ = −[10(10 + 49 990) + 100] = −500 100 A/s²
Mesh 1 differentiated twice:  0 = i₁′/C + 10(i₁″ − i₂″)
        i₁″ = i₂″ − i₁′/(10C) = −500 100 + 49 990/(2×10⁻⁵)
            = 2.499×10⁹ A/s²   (2 498 999 900)
QuantityValue at 0⁺
i₁1 A
i₂0
di₁/dt−49 990 A/s
di₂/dt10 A/s
d²i₁/dt²2.499×10⁹ A/s²
d²i₂/dt²−5.001×10⁵ A/s²

Answer: as tabulated above.

✓ Answer checked by Claude Opus 5.5 (high effort)

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